Calculus Examples

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

Step 1

Write $\frac{x^{4}}{4 - 8 x^{2}}$ as a function.

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

Step 2

Find the first derivative of the function.

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Step 2.1

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 $f (x) = x^{4}$ and $g (x) = 4 - 8 x^{2}$ .

$\frac{(4 - 8 x^{2}) \frac{d}{d x} [x^{4}] - x^{4} \frac{d}{d x} [4 - 8 x^{2}]}{{(4 - 8 x^{2})}^{2}}$

Step 2.2

Differentiate.

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Step 2.2.1

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$\frac{(4 - 8 x^{2}) (4 x^{3}) - x^{4} \frac{d}{d x} [4 - 8 x^{2}]}{{(4 - 8 x^{2})}^{2}}$

Step 2.2.2

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

$\frac{4 \cdot (4 - 8 x^{2}) x^{3} - x^{4} \frac{d}{d x} [4 - 8 x^{2}]}{{(4 - 8 x^{2})}^{2}}$

Step 2.2.3

By the Sum Rule, the derivative of $4 - 8 x^{2}$ with respect to $x$ is $\frac{d}{d x} [4] + \frac{d}{d x} [- 8 x^{2}]$ .

$\frac{4 (4 - 8 x^{2}) x^{3} - x^{4} (\frac{d}{d x} [4] + \frac{d}{d x} [- 8 x^{2}])}{{(4 - 8 x^{2})}^{2}}$

Step 2.2.4

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

$\frac{4 (4 - 8 x^{2}) x^{3} - x^{4} (0 + \frac{d}{d x} [- 8 x^{2}])}{{(4 - 8 x^{2})}^{2}}$

Step 2.2.5

Add $0$ and $\frac{d}{d x} [- 8 x^{2}]$ .

$\frac{4 (4 - 8 x^{2}) x^{3} - x^{4} \frac{d}{d x} [- 8 x^{2}]}{{(4 - 8 x^{2})}^{2}}$

Step 2.2.6

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 x^{2}$ with respect to $x$ is $- 8 \frac{d}{d x} [x^{2}]$ .

$\frac{4 (4 - 8 x^{2}) x^{3} - x^{4} (- 8 \frac{d}{d x} [x^{2}])}{{(4 - 8 x^{2})}^{2}}$

Step 2.2.7

Multiply $- 8$ by $- 1$ .

$\frac{4 (4 - 8 x^{2}) x^{3} + 8 x^{4} \frac{d}{d x} [x^{2}]}{{(4 - 8 x^{2})}^{2}}$

Step 2.2.8

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

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

Step 2.2.9

Multiply $2$ by $8$ .

$\frac{4 (4 - 8 x^{2}) x^{3} + 16 x^{4} x}{{(4 - 8 x^{2})}^{2}}$

Step 2.3

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

$\frac{4 (4 - 8 x^{2}) x^{3} + 16 (x^{1} x^{4})}{{(4 - 8 x^{2})}^{2}}$

Step 2.4

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

$\frac{4 (4 - 8 x^{2}) x^{3} + 16 x^{1 + 4}}{{(4 - 8 x^{2})}^{2}}$

Step 2.5

Add $1$ and $4$ .

$\frac{4 (4 - 8 x^{2}) x^{3} + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 2.6

Simplify.

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Step 2.6.1

Apply the distributive property.

$\frac{(4 \cdot 4 + 4 (- 8 x^{2})) x^{3} + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 2.6.2

Apply the distributive property.

$\frac{4 \cdot 4 x^{3} + 4 (- 8 x^{2}) x^{3} + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

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

Multiply $4$ by $4$ .

$\frac{16 x^{3} + 4 (- 8 x^{2}) x^{3} + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 2.6.3.1.2

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

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Step 2.6.3.1.2.1

Move $x^{3}$ .

$\frac{16 x^{3} + 4 (- 8 (x^{3} x^{2})) + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 2.6.3.1.2.2

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

$\frac{16 x^{3} + 4 (- 8 x^{3 + 2}) + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 2.6.3.1.2.3

Add $3$ and $2$ .

$\frac{16 x^{3} + 4 (- 8 x^{5}) + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 2.6.3.1.3

Multiply $- 8$ by $4$ .

$\frac{16 x^{3} - 32 x^{5} + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 2.6.3.2

Add $- 32 x^{5}$ and $16 x^{5}$ .

$\frac{16 x^{3} - 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 2.6.4

Reorder terms.

$\frac{- 16 x^{5} + 16 x^{3}}{{(- 8 x^{2} + 4)}^{2}}$

Step 2.6.5

Simplify the numerator.

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Step 2.6.5.1

Factor $16 x^{3}$ out of $- 16 x^{5} + 16 x^{3}$ .

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Step 2.6.5.1.1

Factor $16 x^{3}$ out of $- 16 x^{5}$ .

$\frac{16 x^{3} (- x^{2}) + 16 x^{3}}{{(- 8 x^{2} + 4)}^{2}}$

Step 2.6.5.1.2

Factor $16 x^{3}$ out of $16 x^{3}$ .

