Calculus Examples

$f (x) = | 4 - x^{2} |$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = | x |$ and $g (x) = 4 - x^{2}$ .

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Step 1.1.1

To apply the Chain Rule, set $u$ as $4 - x^{2}$ .

$\frac{d}{d u} [| u |] \frac{d}{d x} [4 - x^{2}]$

Step 1.1.2

The derivative of $| u |$ with respect to $u$ is $\frac{u}{| u |}$ .

$\frac{u}{| u |} \frac{d}{d x} [4 - x^{2}]$

Step 1.1.3

Replace all occurrences of $u$ with $4 - x^{2}$ .

$\frac{4 - x^{2}}{| 4 - x^{2} |} \frac{d}{d x} [4 - x^{2}]$

Step 1.2

Differentiate.

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Step 1.2.1

By the Sum Rule, the derivative of $4 - x^{2}$ with respect to $x$ is $\frac{d}{d x} [4] + \frac{d}{d x} [- x^{2}]$ .

$\frac{4 - x^{2}}{| 4 - x^{2} |} (\frac{d}{d x} [4] + \frac{d}{d x} [- x^{2}])$

Step 1.2.2

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$\frac{4 - x^{2}}{| 4 - x^{2} |} (0 + \frac{d}{d x} [- x^{2}])$

Step 1.2.3

Add $0$ and $\frac{d}{d x} [- x^{2}]$ .

$\frac{4 - x^{2}}{| 4 - x^{2} |} \frac{d}{d x} [- x^{2}]$

Step 1.2.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$\frac{4 - x^{2}}{| 4 - x^{2} |} (- \frac{d}{d x} [x^{2}])$

Step 1.2.5

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$\frac{4 - x^{2}}{| 4 - x^{2} |} (- (2 x))$

Step 1.2.6

Combine fractions.

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Step 1.2.6.1

Multiply $2$ by $- 1$ .

$\frac{4 - x^{2}}{| 4 - x^{2} |} (- 2 x)$

Step 1.2.6.2

Combine $- 2$ and $\frac{4 - x^{2}}{| 4 - x^{2} |}$ .

$\frac{- 2 (4 - x^{2})}{| 4 - x^{2} |} x$

Step 1.2.6.3

Combine $\frac{- 2 (4 - x^{2})}{| 4 - x^{2} |}$ and $x$ .

$\frac{- 2 (4 - x^{2}) x}{| 4 - x^{2} |}$

Step 1.2.6.4

Move the negative in front of the fraction.

$- \frac{(2 (4 - x^{2})) x}{| 4 - x^{2} |}$

Step 1.3

Simplify.

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Step 1.3.1

Apply the distributive property.

$- \frac{(2 \cdot 4 + 2 (- x^{2})) x}{| 4 - x^{2} |}$

Step 1.3.2

Apply the distributive property.

$- \frac{2 \cdot 4 x + 2 (- x^{2}) x}{| 4 - x^{2} |}$

Step 1.3.3

Simplify each term.

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Step 1.3.3.1

Multiply $2$ by $4$ .

$- \frac{8 x + 2 (- x^{2}) x}{| 4 - x^{2} |}$

Step 1.3.3.2

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 1.3.3.2.1

Move $x$ .

$- \frac{8 x + 2 (- (x \cdot x^{2}))}{| 4 - x^{2} |}$

Step 1.3.3.2.2

Multiply $x$ by $x^{2}$ .

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Step 1.3.3.2.2.1

Raise $x$ to the power of $1$ .

$- \frac{8 x + 2 (- (x^{1} x^{2}))}{| 4 - x^{2} |}$

Step 1.3.3.2.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- \frac{8 x + 2 (- x^{1 + 2})}{| 4 - x^{2} |}$

Step 1.3.3.2.3

Add $1$ and $2$ .

$- \frac{8 x + 2 (- x^{3})}{| 4 - x^{2} |}$

Step 1.3.3.3

Multiply $- 1$ by $2$ .

$- \frac{8 x - 2 x^{3}}{| 4 - x^{2} |}$

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = - 1$ and $g (x) = \frac{8 x - 2 x^{3}}{| 4 - x^{2} |}$ .

$f'' (x) = - \frac{d}{d x} \cdot \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.2

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = 8 x - 2 x^{3}$ and $g (x) = | 4 - x^{2} |$ .

$f'' (x) = - \frac{| 4 - x^{2} | \frac{d}{d x} (8 x - 2 x^{3}) - (8 x - 2 x^{3}) \frac{d}{d x} | 4 - x^{2} |}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.3

Differentiate.

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Step 2.3.1

By the Sum Rule, the derivative of $8 x - 2 x^{3}$ with respect to $x$ is $\frac{d}{d x} [8 x] + \frac{d}{d x} [- 2 x^{3}]$ .

$f'' (x) = - \frac{| 4 - x^{2} | (\frac{d}{d x} (8 x) + \frac{d}{d x} (- 2 x^{3})) - (8 x - 2 x^{3}) \frac{d}{d x} | 4 - x^{2} |}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.3.2

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 \frac{d}{d x} (x) + \frac{d}{d x} (- 2 x^{3})) - (8 x - 2 x^{3}) \frac{d}{d x} | 4 - x^{2} |}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.3.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 \cdot 1 + \frac{d}{d x} (- 2 x^{3})) - (8 x - 2 x^{3}) \frac{d}{d x} | 4 - x^{2} |}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.3.4

Multiply $8$ by $1$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 + \frac{d}{d x} (- 2 x^{3})) - (8 x - 2 x^{3}) \frac{d}{d x} | 4 - x^{2} |}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.3.5

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x^{3}$ with respect to $x$ is $- 2 \frac{d}{d x} [x^{3}]$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 2 \frac{d}{d x} x^{3}) - (8 x - 2 x^{3}) \frac{d}{d x} | 4 - x^{2} |}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.3.6

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 2 (3 x^{2})) - (8 x - 2 x^{3}) \frac{d}{d x} | 4 - x^{2} |}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.3.7

Multiply $3$ by $- 2$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) - (8 x - 2 x^{3}) \frac{d}{d x} | 4 - x^{2} |}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = | x |$ and $g (x) = 4 - x^{2}$ .

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Step 2.4.1

To apply the Chain Rule, set $u$ as $4 - x^{2}$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) - (8 x - 2 x^{3}) (\frac{d}{d u} (| u |) \frac{d}{d x} (4 - x^{2}))}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.4.2

The derivative of $| u |$ with respect to $u$ is $\frac{u}{| u |}$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) - (8 x - 2 x^{3}) (\frac{u}{| u |} \cdot \frac{d}{d x} (4 - x^{2}))}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.4.3

Replace all occurrences of $u$ with $4 - x^{2}$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) - (8 x - 2 x^{3}) (\frac{4 - x^{2}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (4 - x^{2}))}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.5

Differentiate.

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Step 2.5.1

By the Sum Rule, the derivative of $4 - x^{2}$ with respect to $x$ is $\frac{d}{d x} [4] + \frac{d}{d x} [- x^{2}]$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) - (8 x - 2 x^{3}) (\frac{4 - x^{2}}{| 4 - x^{2} |} \cdot (\frac{d}{d x} (4) + \frac{d}{d x} (- x^{2})))}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.5.2

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) - (8 x - 2 x^{3}) (\frac{4 - x^{2}}{| 4 - x^{2} |} \cdot (0 + \frac{d}{d x} (- x^{2})))}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.5.3

Add $0$ and $\frac{d}{d x} [- x^{2}]$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) - (8 x - 2 x^{3}) (\frac{4 - x^{2}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- x^{2}))}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.5.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) - (8 x - 2 x^{3}) (\frac{4 - x^{2}}{| 4 - x^{2} |} \cdot (- \frac{d}{d x} x^{2}))}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.5.5

Multiply.

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Step 2.5.5.1

Multiply $- 1$ by $- 1$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) + 1 (8 x - 2 x^{3}) (\frac{4 - x^{2}}{| 4 - x^{2} |} \cdot (\frac{d}{d x} (x^{2})))}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.5.5.2

Multiply $8 x - 2 x^{3}$ by $1$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) + (8 x - 2 x^{3}) (\frac{4 - x^{2}}{| 4 - x^{2} |} \cdot (\frac{d}{d x} (x^{2})))}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.5.6

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) + (8 x - 2 x^{3}) (\frac{4 - x^{2}}{| 4 - x^{2} |}) \cdot (2 x)}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.5.7

Combine fractions.

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Step 2.5.7.1

Combine $2$ and $\frac{4 - x^{2}}{| 4 - x^{2} |}$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) + (8 x - 2 x^{3}) (\frac{2 (4 - x^{2})}{| 4 - x^{2} |}) \cdot x}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.5.7.2

Combine $x$ and $\frac{2 (4 - x^{2})}{| 4 - x^{2} |}$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) + (8 x - 2 x^{3}) (\frac{x (2 (4 - x^{2}))}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot \frac{d}{d x} (- 1)$

Step 2.5.8

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) + (8 x - 2 x^{3}) (\frac{x (2 (4 - x^{2}))}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}} + \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} \cdot 0$

Step 2.5.9

Simplify the expression.

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Step 2.5.9.1

Multiply $\frac{8 x - 2 x^{3}}{| 4 - x^{2} |}$ by $0$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) + (8 x - 2 x^{3}) (\frac{x (2 (4 - x^{2}))}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}} + 0$

Step 2.5.9.2

Add $- \frac{| 4 - x^{2} | (8 - 6 x^{2}) + (8 x - 2 x^{3}) \frac{x (2 (4 - x^{2}))}{| 4 - x^{2} |}}{{| 4 - x^{2} |}^{2}}$ and $0$ .

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) + (8 x - 2 x^{3}) (\frac{x (2 (4 - x^{2}))}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6

Simplify.

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Step 2.6.1

Apply the distributive property.

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) + (8 x - 2 x^{3}) (\frac{x (2 \cdot 4 + 2 (- x^{2}))}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.2

Apply the distributive property.

$f'' (x) = - \frac{| 4 - x^{2} | (8 - 6 x^{2}) + (8 x - 2 x^{3}) (\frac{x (2 \cdot 4) + x (2 (- x^{2}))}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3

Simplify the numerator.

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Step 2.6.3.1

Simplify each term.

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Step 2.6.3.1.1

Apply the distributive property.

$f'' (x) = - \frac{| 4 - x^{2} | \cdot 8 + | 4 - x^{2} | (- 6 x^{2}) + (8 x - 2 x^{3}) (\frac{x (2 \cdot 4) + x (2 (- x^{2}))}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.2

Move $8$ to the left of $| 4 - x^{2} |$ .

$f'' (x) = - \frac{8 \cdot | 4 - x^{2} | + | 4 - x^{2} | (- 6 x^{2}) + (8 x - 2 x^{3}) (\frac{x (2 \cdot 4) + x (2 (- x^{2}))}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.3

Rewrite using the commutative property of multiplication.

$f'' (x) = - \frac{8 \cdot | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{x (2 \cdot 4) + x (2 (- x^{2}))}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.4

Simplify the numerator.

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Step 2.6.3.1.4.1

Multiply $2$ by $4$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{x \cdot 8 + x (2 (- x^{2}))}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.4.2

Move $8$ to the left of $x$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{8 \cdot x + x (2 (- x^{2}))}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.4.3

Rewrite using the commutative property of multiplication.

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{8 x + 2 \cdot (- 1 x \cdot x^{2})}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.4.4

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 2.6.3.1.4.4.1

Move $x^{2}$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{8 x + 2 \cdot (- 1 (x^{2} x))}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.4.4.2

Multiply $x^{2}$ by $x$ .

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Step 2.6.3.1.4.4.2.1

Raise $x$ to the power of $1$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{8 x + 2 \cdot (- 1 (x^{2} x))}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.4.4.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{8 x + 2 \cdot (- 1 x^{2 + 1})}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.4.4.3

Add $2$ and $1$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{8 x + 2 \cdot (- 1 x^{3})}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.4.5

Multiply $2$ by $- 1$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{8 x - 2 x^{3}}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.4.6

Rewrite $8 x - 2 x^{3}$ in a factored form.

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Step 2.6.3.1.4.6.1

Factor $2 x$ out of $8 x - 2 x^{3}$ .

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Step 2.6.3.1.4.6.1.1

Factor $2 x$ out of $8 x$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (4) - 2 x^{3}}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.4.6.1.2

Factor $2 x$ out of $- 2 x^{3}$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (4) + 2 x (- x^{2})}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.4.6.1.3

Factor $2 x$ out of $2 x (4) + 2 x (- x^{2})$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (4 - x^{2})}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.4.6.2

Rewrite $4$ as $2^{2}$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2^{2} - x^{2})}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.4.6.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = x$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.5

Simplify the denominator.

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Step 2.6.3.1.5.1

Rewrite $4$ as $2^{2}$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| 2^{2} - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.5.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = x$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| (2 + x) (2 - x) |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.5.3

Expand $(2 + x) (2 - x)$ using the FOIL Method.

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Step 2.6.3.1.5.3.1

Apply the distributive property.

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| 2 (2 - x) + x (2 - x) |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.5.3.2

Apply the distributive property.

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| 2 \cdot 2 + 2 (- x) + x (2 - x) |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.5.3.3

Apply the distributive property.

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| 2 \cdot 2 + 2 (- x) + x \cdot 2 + x (- x) |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.5.4

Simplify and combine like terms.

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Step 2.6.3.1.5.4.1

Simplify each term.

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Step 2.6.3.1.5.4.1.1

Multiply $2$ by $2$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| 4 + 2 (- x) + x \cdot 2 + x (- x) |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.5.4.1.2

Multiply $- 1$ by $2$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| 4 - 2 x + x \cdot 2 + x (- x) |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.5.4.1.3

Move $2$ to the left of $x$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| 4 - 2 x + 2 \cdot x + x (- x) |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.5.4.1.4

Rewrite using the commutative property of multiplication.

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| 4 - 2 x + 2 x - x \cdot x |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.5.4.1.5

Multiply $x$ by $x$ by adding the exponents.

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Step 2.6.3.1.5.4.1.5.1

Move $x$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| 4 - 2 x + 2 x - (x \cdot x) |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.5.4.1.5.2

Multiply $x$ by $x$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| 4 - 2 x + 2 x - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.5.4.2

Add $- 2 x$ and $2 x$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| 4 + 0 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.5.4.3

Add $4$ and $0$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| 4 - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.5.5

Rewrite $4$ as $2^{2}$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| 2^{2} - x^{2} |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.5.6

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = x$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + (8 x - 2 x^{3}) (\frac{2 x (2 + x) (2 - x)}{| (2 + x) (2 - x) |})}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.6

Multiply $8 x - 2 x^{3}$ by $\frac{2 x (2 + x) (2 - x)}{| (2 + x) (2 - x) |}$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{(8 x - 2 x^{3}) (2 x (2 + x) (2 - x))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7

Simplify the numerator.

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Step 2.6.3.1.7.1

Factor $2 x$ out of $8 x - 2 x^{3}$ .

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Step 2.6.3.1.7.1.1

Factor $2 x$ out of $8 x$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{(2 x (4) - 2 x^{3}) \cdot (2 x (2 + x) (2 - x))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.1.2

Factor $2 x$ out of $- 2 x^{3}$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{(2 x (4) + 2 x (- x^{2})) \cdot (2 x (2 + x) (2 - x))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.1.3

Factor $2 x$ out of $2 x (4) + 2 x (- x^{2})$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{2 x (4 - x^{2}) \cdot (2 x (2 + x) (2 - x))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.2

Rewrite $4$ as $2^{2}$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{2 x (2^{2} - x^{2}) \cdot (2 x (2 + x) (2 - x))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = x$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{2 x (2 + x) (2 - x) \cdot (2 x (2 + x) (2 - x))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.4

Combine exponents.

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Step 2.6.3.1.7.4.1

Multiply $2$ by $2$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{4 x (2 + x) (2 - x) x (2 + x) (2 - x)}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.4.2

Raise $x$ to the power of $1$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{4 (x \cdot x) (2 + x) (2 - x) (2 + x) (2 - x)}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.4.3

Raise $x$ to the power of $1$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{4 (x \cdot x) (2 + x) (2 - x) (2 + x) (2 - x)}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.4.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{4 x^{1 + 1} (2 + x) (2 - x) (2 + x) (2 - x)}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.4.5

Add $1$ and $1$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{4 x^{2} (2 + x) (2 - x) (2 + x) (2 - x)}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.4.6

Raise $2 + x$ to the power of $1$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{4 x^{2} ((2 + x) (2 + x)) (2 - x) (2 - x)}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.4.7

Raise $2 + x$ to the power of $1$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{4 x^{2} ((2 + x) (2 + x)) (2 - x) (2 - x)}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.4.8

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{4 x^{2} {(2 + x)}^{1 + 1} (2 - x) (2 - x)}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.4.9

Add $1$ and $1$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{4 x^{2} {(2 + x)}^{2} (2 - x) (2 - x)}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.4.10

Raise $2 - x$ to the power of $1$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{4 x^{2} {(2 + x)}^{2} ((2 - x) (2 - x))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.4.11

Raise $2 - x$ to the power of $1$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{4 x^{2} {(2 + x)}^{2} ((2 - x) (2 - x))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.4.12

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{4 x^{2} {(2 + x)}^{2} {(2 - x)}^{1 + 1}}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.1.7.4.13

Add $1$ and $1$ .

$f'' (x) = - \frac{8 | 4 - x^{2} | - 6 | 4 - x^{2} | x^{2} + \frac{4 x^{2} {(2 + x)}^{2} {(2 - x)}^{2}}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.2

To write $8 | 4 - x^{2} |$ as a fraction with a common denominator, multiply by $\frac{| (2 + x) (2 - x) |}{| (2 + x) (2 - x) |}$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{8 | 4 - x^{2} | | (2 + x) (2 - x) |}{| (2 + x) (2 - x) |} + \frac{4 x^{2} {(2 + x)}^{2} {(2 - x)}^{2}}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.3

Combine the numerators over the common denominator.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{8 | 4 - x^{2} | | (2 + x) (2 - x) | + 4 x^{2} {(2 + x)}^{2} {(2 - x)}^{2}}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4

Simplify the numerator.

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Step 2.6.3.4.1

Factor $4$ out of $8 | 4 - x^{2} | | (2 + x) (2 - x) | + 4 x^{2} {(2 + x)}^{2} {(2 - x)}^{2}$ .

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Step 2.6.3.4.1.1

Factor $4$ out of $8 | 4 - x^{2} | | (2 + x) (2 - x) |$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 - x^{2} | | (2 + x) (2 - x) |) + 4 x^{2} {(2 + x)}^{2} {(2 - x)}^{2}}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.1.2

Factor $4$ out of $4 x^{2} {(2 + x)}^{2} {(2 - x)}^{2}$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 - x^{2} | | (2 + x) (2 - x) |) + 4 (x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.1.3

Factor $4$ out of $4 (2 | 4 - x^{2} | | (2 + x) (2 - x) |) + 4 (x^{2} {(2 + x)}^{2} {(2 - x)}^{2})$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 - x^{2} | | (2 + x) (2 - x) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.2

Expand $(2 + x) (2 - x)$ using the FOIL Method.