$\frac{16 x^{3} (- x^{2}) + 16 x^{3} (1)}{{(- 8 x^{2} + 4)}^{2}}$

Step 2.6.5.1.3

Factor $16 x^{3}$ out of $16 x^{3} (- x^{2}) + 16 x^{3} (1)$ .

$\frac{16 x^{3} (- x^{2} + 1)}{{(- 8 x^{2} + 4)}^{2}}$

Step 2.6.5.2

Rewrite $1$ as $1^{2}$ .

$\frac{16 x^{3} (- x^{2} + 1^{2})}{{(- 8 x^{2} + 4)}^{2}}$

Step 2.6.5.3

Reorder $- x^{2}$ and $1^{2}$ .

$\frac{16 x^{3} (1^{2} - x^{2})}{{(- 8 x^{2} + 4)}^{2}}$

Step 2.6.5.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x$ .

$\frac{16 x^{3} (1 + x) (1 - x)}{{(- 8 x^{2} + 4)}^{2}}$

Step 2.6.6

Simplify the denominator.

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Step 2.6.6.1

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

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Step 2.6.6.1.1

Factor $4$ out of $- 8 x^{2}$ .

$\frac{16 x^{3} (1 + x) (1 - x)}{{(4 (- 2 x^{2}) + 4)}^{2}}$

Step 2.6.6.1.2

Factor $4$ out of $4$ .

$\frac{16 x^{3} (1 + x) (1 - x)}{{(4 (- 2 x^{2}) + 4 (1))}^{2}}$

Step 2.6.6.1.3

Factor $4$ out of $4 (- 2 x^{2}) + 4 (1)$ .

$\frac{16 x^{3} (1 + x) (1 - x)}{{(4 (- 2 x^{2} + 1))}^{2}}$

Step 2.6.6.2

Apply the product rule to $4 (- 2 x^{2} + 1)$ .

$\frac{16 x^{3} (1 + x) (1 - x)}{4^{2} {(- 2 x^{2} + 1)}^{2}}$

Step 2.6.6.3

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

$\frac{16 x^{3} (1 + x) (1 - x)}{16 {(- 2 x^{2} + 1)}^{2}}$

Step 2.6.7

Cancel the common factor of $16$ .

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Step 2.6.7.1

Cancel the common factor.

$\frac{16 x^{3} (1 + x) (1 - x)}{16 {(- 2 x^{2} + 1)}^{2}}$

Step 2.6.7.2

Rewrite the expression.

$\frac{(x^{3} (1 + x)) (1 - x)}{{(- 2 x^{2} + 1)}^{2}}$

$\frac{x^{3} (1 + x) (1 - x)}{{(- 2 x^{2} + 1)}^{2}}$

Step 3

Find the second derivative of the function.

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Step 3.1

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

Step 3.2

Multiply the exponents in ${({(- 2 x^{2} + 1)}^{2})}^{2}$ .

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Step 3.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 3.2.2

Multiply $2$ by $2$ .

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

Step 3.3

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 $f (x) = x^{3} (1 + x)$ and $g (x) = 1 - x$ .

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

Step 3.4

Differentiate.

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Step 3.4.1

By the Sum Rule, the derivative of $1 - x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- x]$ .

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

Step 3.4.2

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

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

Step 3.4.3

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

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

Step 3.4.4

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

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

Step 3.4.5

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

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

Step 3.4.6

Simplify the expression.

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Step 3.4.6.1

Multiply $- 1$ by $1$ .

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

Step 3.4.6.2

Move $- 1$ to the left of $x^{3} (1 + x)$ .

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

Step 3.4.6.3

Rewrite $- 1 x^{3}$ as $- x^{3}$ .

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

Step 3.5

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 $f (x) = x^{3}$ and $g (x) = 1 + x$ .

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

Step 3.6

Differentiate.

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Step 3.6.1

By the Sum Rule, the derivative of $1 + x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [x]$ .

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

Step 3.6.2

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

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

Step 3.6.3

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

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

Step 3.6.4

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

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

Step 3.6.5

Multiply $x^{3}$ by $1$ .

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

Step 3.6.6

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

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

Step 3.6.7

Move $3$ to the left of $1 + x$ .

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

Step 3.7

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = - 2 x^{2} + 1$ .

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Step 3.7.1

To apply the Chain Rule, set $u$ as $- 2 x^{2} + 1$ .

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

Step 3.7.2

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

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

Step 3.7.3

Replace all occurrences of $u$ with $- 2 x^{2} + 1$ .

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

Step 3.8

Simplify with factoring out.

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Step 3.8.1

Multiply $2$ by $- 1$ .

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

Step 3.8.2

Factor $- 2 x^{2} + 1$ out of ${(- 2 x^{2} + 1)}^{2} (- x^{3} (1 + x) + (1 - x) (x^{3} + 3 (1 + x) x^{2})) - 2 x^{3} (1 + x) (1 - x) ((- 2 x^{2} + 1) \frac{d}{d x} [- 2 x^{2} + 1])$ .

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Step 3.8.2.1

Factor $- 2 x^{2} + 1$ out of ${(- 2 x^{2} + 1)}^{2} (- x^{3} (1 + x) + (1 - x) (x^{3} + 3 (1 + x) x^{2}))$ .