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Step 2.6.3.4.2.1

Apply the distributive property.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 - x^{2} | | 2 (2 - x) + x (2 - x) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.2.2

Apply the distributive property.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 - x^{2} | | 2 \cdot 2 + 2 (- x) + x (2 - x) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.2.3

Apply the distributive property.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 - x^{2} | | 2 \cdot 2 + 2 (- x) + x \cdot 2 + x (- x) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.3

Simplify and combine like terms.

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Step 2.6.3.4.3.1

Simplify each term.

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Step 2.6.3.4.3.1.1

Multiply $2$ by $2$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 - x^{2} | | 4 + 2 (- x) + x \cdot 2 + x (- x) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.3.1.2

Multiply $- 1$ by $2$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 - x^{2} | | 4 - 2 x + x \cdot 2 + x (- x) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.3.1.3

Move $2$ to the left of $x$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 - x^{2} | | 4 - 2 x + 2 \cdot x + x (- x) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.3.1.4

Rewrite using the commutative property of multiplication.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 - x^{2} | | 4 - 2 x + 2 x - x \cdot x | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.3.1.5

Multiply $x$ by $x$ by adding the exponents.

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Step 2.6.3.4.3.1.5.1

Move $x$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 - x^{2} | | 4 - 2 x + 2 x - (x \cdot x) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.3.1.5.2

Multiply $x$ by $x$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 - x^{2} | | 4 - 2 x + 2 x - x^{2} | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.3.2

Add $- 2 x$ and $2 x$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 - x^{2} | | 4 + 0 - x^{2} | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.3.3

Add $4$ and $0$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 - x^{2} | | 4 - x^{2} | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.4

Multiply $2 | 4 - x^{2} | | 4 - x^{2} |$ .

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Step 2.6.3.4.4.1

To multiply absolute values, multiply the terms inside each absolute value.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | (4 - x^{2}) (4 - x^{2}) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.4.2

Raise $4 - x^{2}$ to the power of $1$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | (4 - x^{2}) (4 - x^{2}) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.4.3

Raise $4 - x^{2}$ to the power of $1$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | (4 - x^{2}) (4 - x^{2}) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.4.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | {(4 - x^{2})}^{1 + 1} | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.4.5

Add $1$ and $1$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | {(4 - x^{2})}^{2} | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.5

Rewrite ${(4 - x^{2})}^{2}$ as $(4 - x^{2}) (4 - x^{2})$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | (4 - x^{2}) (4 - x^{2}) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.6

Expand $(4 - x^{2}) (4 - x^{2})$ using the FOIL Method.

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Step 2.6.3.4.6.1

Apply the distributive property.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 (4 - x^{2}) - x^{2} (4 - x^{2}) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.6.2

Apply the distributive property.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 \cdot 4 + 4 (- x^{2}) - x^{2} (4 - x^{2}) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.6.3

Apply the distributive property.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 4 \cdot 4 + 4 (- x^{2}) - x^{2} \cdot 4 - x^{2} (- x^{2}) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.7

Simplify and combine like terms.

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Step 2.6.3.4.7.1

Simplify each term.

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Step 2.6.3.4.7.1.1

Multiply $4$ by $4$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 + 4 (- x^{2}) - x^{2} \cdot 4 - x^{2} (- x^{2}) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.7.1.2

Multiply $- 1$ by $4$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 4 x^{2} - x^{2} \cdot 4 - x^{2} (- x^{2}) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.7.1.3

Multiply $4$ by $- 1$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 4 x^{2} - 4 x^{2} - x^{2} (- x^{2}) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.7.1.4

Rewrite using the commutative property of multiplication.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 4 x^{2} - 4 x^{2} - 1 \cdot (- 1 x^{2} x^{2}) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.7.1.5

Multiply $x^{2}$ by $x^{2}$ by adding the exponents.

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Step 2.6.3.4.7.1.5.1

Move $x^{2}$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 4 x^{2} - 4 x^{2} - 1 \cdot (- 1 (x^{2} x^{2})) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.7.1.5.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 4 x^{2} - 4 x^{2} - 1 \cdot (- 1 x^{2 + 2}) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.7.1.5.3

Add $2$ and $2$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 4 x^{2} - 4 x^{2} - 1 \cdot (- 1 x^{4}) | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.7.1.6

Multiply $- 1$ by $- 1$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 4 x^{2} - 4 x^{2} + 1 x^{4} | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.7.1.7

Multiply $x^{4}$ by $1$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 4 x^{2} - 4 x^{2} + x^{4} | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.7.2

Subtract $4 x^{2}$ from $- 4 x^{2}$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + x^{2} {(2 + x)}^{2} {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.8

Rewrite ${(2 + x)}^{2}$ as $(2 + x) (2 + x)$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + x^{2} ((2 + x) (2 + x)) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.9

Expand $(2 + x) (2 + x)$ using the FOIL Method.

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Step 2.6.3.4.9.1

Apply the distributive property.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + x^{2} (2 (2 + x) + x (2 + x)) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.9.2

Apply the distributive property.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + x^{2} (2 \cdot 2 + 2 x + x (2 + x)) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.9.3

Apply the distributive property.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + x^{2} (2 \cdot 2 + 2 x + x \cdot 2 + x \cdot x) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.10

Simplify and combine like terms.

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Step 2.6.3.4.10.1

Simplify each term.

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Step 2.6.3.4.10.1.1

Multiply $2$ by $2$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + x^{2} (4 + 2 x + x \cdot 2 + x \cdot x) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.10.1.2

Move $2$ to the left of $x$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + x^{2} (4 + 2 x + 2 \cdot x + x \cdot x) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.10.1.3

Multiply $x$ by $x$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + x^{2} (4 + 2 x + 2 x + x^{2}) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.10.2

Add $2 x$ and $2 x$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + x^{2} (4 + 4 x + x^{2}) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.11

Apply the distributive property.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (x^{2} \cdot 4 + x^{2} (4 x) + x^{2} x^{2}) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.12

Simplify.

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Step 2.6.3.4.12.1

Move $4$ to the left of $x^{2}$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 \cdot x^{2} + x^{2} (4 x) + x^{2} x^{2}) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.12.2

Rewrite using the commutative property of multiplication.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 \cdot x^{2} + 4 x^{2} x + x^{2} x^{2}) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.12.3

Multiply $x^{2}$ by $x^{2}$ by adding the exponents.

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Step 2.6.3.4.12.3.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 \cdot x^{2} + 4 x^{2} x + x^{2 + 2}) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.12.3.2

Add $2$ and $2$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 \cdot x^{2} + 4 x^{2} x + x^{4}) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.13

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 2.6.3.4.13.1

Move $x$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 (x \cdot x^{2}) + x^{4}) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.13.2

Multiply $x$ by $x^{2}$ .

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Step 2.6.3.4.13.2.1

Raise $x$ to the power of $1$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 (x \cdot x^{2}) + x^{4}) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.13.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 x^{1 + 2} + x^{4}) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.13.3

Add $1$ and $2$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 x^{3} + x^{4}) {(2 - x)}^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.14

Rewrite ${(2 - x)}^{2}$ as $(2 - x) (2 - x)$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 x^{3} + x^{4}) ((2 - x) (2 - x)))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.15

Expand $(2 - x) (2 - x)$ using the FOIL Method.

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Step 2.6.3.4.15.1

Apply the distributive property.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 x^{3} + x^{4}) (2 (2 - x) - x (2 - x)))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.15.2

Apply the distributive property.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 x^{3} + x^{4}) (2 \cdot 2 + 2 (- x) - x (2 - x)))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.15.3

Apply the distributive property.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 x^{3} + x^{4}) (2 \cdot 2 + 2 (- x) - x \cdot 2 - x (- x)))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.16

Simplify and combine like terms.

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Step 2.6.3.4.16.1

Simplify each term.

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Step 2.6.3.4.16.1.1

Multiply $2$ by $2$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 x^{3} + x^{4}) (4 + 2 (- x) - x \cdot 2 - x (- x)))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.16.1.2

Multiply $- 1$ by $2$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 x^{3} + x^{4}) (4 - 2 x - x \cdot 2 - x (- x)))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.16.1.3

Multiply $2$ by $- 1$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 x^{3} + x^{4}) (4 - 2 x - 2 x - x (- x)))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.16.1.4

Rewrite using the commutative property of multiplication.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 x^{3} + x^{4}) (4 - 2 x - 2 x - 1 \cdot (- 1 x \cdot x)))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.16.1.5

Multiply $x$ by $x$ by adding the exponents.

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Step 2.6.3.4.16.1.5.1

Move $x$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 x^{3} + x^{4}) (4 - 2 x - 2 x - 1 \cdot (- 1 (x \cdot x))))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.16.1.5.2

Multiply $x$ by $x$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 x^{3} + x^{4}) (4 - 2 x - 2 x - 1 \cdot (- 1 x^{2})))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.16.1.6

Multiply $- 1$ by $- 1$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 x^{3} + x^{4}) (4 - 2 x - 2 x + 1 x^{2}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.16.1.7

Multiply $x^{2}$ by $1$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 x^{3} + x^{4}) (4 - 2 x - 2 x + x^{2}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.16.2

Subtract $2 x$ from $- 2 x$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + (4 x^{2} + 4 x^{3} + x^{4}) (4 - 4 x + x^{2}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.17

Expand $(4 x^{2} + 4 x^{3} + x^{4}) (4 - 4 x + x^{2})$ by multiplying each term in the first expression by each term in the second expression.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 4 x^{2} \cdot 4 + 4 x^{2} (- 4 x) + 4 x^{2} x^{2} + 4 x^{3} \cdot 4 + 4 x^{3} (- 4 x) + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18

Simplify each term.

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Step 2.6.3.4.18.1

Multiply $4$ by $4$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} + 4 x^{2} (- 4 x) + 4 x^{2} x^{2} + 4 x^{3} \cdot 4 + 4 x^{3} (- 4 x) + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.2

Rewrite using the commutative property of multiplication.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} + 4 \cdot (- 4 x^{2} x) + 4 x^{2} x^{2} + 4 x^{3} \cdot 4 + 4 x^{3} (- 4 x) + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.3

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 2.6.3.4.18.3.1

Move $x$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} + 4 \cdot (- 4 (x \cdot x^{2})) + 4 x^{2} x^{2} + 4 x^{3} \cdot 4 + 4 x^{3} (- 4 x) + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.3.2

Multiply $x$ by $x^{2}$ .

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Step 2.6.3.4.18.3.2.1

Raise $x$ to the power of $1$ .

Step 2.6.3.4.18.3.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} + 4 \cdot (- 4 x^{1 + 2}) + 4 x^{2} x^{2} + 4 x^{3} \cdot 4 + 4 x^{3} (- 4 x) + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.3.3

Add $1$ and $2$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} + 4 \cdot (- 4 x^{3}) + 4 x^{2} x^{2} + 4 x^{3} \cdot 4 + 4 x^{3} (- 4 x) + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.4

Multiply $4$ by $- 4$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{2} x^{2} + 4 x^{3} \cdot 4 + 4 x^{3} (- 4 x) + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.5

Multiply $x^{2}$ by $x^{2}$ by adding the exponents.

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Step 2.6.3.4.18.5.1

Move $x^{2}$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 (x^{2} x^{2}) + 4 x^{3} \cdot 4 + 4 x^{3} (- 4 x) + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.5.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{2 + 2} + 4 x^{3} \cdot 4 + 4 x^{3} (- 4 x) + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.5.3

Add $2$ and $2$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 4 x^{3} \cdot 4 + 4 x^{3} (- 4 x) + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.6

Multiply $4$ by $4$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} + 4 x^{3} (- 4 x) + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.7

Rewrite using the commutative property of multiplication.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} + 4 \cdot (- 4 x^{3} x) + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.8

Multiply $x^{3}$ by $x$ by adding the exponents.

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Step 2.6.3.4.18.8.1

Move $x$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} + 4 \cdot (- 4 (x \cdot x^{3})) + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.8.2

Multiply $x$ by $x^{3}$ .

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Step 2.6.3.4.18.8.2.1

Raise $x$ to the power of $1$ .

Step 2.6.3.4.18.8.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} + 4 \cdot (- 4 x^{1 + 3}) + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.8.3

Add $1$ and $3$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} + 4 \cdot (- 4 x^{4}) + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.9

Multiply $4$ by $- 4$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} - 16 x^{4} + 4 x^{3} x^{2} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.10

Multiply $x^{3}$ by $x^{2}$ by adding the exponents.

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Step 2.6.3.4.18.10.1

Move $x^{2}$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} - 16 x^{4} + 4 (x^{2} x^{3}) + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.10.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} - 16 x^{4} + 4 x^{2 + 3} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.10.3

Add $2$ and $3$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} - 16 x^{4} + 4 x^{5} + x^{4} \cdot 4 + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.11

Move $4$ to the left of $x^{4}$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} - 16 x^{4} + 4 x^{5} + 4 \cdot x^{4} + x^{4} (- 4 x) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.12

Rewrite using the commutative property of multiplication.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} - 16 x^{4} + 4 x^{5} + 4 x^{4} - 4 x^{4} x + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.13

Multiply $x^{4}$ by $x$ by adding the exponents.

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Step 2.6.3.4.18.13.1

Move $x$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} - 16 x^{4} + 4 x^{5} + 4 x^{4} - 4 (x \cdot x^{4}) + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.13.2

Multiply $x$ by $x^{4}$ .

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Step 2.6.3.4.18.13.2.1

Raise $x$ to the power of $1$ .

Step 2.6.3.4.18.13.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} - 16 x^{4} + 4 x^{5} + 4 x^{4} - 4 x^{1 + 4} + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.13.3

Add $1$ and $4$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} - 16 x^{4} + 4 x^{5} + 4 x^{4} - 4 x^{5} + x^{4} x^{2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.14

Multiply $x^{4}$ by $x^{2}$ by adding the exponents.

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Step 2.6.3.4.18.14.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} - 16 x^{4} + 4 x^{5} + 4 x^{4} - 4 x^{5} + x^{4 + 2})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.18.14.2

Add $4$ and $2$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} - 16 x^{4} + 4 x^{5} + 4 x^{4} - 4 x^{5} + x^{6})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.19

Combine the opposite terms in $16 x^{2} - 16 x^{3} + 4 x^{4} + 16 x^{3} - 16 x^{4} + 4 x^{5} + 4 x^{4} - 4 x^{5} + x^{6}$ .

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Step 2.6.3.4.19.1

Add $- 16 x^{3}$ and $16 x^{3}$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} + 4 x^{4} + 0 - 16 x^{4} + 4 x^{5} + 4 x^{4} - 4 x^{5} + x^{6})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.19.2

Add $16 x^{2} + 4 x^{4}$ and $0$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} + 4 x^{4} - 16 x^{4} + 4 x^{5} + 4 x^{4} - 4 x^{5} + x^{6})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.19.3

Subtract $4 x^{5}$ from $4 x^{5}$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} + 4 x^{4} - 16 x^{4} + 4 x^{4} + 0 + x^{6})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.19.4

Add $16 x^{2} + 4 x^{4} - 16 x^{4} + 4 x^{4}$ and $0$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} + 4 x^{4} - 16 x^{4} + 4 x^{4} + x^{6})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.20

Subtract $16 x^{4}$ from $4 x^{4}$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 12 x^{4} + 4 x^{4} + x^{6})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.4.21

Add $- 12 x^{4}$ and $4 x^{4}$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.5

To write $- 6 | 4 - x^{2} | x^{2}$ as a fraction with a common denominator, multiply by $\frac{| (2 + x) (2 - x) |}{| (2 + x) (2 - x) |}$ .

$f'' (x) = - \frac{- 6 | 4 - x^{2} | x^{2} \cdot \frac{| (2 + x) (2 - x) |}{| (2 + x) (2 - x) |} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.6

Combine $- 6 | 4 - x^{2} | x^{2}$ and $\frac{| (2 + x) (2 - x) |}{| (2 + x) (2 - x) |}$ .

$f'' (x) = - \frac{\frac{- 6 | 4 - x^{2} | x^{2} | (2 + x) (2 - x) |}{| (2 + x) (2 - x) |} + \frac{4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.7

Combine the numerators over the common denominator.

$f'' (x) = - \frac{\frac{- 6 | 4 - x^{2} | x^{2} | (2 + x) (2 - x) | + 4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8

Simplify the numerator.

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Step 2.6.3.8.1

Factor $2$ out of $- 6 | 4 - x^{2} | x^{2} | (2 + x) (2 - x) | + 4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6})$ .

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Step 2.6.3.8.1.1

Factor $2$ out of $- 6 | 4 - x^{2} | x^{2} | (2 + x) (2 - x) |$ .

$f'' (x) = - \frac{\frac{2 (- 3 | 4 - x^{2} | x^{2} | (2 + x) (2 - x) |) + 4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.1.2

Factor $2$ out of $4 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6})$ .

$f'' (x) = - \frac{\frac{2 (- 3 | 4 - x^{2} | x^{2} | (2 + x) (2 - x) |) + 2 (2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.1.3

Factor $2$ out of $2 (- 3 | 4 - x^{2} | x^{2} | (2 + x) (2 - x) |) + 2 (2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))$ .

$f'' (x) = - \frac{\frac{2 (- 3 | 4 - x^{2} | x^{2} | (2 + x) (2 - x) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.2

Expand $(2 + x) (2 - x)$ using the FOIL Method.

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Step 2.6.3.8.2.1

Apply the distributive property.

$f'' (x) = - \frac{\frac{2 (- 3 | 4 - x^{2} | x^{2} | 2 (2 - x) + x (2 - x) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.2.2

Apply the distributive property.

$f'' (x) = - \frac{\frac{2 (- 3 | 4 - x^{2} | x^{2} | 2 \cdot 2 + 2 (- x) + x (2 - x) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.2.3

Apply the distributive property.

$f'' (x) = - \frac{\frac{2 (- 3 | 4 - x^{2} | x^{2} | 2 \cdot 2 + 2 (- x) + x \cdot 2 + x (- x) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.3

Simplify and combine like terms.

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Step 2.6.3.8.3.1

Simplify each term.

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Step 2.6.3.8.3.1.1

Multiply $2$ by $2$ .

$f'' (x) = - \frac{\frac{2 (- 3 | 4 - x^{2} | x^{2} | 4 + 2 (- x) + x \cdot 2 + x (- x) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.3.1.2

Multiply $- 1$ by $2$ .

$f'' (x) = - \frac{\frac{2 (- 3 | 4 - x^{2} | x^{2} | 4 - 2 x + x \cdot 2 + x (- x) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.3.1.3

Move $2$ to the left of $x$ .

$f'' (x) = - \frac{\frac{2 (- 3 | 4 - x^{2} | x^{2} | 4 - 2 x + 2 \cdot x + x (- x) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.3.1.4

Rewrite using the commutative property of multiplication.

$f'' (x) = - \frac{\frac{2 (- 3 | 4 - x^{2} | x^{2} | 4 - 2 x + 2 x - x \cdot x | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.3.1.5

Multiply $x$ by $x$ by adding the exponents.