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

Step 3.8.2.2

Factor $- 2 x^{2} + 1$ out of $- 2 x^{3} (1 + x) (1 - x) ((- 2 x^{2} + 1) \frac{d}{d x} [- 2 x^{2} + 1])$ .

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

Step 3.8.2.3

Factor $- 2 x^{2} + 1$ out of $(- 2 x^{2} + 1) ((- 2 x^{2} + 1) (- x^{3} (1 + x) + (1 - x) (x^{3} + 3 (1 + x) x^{2}))) + (- 2 x^{2} + 1) (- 2 x^{3} (1 + x) (1 - x) (\frac{d}{d x} [- 2 x^{2} + 1]))$ .

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

Step 3.9

Cancel the common factors.

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Step 3.9.1

Factor $- 2 x^{2} + 1$ out of ${(- 2 x^{2} + 1)}^{4}$ .

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

Step 3.9.2

Cancel the common factor.

Step 3.9.3

Rewrite the expression.

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

Step 3.10

By the Sum Rule, the derivative of $- 2 x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [- 2 x^{2}] + \frac{d}{d x} [1]$ .

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

Step 3.11

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

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

Step 3.12

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

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

Step 3.13

Multiply $2$ by $- 2$ .

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

Step 3.14

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

$f'' (x) = \frac{(- 2 x^{2} + 1) (- x^{3} (1 + x) + (1 - x) (x^{3} + 3 (1 + x) x^{2})) - 2 x^{3} (1 + x) (1 - x) (- 4 x + 0)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.15

Simplify the expression.

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Step 3.15.1

Add $- 4 x$ and $0$ .

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

Step 3.15.2

Multiply $- 4$ by $- 2$ .

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

Step 3.16

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

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

Step 3.17

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

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

Step 3.18

Add $1$ and $3$ .

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

Step 3.19

Simplify.

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Step 3.19.1

Apply the distributive property.

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

Step 3.19.2

Apply the distributive property.

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

Step 3.19.3

Apply the distributive property.

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

Step 3.19.4

Apply the distributive property.

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

Step 3.19.5

Simplify the numerator.

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Step 3.19.5.1

Simplify each term.

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Step 3.19.5.1.1

Simplify each term.

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Step 3.19.5.1.1.1

Multiply $- 1$ by $1$ .

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

Step 3.19.5.1.1.2

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

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Step 3.19.5.1.1.2.1

Move $x$ .

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

Step 3.19.5.1.1.2.2

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

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Step 3.19.5.1.1.2.2.1

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

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

Step 3.19.5.1.1.2.2.2

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

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

Step 3.19.5.1.1.2.3

Add $1$ and $3$ .

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

Step 3.19.5.1.1.3

Simplify each term.

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Step 3.19.5.1.1.3.1

Multiply $3$ by $1$ .

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

Step 3.19.5.1.1.3.2

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

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Step 3.19.5.1.1.3.2.1

Move $x^{2}$ .

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

Step 3.19.5.1.1.3.2.2

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

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Step 3.19.5.1.1.3.2.2.1

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

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

Step 3.19.5.1.1.3.2.2.2

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

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

Step 3.19.5.1.1.3.2.3

Add $2$ and $1$ .

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

Step 3.19.5.1.1.4

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

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

Step 3.19.5.1.1.5

Expand $(1 - x) (4 x^{3} + 3 x^{2})$ using the FOIL Method.

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Step 3.19.5.1.1.5.1

Apply the distributive property.

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

Step 3.19.5.1.1.5.2

Apply the distributive property.

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

Step 3.19.5.1.1.5.3

Apply the distributive property.

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

Step 3.19.5.1.1.6

Simplify and combine like terms.

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Step 3.19.5.1.1.6.1

Simplify each term.

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Step 3.19.5.1.1.6.1.1

Multiply $4 x^{3}$ by $1$ .

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

Step 3.19.5.1.1.6.1.2

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

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

Step 3.19.5.1.1.6.1.3

Rewrite using the commutative property of multiplication.

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

Step 3.19.5.1.1.6.1.4

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

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Step 3.19.5.1.1.6.1.4.1

Move $x^{3}$ .

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

Step 3.19.5.1.1.6.1.4.2

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

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Step 3.19.5.1.1.6.1.4.2.1

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

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

Step 3.19.5.1.1.6.1.4.2.2

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

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

Step 3.19.5.1.1.6.1.4.3

Add $3$ and $1$ .

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

Step 3.19.5.1.1.6.1.5

Multiply $- 1$ by $4$ .

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

Step 3.19.5.1.1.6.1.6

Rewrite using the commutative property of multiplication.

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

Step 3.19.5.1.1.6.1.7

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

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Step 3.19.5.1.1.6.1.7.1

Move $x^{2}$ .

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

Step 3.19.5.1.1.6.1.7.2

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

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Step 3.19.5.1.1.6.1.7.2.1

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

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

Step 3.19.5.1.1.6.1.7.2.2

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

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

Step 3.19.5.1.1.6.1.7.3

Add $2$ and $1$ .

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

Step 3.19.5.1.1.6.1.8

Multiply $- 1$ by $3$ .

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

Step 3.19.5.1.1.6.2

Subtract $3 x^{3}$ from $4 x^{3}$ .

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

Step 3.19.5.1.2

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

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Step 3.19.5.1.2.1

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

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

Step 3.19.5.1.2.2

Add $- x^{4}$ and $0$ .

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

Step 3.19.5.1.3

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

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

Step 3.19.5.1.4

Expand $(- 2 x^{2} + 1) (- 5 x^{4} + 3 x^{2})$ using the FOIL Method.

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Step 3.19.5.1.4.1

Apply the distributive property.

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

Step 3.19.5.1.4.2

Apply the distributive property.

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

Step 3.19.5.1.4.3

Apply the distributive property.

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

Step 3.19.5.1.5

Simplify and combine like terms.

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Step 3.19.5.1.5.1

Simplify each term.

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Step 3.19.5.1.5.1.1

Rewrite using the commutative property of multiplication.

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

Step 3.19.5.1.5.1.2

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

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Step 3.19.5.1.5.1.2.1

Move $x^{4}$ .

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

Step 3.19.5.1.5.1.2.2

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

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

Step 3.19.5.1.5.1.2.3

Add $4$ and $2$ .

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

Step 3.19.5.1.5.1.3

Multiply $- 2$ by $- 5$ .

$f'' (x) = \frac{10 x^{6} - 2 x^{2} (3 x^{2}) + 1 (- 5 x^{4}) + 1 (3 x^{2}) + (8 x^{4} \cdot 1 + 8 x^{4} x) (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.5.1.4

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{10 x^{6} - 2 \cdot (3 x^{2} x^{2}) + 1 (- 5 x^{4}) + 1 (3 x^{2}) + (8 x^{4} \cdot 1 + 8 x^{4} x) (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.5.1.5

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

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Step 3.19.5.1.5.1.5.1

Move $x^{2}$ .

$f'' (x) = \frac{10 x^{6} - 2 \cdot (3 (x^{2} x^{2})) + 1 (- 5 x^{4}) + 1 (3 x^{2}) + (8 x^{4} \cdot 1 + 8 x^{4} x) (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.5.1.5.2

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

$f'' (x) = \frac{10 x^{6} - 2 \cdot (3 x^{2 + 2}) + 1 (- 5 x^{4}) + 1 (3 x^{2}) + (8 x^{4} \cdot 1 + 8 x^{4} x) (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.5.1.5.3

Add $2$ and $2$ .

$f'' (x) = \frac{10 x^{6} - 2 \cdot (3 x^{4}) + 1 (- 5 x^{4}) + 1 (3 x^{2}) + (8 x^{4} \cdot 1 + 8 x^{4} x) (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.5.1.6

Multiply $- 2$ by $3$ .

$f'' (x) = \frac{10 x^{6} - 6 x^{4} + 1 (- 5 x^{4}) + 1 (3 x^{2}) + (8 x^{4} \cdot 1 + 8 x^{4} x) (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.5.1.7

Multiply $- 5 x^{4}$ by $1$ .

$f'' (x) = \frac{10 x^{6} - 6 x^{4} - 5 x^{4} + 1 (3 x^{2}) + (8 x^{4} \cdot 1 + 8 x^{4} x) (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.5.1.8

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

$f'' (x) = \frac{10 x^{6} - 6 x^{4} - 5 x^{4} + 3 x^{2} + (8 x^{4} \cdot 1 + 8 x^{4} x) (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.5.2

Subtract $5 x^{4}$ from $- 6 x^{4}$ .

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + (8 x^{4} \cdot 1 + 8 x^{4} x) (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.6

Simplify each term.

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Step 3.19.5.1.6.1

Multiply $8$ by $1$ .

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + (8 x^{4} + 8 x^{4} x) (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.6.2

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

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Step 3.19.5.1.6.2.1

Move $x$ .

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + (8 x^{4} + 8 (x \cdot x^{4})) (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.6.2.2

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

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Step 3.19.5.1.6.2.2.1

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

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + (8 x^{4} + 8 (x \cdot x^{4})) (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.6.2.2.2

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

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + (8 x^{4} + 8 x^{1 + 4}) (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.6.2.3

Add $1$ and $4$ .

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + (8 x^{4} + 8 x^{5}) (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.7

Expand $(8 x^{4} + 8 x^{5}) (1 - x)$ using the FOIL Method.

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Step 3.19.5.1.7.1

Apply the distributive property.

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} (1 - x) + 8 x^{5} (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.7.2

Apply the distributive property.

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} \cdot 1 + 8 x^{4} (- x) + 8 x^{5} (1 - x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.7.3

Apply the distributive property.

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} \cdot 1 + 8 x^{4} (- x) + 8 x^{5} \cdot 1 + 8 x^{5} (- x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8

Simplify and combine like terms.

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Step 3.19.5.1.8.1

Simplify each term.

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Step 3.19.5.1.8.1.1

Multiply $8$ by $1$ .

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} + 8 x^{4} (- x) + 8 x^{5} \cdot 1 + 8 x^{5} (- x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8.1.2

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} + 8 \cdot (- 1 x^{4} x) + 8 x^{5} \cdot 1 + 8 x^{5} (- x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8.1.3

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

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Step 3.19.5.1.8.1.3.1

Move $x$ .