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Step 2.6.3.8.3.1.5.1

Move $x$ .

$f'' (x) = - \frac{\frac{2 (- 3 | 4 - x^{2} | x^{2} | 4 - 2 x + 2 x - (x \cdot x) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.3.1.5.2

Multiply $x$ by $x$ .

$f'' (x) = - \frac{\frac{2 (- 3 | 4 - x^{2} | x^{2} | 4 - 2 x + 2 x - x^{2} | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.3.2

Add $- 2 x$ and $2 x$ .

$f'' (x) = - \frac{\frac{2 (- 3 | 4 - x^{2} | x^{2} | 4 + 0 - x^{2} | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.3.3

Add $4$ and $0$ .

$f'' (x) = - \frac{\frac{2 (- 3 | 4 - x^{2} | x^{2} | 4 - x^{2} | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.4

Multiply $- 3 | 4 - x^{2} | x^{2} | 4 - x^{2} |$ .

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Step 2.6.3.8.4.1

To multiply absolute values, multiply the terms inside each absolute value.

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | (4 - x^{2}) (4 - x^{2}) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.4.2

Raise $4 - x^{2}$ to the power of $1$ .

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | (4 - x^{2}) (4 - x^{2}) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.4.3

Raise $4 - x^{2}$ to the power of $1$ .

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | (4 - x^{2}) (4 - x^{2}) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.4.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | {(4 - x^{2})}^{1 + 1} | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.4.5

Add $1$ and $1$ .

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | {(4 - x^{2})}^{2} | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.5

Rewrite ${(4 - x^{2})}^{2}$ as $(4 - x^{2}) (4 - x^{2})$ .

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | (4 - x^{2}) (4 - x^{2}) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.6

Expand $(4 - x^{2}) (4 - x^{2})$ using the FOIL Method.

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Step 2.6.3.8.6.1

Apply the distributive property.

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 4 (4 - x^{2}) - x^{2} (4 - x^{2}) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.6.2

Apply the distributive property.

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 4 \cdot 4 + 4 (- x^{2}) - x^{2} (4 - x^{2}) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.6.3

Apply the distributive property.

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 4 \cdot 4 + 4 (- x^{2}) - x^{2} \cdot 4 - x^{2} (- x^{2}) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.7

Simplify and combine like terms.

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Step 2.6.3.8.7.1

Simplify each term.

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Step 2.6.3.8.7.1.1

Multiply $4$ by $4$ .

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 16 + 4 (- x^{2}) - x^{2} \cdot 4 - x^{2} (- x^{2}) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.7.1.2

Multiply $- 1$ by $4$ .

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 16 - 4 x^{2} - x^{2} \cdot 4 - x^{2} (- x^{2}) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.7.1.3

Multiply $4$ by $- 1$ .

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 16 - 4 x^{2} - 4 x^{2} - x^{2} (- x^{2}) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.7.1.4

Rewrite using the commutative property of multiplication.

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 16 - 4 x^{2} - 4 x^{2} - 1 \cdot (- 1 x^{2} x^{2}) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.7.1.5

Multiply $x^{2}$ by $x^{2}$ by adding the exponents.

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Step 2.6.3.8.7.1.5.1

Move $x^{2}$ .

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 16 - 4 x^{2} - 4 x^{2} - 1 \cdot (- 1 (x^{2} x^{2})) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.7.1.5.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 16 - 4 x^{2} - 4 x^{2} - 1 \cdot (- 1 x^{2 + 2}) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.7.1.5.3

Add $2$ and $2$ .

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 16 - 4 x^{2} - 4 x^{2} - 1 \cdot (- 1 x^{4}) | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.7.1.6

Multiply $- 1$ by $- 1$ .

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 16 - 4 x^{2} - 4 x^{2} + 1 x^{4} | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.7.1.7

Multiply $x^{4}$ by $1$ .

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 16 - 4 x^{2} - 4 x^{2} + x^{4} | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.7.2

Subtract $4 x^{2}$ from $- 4 x^{2}$ .

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 16 - 8 x^{2} + x^{4} | + 2 (2 | 16 - 8 x^{2} + x^{4} | + 16 x^{2} - 8 x^{4} + x^{6}))}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.8

Apply the distributive property.

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 16 - 8 x^{2} + x^{4} | + 2 (2 | 16 - 8 x^{2} + x^{4} |) + 2 (16 x^{2}) + 2 (- 8 x^{4}) + 2 x^{6})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.9

Simplify.

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Step 2.6.3.8.9.1

Multiply $2$ by $2$ .

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 16 - 8 x^{2} + x^{4} | + 4 | 16 - 8 x^{2} + x^{4} | + 2 (16 x^{2}) + 2 (- 8 x^{4}) + 2 x^{6})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.9.2

Multiply $16$ by $2$ .

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 16 - 8 x^{2} + x^{4} | + 4 | 16 - 8 x^{2} + x^{4} | + 32 x^{2} + 2 (- 8 x^{4}) + 2 x^{6})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.9.3

Multiply $- 8$ by $2$ .

$f'' (x) = - \frac{\frac{2 (- 3 x^{2} | 16 - 8 x^{2} + x^{4} | + 4 | 16 - 8 x^{2} + x^{4} | + 32 x^{2} - 16 x^{4} + 2 x^{6})}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.3.8.10

Reorder terms.

$f'' (x) = - \frac{\frac{2 (2 x^{6} - 16 x^{4} + 32 x^{2} - 3 x^{2} | 16 - 8 x^{2} + x^{4} | + 4 | 16 - 8 x^{2} + x^{4} |)}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$

Step 2.6.4

Combine terms.

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Step 2.6.4.1

Rewrite $\frac{\frac{2 (2 x^{6} - 16 x^{4} + 32 x^{2} - 3 x^{2} | 16 - 8 x^{2} + x^{4} | + 4 | 16 - 8 x^{2} + x^{4} |)}{| (2 + x) (2 - x) |}}{{| 4 - x^{2} |}^{2}}$ as a product.

$f'' (x) = - (\frac{2 (2 x^{6} - 16 x^{4} + 32 x^{2} - 3 x^{2} | 16 - 8 x^{2} + x^{4} | + 4 | 16 - 8 x^{2} + x^{4} |)}{| (2 + x) (2 - x) |} \cdot \frac{1}{{| 4 - x^{2} |}^{2}})$

Step 2.6.4.2

Multiply $\frac{2 (2 x^{6} - 16 x^{4} + 32 x^{2} - 3 x^{2} | 16 - 8 x^{2} + x^{4} | + 4 | 16 - 8 x^{2} + x^{4} |)}{| (2 + x) (2 - x) |}$ by $\frac{1}{{| 4 - x^{2} |}^{2}}$ .

$f'' (x) = - \frac{2 (2 x^{6} - 16 x^{4} + 32 x^{2} - 3 x^{2} | 16 - 8 x^{2} + x^{4} | + 4 | 16 - 8 x^{2} + x^{4} |)}{| (2 + x) (2 - x) | {| 4 - x^{2} |}^{2}}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$- \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = | x |$ and $g (x) = 4 - x^{2}$ .

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Step 4.1.1.1

To apply the Chain Rule, set $u$ as $4 - x^{2}$ .

$\frac{d}{d u} [| u |] \frac{d}{d x} [4 - x^{2}]$

Step 4.1.1.2

The derivative of $| u |$ with respect to $u$ is $\frac{u}{| u |}$ .

$\frac{u}{| u |} \frac{d}{d x} [4 - x^{2}]$

Step 4.1.1.3

Replace all occurrences of $u$ with $4 - x^{2}$ .

$\frac{4 - x^{2}}{| 4 - x^{2} |} \frac{d}{d x} [4 - x^{2}]$

Step 4.1.2

Differentiate.

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Step 4.1.2.1

By the Sum Rule, the derivative of $4 - x^{2}$ with respect to $x$ is $\frac{d}{d x} [4] + \frac{d}{d x} [- x^{2}]$ .

$\frac{4 - x^{2}}{| 4 - x^{2} |} (\frac{d}{d x} [4] + \frac{d}{d x} [- x^{2}])$

Step 4.1.2.2

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$\frac{4 - x^{2}}{| 4 - x^{2} |} (0 + \frac{d}{d x} [- x^{2}])$

Step 4.1.2.3

Add $0$ and $\frac{d}{d x} [- x^{2}]$ .

$\frac{4 - x^{2}}{| 4 - x^{2} |} \frac{d}{d x} [- x^{2}]$

Step 4.1.2.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$\frac{4 - x^{2}}{| 4 - x^{2} |} (- \frac{d}{d x} [x^{2}])$

Step 4.1.2.5

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$\frac{4 - x^{2}}{| 4 - x^{2} |} (- (2 x))$

Step 4.1.2.6

Combine fractions.

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Step 4.1.2.6.1

Multiply $2$ by $- 1$ .

$\frac{4 - x^{2}}{| 4 - x^{2} |} (- 2 x)$

Step 4.1.2.6.2

Combine $- 2$ and $\frac{4 - x^{2}}{| 4 - x^{2} |}$ .

$\frac{- 2 (4 - x^{2})}{| 4 - x^{2} |} x$

Step 4.1.2.6.3

Combine $\frac{- 2 (4 - x^{2})}{| 4 - x^{2} |}$ and $x$ .

$\frac{- 2 (4 - x^{2}) x}{| 4 - x^{2} |}$

Step 4.1.2.6.4

Move the negative in front of the fraction.

$- \frac{(2 (4 - x^{2})) x}{| 4 - x^{2} |}$

Step 4.1.3

Simplify.

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Step 4.1.3.1

Apply the distributive property.

$- \frac{(2 \cdot 4 + 2 (- x^{2})) x}{| 4 - x^{2} |}$

Step 4.1.3.2

Apply the distributive property.

$- \frac{2 \cdot 4 x + 2 (- x^{2}) x}{| 4 - x^{2} |}$

Step 4.1.3.3

Simplify each term.

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Step 4.1.3.3.1

Multiply $2$ by $4$ .

$- \frac{8 x + 2 (- x^{2}) x}{| 4 - x^{2} |}$

Step 4.1.3.3.2

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 4.1.3.3.2.1

Move $x$ .

$- \frac{8 x + 2 (- (x \cdot x^{2}))}{| 4 - x^{2} |}$

Step 4.1.3.3.2.2

Multiply $x$ by $x^{2}$ .

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Step 4.1.3.3.2.2.1

Raise $x$ to the power of $1$ .

$- \frac{8 x + 2 (- (x^{1} x^{2}))}{| 4 - x^{2} |}$

Step 4.1.3.3.2.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- \frac{8 x + 2 (- x^{1 + 2})}{| 4 - x^{2} |}$

Step 4.1.3.3.2.3

Add $1$ and $2$ .

$- \frac{8 x + 2 (- x^{3})}{| 4 - x^{2} |}$

Step 4.1.3.3.3

Multiply $- 1$ by $2$ .

$f' (x) = - \frac{8 x - 2 x^{3}}{| 4 - x^{2} |}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{8 x - 2 x^{3}}{| 4 - x^{2} |}$ .

$- \frac{8 x - 2 x^{3}}{| 4 - x^{2} |}$

Step 5

Set the first derivative equal to $0$ then solve the equation $- \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$- \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} = 0$

Step 5.2

Set the numerator equal to zero.

$8 x - 2 x^{3} = 0$

Step 5.3

Solve the equation for $x$ .

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Step 5.3.1

Factor the left side of the equation.

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Step 5.3.1.1

Factor $2 x$ out of $8 x - 2 x^{3}$ .

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Step 5.3.1.1.1

Factor $2 x$ out of $8 x$ .

$2 x (4) - 2 x^{3} = 0$

Step 5.3.1.1.2

Factor $2 x$ out of $- 2 x^{3}$ .

$2 x (4) + 2 x (- x^{2}) = 0$

Step 5.3.1.1.3

Factor $2 x$ out of $2 x (4) + 2 x (- x^{2})$ .

$2 x (4 - x^{2}) = 0$

Step 5.3.1.2

Rewrite $4$ as $2^{2}$ .

$2 x (2^{2} - x^{2}) = 0$

Step 5.3.1.3

Factor.

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Step 5.3.1.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = x$ .

$2 x ((2 + x) (2 - x)) = 0$

Step 5.3.1.3.2

Remove unnecessary parentheses.

$2 x (2 + x) (2 - x) = 0$

Step 5.3.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$2 + x = 0$

$2 - x = 0$

Step 5.3.3

Set $x$ equal to $0$ .

$x = 0$

Step 5.3.4

Set $2 + x$ equal to $0$ and solve for $x$ .

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Step 5.3.4.1

Set $2 + x$ equal to $0$ .

$2 + x = 0$

Step 5.3.4.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5.3.5

Set $2 - x$ equal to $0$ and solve for $x$ .

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Step 5.3.5.1

Set $2 - x$ equal to $0$ .

$2 - x = 0$

Step 5.3.5.2

Solve $2 - x = 0$ for $x$ .

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Step 5.3.5.2.1

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 5.3.5.2.2

Divide each term in $- x = - 2$ by $- 1$ and simplify.

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Step 5.3.5.2.2.1

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Step 5.3.5.2.2.2

Simplify the left side.

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Step 5.3.5.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Step 5.3.5.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Step 5.3.5.2.2.3

Simplify the right side.

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Step 5.3.5.2.2.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 5.3.6

The final solution is all the values that make $2 x (2 + x) (2 - x) = 0$ true.

$x = 0, - 2, 2$

Step 5.4

Exclude the solutions that do not make $- \frac{8 x - 2 x^{3}}{| 4 - x^{2} |} = 0$ true.

$x = 0$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in $\frac{8 x - 2 x^{3}}{| 4 - x^{2} |}$ equal to $0$ to find where the expression is undefined.

$| 4 - x^{2} | = 0$

Step 6.2

Solve for $x$ .

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Step 6.2.1

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$4 - x^{2} = \pm 0$

Step 6.2.2

Plus or minus $0$ is $0$ .

$4 - x^{2} = 0$

Step 6.2.3

Subtract $4$ from both sides of the equation.

$- x^{2} = - 4$

Step 6.2.4

Divide each term in $- x^{2} = - 4$ by $- 1$ and simplify.

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Step 6.2.4.1

Divide each term in $- x^{2} = - 4$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{- 4}{- 1}$

Step 6.2.4.2

Simplify the left side.

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Step 6.2.4.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{- 4}{- 1}$

Step 6.2.4.2.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 4}{- 1}$

Step 6.2.4.3

Simplify the right side.

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Step 6.2.4.3.1

Divide $- 4$ by $- 1$ .

$x^{2} = 4$

Step 6.2.5

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Step 6.2.6

Simplify $\pm \sqrt{4}$ .

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Step 6.2.6.1

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Step 6.2.6.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

Step 6.2.7

The complete solution is the result of both the positive and negative portions of the solution.

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Step 6.2.7.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Step 6.2.7.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

Step 6.2.7.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

Step 6.3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - 2, x = 2$

Step 7

Critical points to evaluate.

$x = 0, - 2, 2$

Step 8

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{2 (2 {(0)}^{6} - 16 {(0)}^{4} + 32 {(0)}^{2} - 3 {(0)}^{2} | 16 - 8 {(0)}^{2} + {(0)}^{4} | + 4 | 16 - 8 {(0)}^{2} + {(0)}^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - {(0)}^{2} |}^{2}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Remove parentheses.

$- \frac{2 (2 {(0)}^{6} - 16 {(0)}^{4} + 32 {(0)}^{2} - 3 {(0)}^{2} | 16 - 8 {(0)}^{2} + {(0)}^{4} | + 4 | 16 - 8 {(0)}^{2} + {(0)}^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - {(0)}^{2} |}^{2}}$

Step 9.2

Simplify the numerator.

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Step 9.2.1

Raising $0$ to any positive power yields $0$ .

$- \frac{2 (2 \cdot 0 - 16 \cdot 0^{4} + 32 \cdot 0^{2} - 3 \cdot 0^{2} | 16 - 8 \cdot 0^{2} + 0^{4} | + 4 | 16 - 8 \cdot 0^{2} + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.2

Multiply $2$ by $0$ .

$- \frac{2 (0 - 16 \cdot 0^{4} + 32 \cdot 0^{2} - 3 \cdot 0^{2} | 16 - 8 \cdot 0^{2} + 0^{4} | + 4 | 16 - 8 \cdot 0^{2} + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.3

Raising $0$ to any positive power yields $0$ .

$- \frac{2 (0 - 16 \cdot 0 + 32 \cdot 0^{2} - 3 \cdot 0^{2} | 16 - 8 \cdot 0^{2} + 0^{4} | + 4 | 16 - 8 \cdot 0^{2} + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.4

Multiply $- 16$ by $0$ .

$- \frac{2 (0 + 0 + 32 \cdot 0^{2} - 3 \cdot 0^{2} | 16 - 8 \cdot 0^{2} + 0^{4} | + 4 | 16 - 8 \cdot 0^{2} + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.5

Raising $0$ to any positive power yields $0$ .

$- \frac{2 (0 + 0 + 32 \cdot 0 - 3 \cdot 0^{2} | 16 - 8 \cdot 0^{2} + 0^{4} | + 4 | 16 - 8 \cdot 0^{2} + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.6

Multiply $32$ by $0$ .

$- \frac{2 (0 + 0 + 0 - 3 \cdot 0^{2} | 16 - 8 \cdot 0^{2} + 0^{4} | + 4 | 16 - 8 \cdot 0^{2} + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.7

Raising $0$ to any positive power yields $0$ .

$- \frac{2 (0 + 0 + 0 - 3 \cdot 0 | 16 - 8 \cdot 0^{2} + 0^{4} | + 4 | 16 - 8 \cdot 0^{2} + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.8

Multiply $- 3$ by $0$ .

$- \frac{2 (0 + 0 + 0 + 0 | 16 - 8 \cdot 0^{2} + 0^{4} | + 4 | 16 - 8 \cdot 0^{2} + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.9

Simplify each term.

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Step 9.2.9.1

Raising $0$ to any positive power yields $0$ .

$- \frac{2 (0 + 0 + 0 + 0 | 16 - 8 \cdot 0 + 0^{4} | + 4 | 16 - 8 \cdot 0^{2} + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.9.2

Multiply $- 8$ by $0$ .

$- \frac{2 (0 + 0 + 0 + 0 | 16 + 0 + 0^{4} | + 4 | 16 - 8 \cdot 0^{2} + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.9.3

Raising $0$ to any positive power yields $0$ .

$- \frac{2 (0 + 0 + 0 + 0 | 16 + 0 + 0 | + 4 | 16 - 8 \cdot 0^{2} + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.10

Add $16$ and $0$ .

$- \frac{2 (0 + 0 + 0 + 0 | 16 + 0 | + 4 | 16 - 8 \cdot 0^{2} + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.11

Add $16$ and $0$ .

$- \frac{2 (0 + 0 + 0 + 0 | 16 | + 4 | 16 - 8 \cdot 0^{2} + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.12

The absolute value is the distance between a number and zero. The distance between $0$ and $16$ is $16$ .