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} + 8 \cdot (- 1 (x \cdot x^{4})) + 8 x^{5} \cdot 1 + 8 x^{5} (- x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8.1.3.2

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

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Step 3.19.5.1.8.1.3.2.1

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

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} + 8 \cdot (- 1 (x \cdot x^{4})) + 8 x^{5} \cdot 1 + 8 x^{5} (- x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8.1.3.2.2

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

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} + 8 \cdot (- 1 x^{1 + 4}) + 8 x^{5} \cdot 1 + 8 x^{5} (- x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8.1.3.3

Add $1$ and $4$ .

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} + 8 \cdot (- 1 x^{5}) + 8 x^{5} \cdot 1 + 8 x^{5} (- x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8.1.4

Multiply $8$ by $- 1$ .

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} - 8 x^{5} + 8 x^{5} \cdot 1 + 8 x^{5} (- x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8.1.5

Multiply $8$ by $1$ .

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} - 8 x^{5} + 8 x^{5} + 8 x^{5} (- x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8.1.6

Rewrite using the commutative property of multiplication.

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} - 8 x^{5} + 8 x^{5} + 8 \cdot (- 1 x^{5} x)}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8.1.7

Multiply $x^{5}$ by $x$ by adding the exponents.

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Step 3.19.5.1.8.1.7.1

Move $x$ .

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} - 8 x^{5} + 8 x^{5} + 8 \cdot (- 1 (x \cdot x^{5}))}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8.1.7.2

Multiply $x$ by $x^{5}$ .

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Step 3.19.5.1.8.1.7.2.1

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

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} - 8 x^{5} + 8 x^{5} + 8 \cdot (- 1 (x \cdot x^{5}))}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8.1.7.2.2

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

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} - 8 x^{5} + 8 x^{5} + 8 \cdot (- 1 x^{1 + 5})}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8.1.7.3

Add $1$ and $5$ .

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} - 8 x^{5} + 8 x^{5} + 8 \cdot (- 1 x^{6})}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8.1.8

Multiply $8$ by $- 1$ .

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} - 8 x^{5} + 8 x^{5} - 8 x^{6}}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8.2

Add $- 8 x^{5}$ and $8 x^{5}$ .

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} + 0 - 8 x^{6}}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.1.8.3

Add $8 x^{4}$ and $0$ .

$f'' (x) = \frac{10 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4} - 8 x^{6}}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.2

Subtract $8 x^{6}$ from $10 x^{6}$ .

$f'' (x) = \frac{2 x^{6} - 11 x^{4} + 3 x^{2} + 8 x^{4}}{{(- 2 x^{2} + 1)}^{3}}$

Step 3.19.5.3

Add $- 11 x^{4}$ and $8 x^{4}$ .

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

Step 3.19.6

Factor $x^{2}$ out of $2 x^{6} - 3 x^{4} + 3 x^{2}$ .

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Step 3.19.6.1

Factor $x^{2}$ out of $2 x^{6}$ .

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

Step 3.19.6.2

Factor $x^{2}$ out of $- 3 x^{4}$ .

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

Step 3.19.6.3

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

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

Step 3.19.6.4

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

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

Step 3.19.6.5

Factor $x^{2}$ out of $x^{2} (2 x^{4} - 3 x^{2}) + x^{2} \cdot 3$ .

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

Step 4

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

$\frac{x^{3} (1 + x) (1 - x)}{{(- 2 x^{2} + 1)}^{2}} = 0$

Step 5

Find the first derivative.

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Step 5.1

Find the first derivative.

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Step 5.1.1

$\frac{(4 - 8 x^{2}) \frac{d}{d x} [x^{4}] - x^{4} \frac{d}{d x} [4 - 8 x^{2}]}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.2

Differentiate.

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Step 5.1.2.1

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$\frac{(4 - 8 x^{2}) (4 x^{3}) - x^{4} \frac{d}{d x} [4 - 8 x^{2}]}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.2.2

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

$\frac{4 \cdot (4 - 8 x^{2}) x^{3} - x^{4} \frac{d}{d x} [4 - 8 x^{2}]}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.2.3

By the Sum Rule, the derivative of $4 - 8 x^{2}$ with respect to $x$ is $\frac{d}{d x} [4] + \frac{d}{d x} [- 8 x^{2}]$ .

$\frac{4 (4 - 8 x^{2}) x^{3} - x^{4} (\frac{d}{d x} [4] + \frac{d}{d x} [- 8 x^{2}])}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.2.4

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

$\frac{4 (4 - 8 x^{2}) x^{3} - x^{4} (0 + \frac{d}{d x} [- 8 x^{2}])}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.2.5

Add $0$ and $\frac{d}{d x} [- 8 x^{2}]$ .

$\frac{4 (4 - 8 x^{2}) x^{3} - x^{4} \frac{d}{d x} [- 8 x^{2}]}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.2.6

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 x^{2}$ with respect to $x$ is $- 8 \frac{d}{d x} [x^{2}]$ .

$\frac{4 (4 - 8 x^{2}) x^{3} - x^{4} (- 8 \frac{d}{d x} [x^{2}])}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.2.7

Multiply $- 8$ by $- 1$ .

$\frac{4 (4 - 8 x^{2}) x^{3} + 8 x^{4} \frac{d}{d x} [x^{2}]}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.2.8

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

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

Step 5.1.2.9

Multiply $2$ by $8$ .

$\frac{4 (4 - 8 x^{2}) x^{3} + 16 x^{4} x}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.3

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

$\frac{4 (4 - 8 x^{2}) x^{3} + 16 (x^{1} x^{4})}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.4

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

$\frac{4 (4 - 8 x^{2}) x^{3} + 16 x^{1 + 4}}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.5

Add $1$ and $4$ .

$\frac{4 (4 - 8 x^{2}) x^{3} + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.6

Simplify.

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Step 5.1.6.1

Apply the distributive property.

$\frac{(4 \cdot 4 + 4 (- 8 x^{2})) x^{3} + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.6.2

Apply the distributive property.

$\frac{4 \cdot 4 x^{3} + 4 (- 8 x^{2}) x^{3} + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.6.3

Simplify the numerator.

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Step 5.1.6.3.1

Simplify each term.

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Step 5.1.6.3.1.1

Multiply $4$ by $4$ .

$\frac{16 x^{3} + 4 (- 8 x^{2}) x^{3} + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.6.3.1.2

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

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Step 5.1.6.3.1.2.1

Move $x^{3}$ .

$\frac{16 x^{3} + 4 (- 8 (x^{3} x^{2})) + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.6.3.1.2.2

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

$\frac{16 x^{3} + 4 (- 8 x^{3 + 2}) + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.6.3.1.2.3

Add $3$ and $2$ .

$\frac{16 x^{3} + 4 (- 8 x^{5}) + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.6.3.1.3

Multiply $- 8$ by $4$ .

$\frac{16 x^{3} - 32 x^{5} + 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.6.3.2

Add $- 32 x^{5}$ and $16 x^{5}$ .

$\frac{16 x^{3} - 16 x^{5}}{{(4 - 8 x^{2})}^{2}}$

Step 5.1.6.4

Reorder terms.

$\frac{- 16 x^{5} + 16 x^{3}}{{(- 8 x^{2} + 4)}^{2}}$

Step 5.1.6.5

Simplify the numerator.

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Step 5.1.6.5.1

Factor $16 x^{3}$ out of $- 16 x^{5} + 16 x^{3}$ .

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Step 5.1.6.5.1.1

Factor $16 x^{3}$ out of $- 16 x^{5}$ .

$\frac{16 x^{3} (- x^{2}) + 16 x^{3}}{{(- 8 x^{2} + 4)}^{2}}$

Step 5.1.6.5.1.2

Factor $16 x^{3}$ out of $16 x^{3}$ .

$\frac{16 x^{3} (- x^{2}) + 16 x^{3} (1)}{{(- 8 x^{2} + 4)}^{2}}$

Step 5.1.6.5.1.3

Factor $16 x^{3}$ out of $16 x^{3} (- x^{2}) + 16 x^{3} (1)$ .

$\frac{16 x^{3} (- x^{2} + 1)}{{(- 8 x^{2} + 4)}^{2}}$

Step 5.1.6.5.2

Rewrite $1$ as $1^{2}$ .

$\frac{16 x^{3} (- x^{2} + 1^{2})}{{(- 8 x^{2} + 4)}^{2}}$

Step 5.1.6.5.3

Reorder $- x^{2}$ and $1^{2}$ .

$\frac{16 x^{3} (1^{2} - x^{2})}{{(- 8 x^{2} + 4)}^{2}}$

Step 5.1.6.5.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x$ .

$\frac{16 x^{3} (1 + x) (1 - x)}{{(- 8 x^{2} + 4)}^{2}}$

Step 5.1.6.6

Simplify the denominator.

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Step 5.1.6.6.1

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

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Step 5.1.6.6.1.1

Factor $4$ out of $- 8 x^{2}$ .

$\frac{16 x^{3} (1 + x) (1 - x)}{{(4 (- 2 x^{2}) + 4)}^{2}}$

Step 5.1.6.6.1.2

Factor $4$ out of $4$ .

$\frac{16 x^{3} (1 + x) (1 - x)}{{(4 (- 2 x^{2}) + 4 (1))}^{2}}$

Step 5.1.6.6.1.3

Factor $4$ out of $4 (- 2 x^{2}) + 4 (1)$ .

$\frac{16 x^{3} (1 + x) (1 - x)}{{(4 (- 2 x^{2} + 1))}^{2}}$

Step 5.1.6.6.2

Apply the product rule to $4 (- 2 x^{2} + 1)$ .

$\frac{16 x^{3} (1 + x) (1 - x)}{4^{2} {(- 2 x^{2} + 1)}^{2}}$

Step 5.1.6.6.3

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

$\frac{16 x^{3} (1 + x) (1 - x)}{16 {(- 2 x^{2} + 1)}^{2}}$

Step 5.1.6.7

Cancel the common factor of $16$ .