$- \frac{2 (0 + 0 + 0 + 0 \cdot 16 + 4 | 16 - 8 \cdot 0^{2} + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.13

Multiply $0$ by $16$ .

$- \frac{2 (0 + 0 + 0 + 0 + 4 | 16 - 8 \cdot 0^{2} + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.14

Simplify each term.

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Step 9.2.14.1

Raising $0$ to any positive power yields $0$ .

$- \frac{2 (0 + 0 + 0 + 0 + 4 | 16 - 8 \cdot 0 + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.14.2

Multiply $- 8$ by $0$ .

$- \frac{2 (0 + 0 + 0 + 0 + 4 | 16 + 0 + 0^{4} |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.14.3

Raising $0$ to any positive power yields $0$ .

$- \frac{2 (0 + 0 + 0 + 0 + 4 | 16 + 0 + 0 |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.15

Add $16$ and $0$ .

$- \frac{2 (0 + 0 + 0 + 0 + 4 | 16 + 0 |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.16

Add $16$ and $0$ .

$- \frac{2 (0 + 0 + 0 + 0 + 4 | 16 |)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.17

The absolute value is the distance between a number and zero. The distance between $0$ and $16$ is $16$ .

$- \frac{2 (0 + 0 + 0 + 0 + 4 \cdot 16)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.18

Multiply $4$ by $16$ .

$- \frac{2 (0 + 0 + 0 + 0 + 64)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.19

Add $0$ and $0$ .

$- \frac{2 (0 + 0 + 0 + 64)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.20

Add $0$ and $0$ .

$- \frac{2 (0 + 0 + 64)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.21

Add $0$ and $0$ .

$- \frac{2 (0 + 64)}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.2.22

Add $0$ and $64$ .

$- \frac{2 \cdot 64}{| (2 + 0) (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.3

Simplify the denominator.

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Step 9.3.1

Add $2$ and $0$ .

$- \frac{2 \cdot 64}{| 2 (2 - (0)) | {| 4 - 0^{2} |}^{2}}$

Step 9.3.2

Multiply $- 1$ by $0$ .

$- \frac{2 \cdot 64}{| 2 (2 + 0) | {| 4 - 0^{2} |}^{2}}$

Step 9.3.3

Add $2$ and $0$ .

$- \frac{2 \cdot 64}{| 2 \cdot 2 | {| 4 - 0^{2} |}^{2}}$

Step 9.3.4

Raising $0$ to any positive power yields $0$ .

$- \frac{2 \cdot 64}{| 2 \cdot 2 | {| 4 - 0 |}^{2}}$

Step 9.3.5

Multiply $- 1$ by $0$ .

$- \frac{2 \cdot 64}{| 2 \cdot 2 | {| 4 + 0 |}^{2}}$

Step 9.3.6

Add $4$ and $0$ .

$- \frac{2 \cdot 64}{| 2 \cdot 2 | {| 4 |}^{2}}$

Step 9.3.7

Multiply $2$ by $2$ .

$- \frac{2 \cdot 64}{| 4 | {| 4 |}^{2}}$

Step 9.3.8

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$- \frac{2 \cdot 64}{4 {| 4 |}^{2}}$

Step 9.3.9

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$- \frac{2 \cdot 64}{4 \cdot 4^{2}}$

Step 9.3.10

Raise $4$ to the power of $2$ .

$- \frac{2 \cdot 64}{4 \cdot 16}$

Step 9.4

Simplify the expression.

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Step 9.4.1

Multiply $2$ by $64$ .

$- \frac{128}{4 \cdot 16}$

Step 9.4.2

Multiply $4$ by $16$ .

$- \frac{128}{64}$

Step 9.4.3

Divide $128$ by $64$ .

$- 1 \cdot 2$

Step 9.4.4

Multiply $- 1$ by $2$ .

$- 2$

Step 10

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

Step 11

Find the y-value when $x = 0$ .

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Step 11.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = | 4 - {(0)}^{2} |$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

Raising $0$ to any positive power yields $0$ .

$f (0) = | 4 - 0 |$

Step 11.2.1.2

Multiply $- 1$ by $0$ .

$f (0) = | 4 + 0 |$

Step 11.2.2

Add $4$ and $0$ .

$f (0) = | 4 |$

Step 11.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$f (0) = 4$

Step 11.2.4

The final answer is $4$ .

$y = 4$

Step 12

Evaluate the second derivative at $x = - 2$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{2 (2 {(- 2)}^{6} - 16 {(- 2)}^{4} + 32 {(- 2)}^{2} - 3 {(- 2)}^{2} | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} | + 4 | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} |)}{| (2 - 2) (2 - (- 2)) | {| 4 - {(- 2)}^{2} |}^{2}}$

Step 13

Evaluate the second derivative.

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Step 13.1

Remove parentheses.

Step 13.2

Subtract $2$ from $2$ .

$- \frac{2 (2 {(- 2)}^{6} - 16 {(- 2)}^{4} + 32 {(- 2)}^{2} - 3 {(- 2)}^{2} | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} | + 4 | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} |)}{| 0 (2 - (- 2)) | {| 4 - {(- 2)}^{2} |}^{2}}$

Step 13.3

Multiply $- 1$ by $- 2$ .

$- \frac{2 (2 {(- 2)}^{6} - 16 {(- 2)}^{4} + 32 {(- 2)}^{2} - 3 {(- 2)}^{2} | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} | + 4 | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} |)}{| 0 (2 + 2) | {| 4 - {(- 2)}^{2} |}^{2}}$

Step 13.4

Add $2$ and $2$ .

$- \frac{2 (2 {(- 2)}^{6} - 16 {(- 2)}^{4} + 32 {(- 2)}^{2} - 3 {(- 2)}^{2} | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} | + 4 | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} |)}{| 0 \cdot 4 | {| 4 - {(- 2)}^{2} |}^{2}}$

Step 13.5

Multiply $0$ by $4$ .

$- \frac{2 (2 {(- 2)}^{6} - 16 {(- 2)}^{4} + 32 {(- 2)}^{2} - 3 {(- 2)}^{2} | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} | + 4 | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} |)}{| 0 | {| 4 - {(- 2)}^{2} |}^{2}}$

Step 13.6

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$- \frac{2 (2 {(- 2)}^{6} - 16 {(- 2)}^{4} + 32 {(- 2)}^{2} - 3 {(- 2)}^{2} | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} | + 4 | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} |)}{0 {| 4 - {(- 2)}^{2} |}^{2}}$

Step 13.7

Simplify each term.

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Step 13.7.1

Raise $- 2$ to the power of $2$ .

$- \frac{2 (2 {(- 2)}^{6} - 16 {(- 2)}^{4} + 32 {(- 2)}^{2} - 3 {(- 2)}^{2} | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} | + 4 | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} |)}{0 {| 4 - 1 \cdot 4 |}^{2}}$

Step 13.7.2

Multiply $- 1$ by $4$ .

$- \frac{2 (2 {(- 2)}^{6} - 16 {(- 2)}^{4} + 32 {(- 2)}^{2} - 3 {(- 2)}^{2} | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} | + 4 | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} |)}{0 {| 4 - 4 |}^{2}}$

Step 13.8

Subtract $4$ from $4$ .

$- \frac{2 (2 {(- 2)}^{6} - 16 {(- 2)}^{4} + 32 {(- 2)}^{2} - 3 {(- 2)}^{2} | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} | + 4 | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} |)}{0 {| 0 |}^{2}}$

Step 13.9

The absolute value is the distance between a number and zero. The distance between $0$ and $0$ is $0$ .

$- \frac{2 (2 {(- 2)}^{6} - 16 {(- 2)}^{4} + 32 {(- 2)}^{2} - 3 {(- 2)}^{2} | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} | + 4 | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} |)}{0 \cdot 0^{2}}$

Step 13.10

Multiply $0$ by $0^{2}$ by adding the exponents.

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Step 13.10.1

Multiply $0$ by $0^{2}$ .

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Step 13.10.1.1

Raise $0$ to the power of $1$ .

$- \frac{2 (2 {(- 2)}^{6} - 16 {(- 2)}^{4} + 32 {(- 2)}^{2} - 3 {(- 2)}^{2} | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} | + 4 | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} |)}{0^{1} \cdot 0^{2}}$

Step 13.10.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- \frac{2 (2 {(- 2)}^{6} - 16 {(- 2)}^{4} + 32 {(- 2)}^{2} - 3 {(- 2)}^{2} | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} | + 4 | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} |)}{0^{1 + 2}}$

Step 13.10.2

Add $1$ and $2$ .

$- \frac{2 (2 {(- 2)}^{6} - 16 {(- 2)}^{4} + 32 {(- 2)}^{2} - 3 {(- 2)}^{2} | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} | + 4 | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} |)}{0^{3}}$

Step 13.11

Raising $0$ to any positive power yields $0$ .

$- \frac{2 (2 {(- 2)}^{6} - 16 {(- 2)}^{4} + 32 {(- 2)}^{2} - 3 {(- 2)}^{2} | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} | + 4 | 16 - 8 {(- 2)}^{2} + {(- 2)}^{4} |)}{0}$

Step 13.12

The expression contains a division by $0$ . The expression is undefined.

Undefined

Step 14

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

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Step 14.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, - 2) \cup (- 2, 0) \cup (0, 2) \cup (2, \infty)$

Step 14.2

Substitute any number, such as $- 4$ , from the interval $(- \infty, - 2)$ in the first derivative $- \frac{8 x - 2 x^{3}}{| 4 - x^{2} |}$ to check if the result is negative or positive.

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Step 14.2.1

Replace the variable $x$ with $- 4$ in the expression.

$f' (- 4) = - \frac{8 (- 4) - 2 {(- 4)}^{3}}{| 4 - {(- 4)}^{2} |}$

Step 14.2.2

Simplify the result.