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Step 5.1.6.7.1

Cancel the common factor.

$\frac{16 x^{3} (1 + x) (1 - x)}{16 {(- 2 x^{2} + 1)}^{2}}$

Step 5.1.6.7.2

Rewrite the expression.

$f' (x) = \frac{(x^{3} (1 + x)) (1 - x)}{{(- 2 x^{2} + 1)}^{2}}$

$f' (x) = \frac{x^{3} (1 + x) (1 - x)}{{(- 2 x^{2} + 1)}^{2}}$

Step 5.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{x^{3} (1 + x) (1 - x)}{{(- 2 x^{2} + 1)}^{2}}$ .

$\frac{x^{3} (1 + x) (1 - x)}{{(- 2 x^{2} + 1)}^{2}}$

Step 6

Set the first derivative equal to $0$ then solve the equation $\frac{x^{3} (1 + x) (1 - x)}{{(- 2 x^{2} + 1)}^{2}} = 0$ .

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Step 6.1

Set the first derivative equal to $0$ .

$\frac{x^{3} (1 + x) (1 - x)}{{(- 2 x^{2} + 1)}^{2}} = 0$

Step 6.2

Set the numerator equal to zero.

$x^{3} (1 + x) (1 - x) = 0$

Step 6.3

Solve the equation for $x$ .

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Step 6.3.1

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

$x^{3} = 0$

$1 + x = 0$

$1 - x = 0$

Step 6.3.2

Set $x^{3}$ equal to $0$ and solve for $x$ .

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Step 6.3.2.1

Set $x^{3}$ equal to $0$ .

$x^{3} = 0$

Step 6.3.2.2

Solve $x^{3} = 0$ for $x$ .

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Step 6.3.2.2.1

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

$x = \sqrt[3]{0}$

Step 6.3.2.2.2

Simplify $\sqrt[3]{0}$ .

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Step 6.3.2.2.2.1

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Step 6.3.2.2.2.2

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

$x = 0$

Step 6.3.3

Set $1 + x$ equal to $0$ and solve for $x$ .

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Step 6.3.3.1

Set $1 + x$ equal to $0$ .

$1 + x = 0$

Step 6.3.3.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6.3.4

Set $1 - x$ equal to $0$ and solve for $x$ .

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Step 6.3.4.1

Set $1 - x$ equal to $0$ .

$1 - x = 0$

Step 6.3.4.2

Solve $1 - x = 0$ for $x$ .

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Step 6.3.4.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 6.3.4.2.2

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

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Step 6.3.4.2.2.1

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

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

Step 6.3.4.2.2.2

Simplify the left side.

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Step 6.3.4.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 6.3.4.2.2.2.2

Divide $x$ by $1$ .

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

Step 6.3.4.2.2.3

Simplify the right side.

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Step 6.3.4.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 6.3.5

The final solution is all the values that make $x^{3} (1 + x) (1 - x) = 0$ true.

$x = 0, - 1, 1$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

Set the denominator in $\frac{x^{3} (1 + x) (1 - x)}{{(- 2 x^{2} + 1)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(- 2 x^{2} + 1)}^{2} = 0$

Step 7.2

Solve for $x$ .

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Step 7.2.1

Factor the left side of the equation.

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Step 7.2.1.1

Factor $- 1$ out of $- 2 x^{2} + 1$ .

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Step 7.2.1.1.1

Factor $- 1$ out of $- 2 x^{2}$ .

${(- (2 x^{2}) + 1)}^{2} = 0$

Step 7.2.1.1.2

Rewrite $1$ as $- 1 (- 1)$ .

${(- (2 x^{2}) - 1 \cdot - 1)}^{2} = 0$

Step 7.2.1.1.3

Factor $- 1$ out of $- (2 x^{2}) - 1 (- 1)$ .

${(- (2 x^{2} - 1))}^{2} = 0$

Step 7.2.1.2

Apply the product rule to $- (2 x^{2} - 1)$ .

${(- 1)}^{2} {(2 x^{2} - 1)}^{2} = 0$

Step 7.2.2

Divide each term in ${(- 1)}^{2} {(2 x^{2} - 1)}^{2} = 0$ by ${(- 1)}^{2}$ and simplify.

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Step 7.2.2.1

Divide each term in ${(- 1)}^{2} {(2 x^{2} - 1)}^{2} = 0$ by ${(- 1)}^{2}$ .

$\frac{{(- 1)}^{2} {(2 x^{2} - 1)}^{2}}{{(- 1)}^{2}} = \frac{0}{{(- 1)}^{2}}$

Step 7.2.2.2

Simplify the left side.

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Step 7.2.2.2.1

Cancel the common factor of ${(- 1)}^{2}$ .

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Step 7.2.2.2.1.1

Cancel the common factor.

$\frac{{(- 1)}^{2} {(2 x^{2} - 1)}^{2}}{{(- 1)}^{2}} = \frac{0}{{(- 1)}^{2}}$

Step 7.2.2.2.1.2

Divide ${(2 x^{2} - 1)}^{2}$ by $1$ .

${(2 x^{2} - 1)}^{2} = \frac{0}{{(- 1)}^{2}}$

Step 7.2.2.3

Simplify the right side.