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Step 14.2.2.1

Simplify the numerator.

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Step 14.2.2.1.1

Multiply $8$ by $- 4$ .

$f' (- 4) = - \frac{- 32 - 2 {(- 4)}^{3}}{| 4 - {(- 4)}^{2} |}$

Step 14.2.2.1.2

Raise $- 4$ to the power of $3$ .

$f' (- 4) = - \frac{- 32 - 2 \cdot - 64}{| 4 - {(- 4)}^{2} |}$

Step 14.2.2.1.3

Multiply $- 2$ by $- 64$ .

$f' (- 4) = - \frac{- 32 + 128}{| 4 - {(- 4)}^{2} |}$

Step 14.2.2.1.4

Add $- 32$ and $128$ .

$f' (- 4) = - \frac{96}{| 4 - {(- 4)}^{2} |}$

Step 14.2.2.2

Simplify the denominator.

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Step 14.2.2.2.1

Raise $- 4$ to the power of $2$ .

$f' (- 4) = - \frac{96}{| 4 - 1 \cdot 16 |}$

Step 14.2.2.2.2

Multiply $- 1$ by $16$ .

$f' (- 4) = - \frac{96}{| 4 - 16 |}$

Step 14.2.2.2.3

Subtract $16$ from $4$ .

$f' (- 4) = - \frac{96}{| - 12 |}$

Step 14.2.2.2.4

The absolute value is the distance between a number and zero. The distance between $- 12$ and $0$ is $12$ .

$f' (- 4) = - \frac{96}{12}$

Step 14.2.2.3

Simplify the expression.

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Step 14.2.2.3.1

Divide $96$ by $12$ .

$f' (- 4) = - 1 \cdot 8$

Step 14.2.2.3.2

Multiply $- 1$ by $8$ .

$f' (- 4) = - 8$

Step 14.2.2.4

The final answer is $- 8$ .

$- 8$

Step 14.3

Substitute any number, such as $- 1$ , from the interval $(- 2, 0)$ in the first derivative $- \frac{8 x - 2 x^{3}}{| 4 - x^{2} |}$ to check if the result is negative or positive.

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Step 14.3.1

Replace the variable $x$ with $- 1$ in the expression.

$f' (- 1) = - \frac{8 (- 1) - 2 {(- 1)}^{3}}{| 4 - {(- 1)}^{2} |}$

Step 14.3.2

Simplify the result.

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Step 14.3.2.1

Simplify the numerator.

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Step 14.3.2.1.1

Multiply $8$ by $- 1$ .

$f' (- 1) = - \frac{- 8 - 2 {(- 1)}^{3}}{| 4 - {(- 1)}^{2} |}$

Step 14.3.2.1.2

Raise $- 1$ to the power of $3$ .

$f' (- 1) = - \frac{- 8 - 2 \cdot - 1}{| 4 - {(- 1)}^{2} |}$

Step 14.3.2.1.3

Multiply $- 2$ by $- 1$ .

$f' (- 1) = - \frac{- 8 + 2}{| 4 - {(- 1)}^{2} |}$

Step 14.3.2.1.4

Add $- 8$ and $2$ .

$f' (- 1) = - \frac{- 6}{| 4 - {(- 1)}^{2} |}$

Step 14.3.2.2

Simplify the denominator.

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Step 14.3.2.2.1

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

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Step 14.3.2.2.1.1

Multiply $- 1$ by ${(- 1)}^{2}$ .

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Step 14.3.2.2.1.1.1

Raise $- 1$ to the power of $1$ .

$f' (- 1) = - \frac{- 6}{| 4 + (- 1) {(- 1)}^{2} |}$

Step 14.3.2.2.1.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f' (- 1) = - \frac{- 6}{| 4 + {(- 1)}^{1 + 2} |}$

Step 14.3.2.2.1.2

Add $1$ and $2$ .

$f' (- 1) = - \frac{- 6}{| 4 + {(- 1)}^{3} |}$

Step 14.3.2.2.2

Raise $- 1$ to the power of $3$ .

$f' (- 1) = - \frac{- 6}{| 4 - 1 |}$

Step 14.3.2.2.3

Subtract $1$ from $4$ .

$f' (- 1) = - \frac{- 6}{| 3 |}$

Step 14.3.2.2.4

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$f' (- 1) = - \frac{- 6}{3}$

Step 14.3.2.3

Simplify the expression.

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Step 14.3.2.3.1

Divide $- 6$ by $3$ .

$f' (- 1) = 2$

Step 14.3.2.3.2

Multiply $- 1$ by $- 2$ .

$f' (- 1) = 2$

Step 14.3.2.4

The final answer is $2$ .

$2$

Step 14.4

Substitute any number, such as $1$ , from the interval $(0, 2)$ in the first derivative $- \frac{8 x - 2 x^{3}}{| 4 - x^{2} |}$ to check if the result is negative or positive.

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Step 14.4.1

Replace the variable $x$ with $1$ in the expression.

$f' (1) = - \frac{8 (1) - 2 {(1)}^{3}}{| 4 - {(1)}^{2} |}$

Step 14.4.2

Simplify the result.

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Step 14.4.2.1

Simplify the numerator.

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Step 14.4.2.1.1

Multiply $8$ by $1$ .

$f' (1) = - \frac{8 - 2 \cdot 1^{3}}{| 4 - 1^{2} |}$

Step 14.4.2.1.2

One to any power is one.

$f' (1) = - \frac{8 - 2 \cdot 1}{| 4 - 1^{2} |}$

Step 14.4.2.1.3

Multiply $- 2$ by $1$ .

$f' (1) = - \frac{8 - 2}{| 4 - 1^{2} |}$

Step 14.4.2.1.4

Subtract $2$ from $8$ .

$f' (1) = - \frac{6}{| 4 - 1^{2} |}$

Step 14.4.2.2

Simplify the denominator.

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Step 14.4.2.2.1

One to any power is one.

$f' (1) = - \frac{6}{| 4 - 1 \cdot 1 |}$

Step 14.4.2.2.2

Multiply $- 1$ by $1$ .

$f' (1) = - \frac{6}{| 4 - 1 |}$

Step 14.4.2.2.3

Subtract $1$ from $4$ .

$f' (1) = - \frac{6}{| 3 |}$

Step 14.4.2.2.4

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$f' (1) = - \frac{6}{3}$

Step 14.4.2.3

Simplify the expression.

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Step 14.4.2.3.1

Divide $6$ by $3$ .

$f' (1) = - 1 \cdot 2$

Step 14.4.2.3.2

Multiply $- 1$ by $2$ .

$f' (1) = - 2$

Step 14.4.2.4

The final answer is $- 2$ .

$- 2$

Step 14.5

Substitute any number, such as $4$ , from the interval $(2, \infty)$ in the first derivative $- \frac{8 x - 2 x^{3}}{| 4 - x^{2} |}$ to check if the result is negative or positive.

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Step 14.5.1

Replace the variable $x$ with $4$ in the expression.

$f' (4) = - \frac{8 (4) - 2 {(4)}^{3}}{| 4 - {(4)}^{2} |}$

Step 14.5.2

Simplify the result.

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Step 14.5.2.1

Simplify the numerator.

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Step 14.5.2.1.1

Multiply $8$ by $4$ .

$f' (4) = - \frac{32 - 2 \cdot 4^{3}}{| 4 - 4^{2} |}$

Step 14.5.2.1.2

Raise $4$ to the power of $3$ .

$f' (4) = - \frac{32 - 2 \cdot 64}{| 4 - 4^{2} |}$

Step 14.5.2.1.3

Multiply $- 2$ by $64$ .

$f' (4) = - \frac{32 - 128}{| 4 - 4^{2} |}$

Step 14.5.2.1.4

Subtract $128$ from $32$ .

$f' (4) = - \frac{- 96}{| 4 - 4^{2} |}$

Step 14.5.2.2

Simplify the denominator.

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Step 14.5.2.2.1

Raise $4$ to the power of $2$ .

$f' (4) = - \frac{- 96}{| 4 - 1 \cdot 16 |}$

Step 14.5.2.2.2

Multiply $- 1$ by $16$ .

$f' (4) = - \frac{- 96}{| 4 - 16 |}$

Step 14.5.2.2.3

Subtract $16$ from $4$ .

$f' (4) = - \frac{- 96}{| - 12 |}$

Step 14.5.2.2.4

The absolute value is the distance between a number and zero. The distance between $- 12$ and $0$ is $12$ .

$f' (4) = - \frac{- 96}{12}$

Step 14.5.2.3

Simplify the expression.

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Step 14.5.2.3.1

Divide $- 96$ by $12$ .

$f' (4) = 8$

Step 14.5.2.3.2

Multiply $- 1$ by $- 8$ .

$f' (4) = 8$

Step 14.5.2.4

The final answer is $8$ .

$8$

Step 14.6

Since the first derivative changed signs from negative to positive around $x = - 2$ , then $x = - 2$ is a local minimum.

$x = - 2$ is a local minimum

Step 14.7

Since the first derivative changed signs from positive to negative around $x = 0$ , then $x = 0$ is a local maximum.

$x = 0$ is a local maximum

Step 14.8

Since the first derivative changed signs from negative to positive around $x = 2$ , then $x = 2$ is a local minimum.

$x = 2$ is a local minimum

Step 14.9

These are the local extrema for $f (x) = | 4 - x^{2} |$ .

$x = - 2$ is a local minimum

$x = 0$ is a local maximum

$x = 2$ is a local minimum

$x = - 2$ is a local minimum

$x = 0$ is a local maximum

$x = 2$ is a local minimum

Step 15