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Step 7.2.2.3.1

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

${(2 x^{2} - 1)}^{2} = \frac{0}{1}$

Step 7.2.2.3.2

Divide $0$ by $1$ .

${(2 x^{2} - 1)}^{2} = 0$

Step 7.2.3

Set the $2 x^{2} - 1$ equal to $0$ .

$2 x^{2} - 1 = 0$

Step 7.2.4

Solve for $x$ .

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Step 7.2.4.1

Add $1$ to both sides of the equation.

$2 x^{2} = 1$

Step 7.2.4.2

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

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Step 7.2.4.2.1

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

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

Step 7.2.4.2.2

Simplify the left side.

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Step 7.2.4.2.2.1

Cancel the common factor of $2$ .

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Step 7.2.4.2.2.1.1

Cancel the common factor.

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

Step 7.2.4.2.2.1.2

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

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

Step 7.2.4.3

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

$x = \pm \sqrt{\frac{1}{2}}$

Step 7.2.4.4

Simplify $\pm \sqrt{\frac{1}{2}}$ .

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Step 7.2.4.4.1

Rewrite $\sqrt{\frac{1}{2}}$ as $\frac{\sqrt{1}}{\sqrt{2}}$ .

$x = \pm \frac{\sqrt{1}}{\sqrt{2}}$

Step 7.2.4.4.2

Any root of $1$ is $1$ .

$x = \pm \frac{1}{\sqrt{2}}$

Step 7.2.4.4.3

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$x = \pm \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Step 7.2.4.4.4

Combine and simplify the denominator.

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Step 7.2.4.4.4.1

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

Step 7.2.4.4.4.2

Raise $\sqrt{2}$ to the power of $1$ .

$x = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}}$

Step 7.2.4.4.4.3

Raise $\sqrt{2}$ to the power of $1$ .

$x = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}}$

Step 7.2.4.4.4.4

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

$x = \pm \frac{\sqrt{2}}{{\sqrt{2}}^{1 + 1}}$

Step 7.2.4.4.4.5

Add $1$ and $1$ .

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

Step 7.2.4.4.4.6

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Step 7.2.4.4.4.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$x = \pm \frac{\sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

Step 7.2.4.4.4.6.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x = \pm \frac{\sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

Step 7.2.4.4.4.6.3

Combine $\frac{1}{2}$ and $2$ .

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

Step 7.2.4.4.4.6.4

Cancel the common factor of $2$ .

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Step 7.2.4.4.4.6.4.1

Cancel the common factor.

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

Step 7.2.4.4.4.6.4.2

Rewrite the expression.

$x = \pm \frac{\sqrt{2}}{2^{1}}$

Step 7.2.4.4.4.6.5

Evaluate the exponent.

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

Step 7.2.4.5

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

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Step 7.2.4.5.1

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

$x = \frac{\sqrt{2}}{2}$

Step 7.2.4.5.2

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

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

Step 7.2.4.5.3

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

$x = \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}$

Step 7.3

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 $0$ .

$x = - \frac{\sqrt{2}}{2}, x = \frac{\sqrt{2}}{2}$

Step 8

Critical points to evaluate.

$x = 0, - 1, 1$

Step 9

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.

$\frac{{(0)}^{2} (2 {(0)}^{4} - 3 {(0)}^{2} + 3)}{{(- 2 {(0)}^{2} + 1)}^{3}}$

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify the numerator.

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Step 10.1.1

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

$\frac{0^{2} (2 \cdot 0 - 3 \cdot 0^{2} + 3)}{{(- 2 \cdot 0^{2} + 1)}^{3}}$

Step 10.1.2

Multiply $2$ by $0$ .

$\frac{0^{2} (0 - 3 \cdot 0^{2} + 3)}{{(- 2 \cdot 0^{2} + 1)}^{3}}$

Step 10.1.3

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

$\frac{0^{2} (0 - 3 \cdot 0 + 3)}{{(- 2 \cdot 0^{2} + 1)}^{3}}$

Step 10.1.4

Multiply $- 3$ by $0$ .

$\frac{0^{2} (0 + 0 + 3)}{{(- 2 \cdot 0^{2} + 1)}^{3}}$

Step 10.1.5

Add $0$ and $0$ .

$\frac{0^{2} (0 + 3)}{{(- 2 \cdot 0^{2} + 1)}^{3}}$

Step 10.1.6

Add $0$ and $3$ .

$\frac{0^{2} \cdot 3}{{(- 2 \cdot 0^{2} + 1)}^{3}}$

Step 10.1.7

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

$\frac{0 \cdot 3}{{(- 2 \cdot 0^{2} + 1)}^{3}}$

Step 10.2

Simplify the denominator.

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Step 10.2.1

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

$\frac{0 \cdot 3}{{(- 2 \cdot 0 + 1)}^{3}}$

Step 10.2.2

Multiply $- 2$ by $0$ .

$\frac{0 \cdot 3}{{(0 + 1)}^{3}}$

Step 10.2.3

Add $0$ and $1$ .

$\frac{0 \cdot 3}{1^{3}}$

Step 10.2.4

One to any power is one.

$\frac{0 \cdot 3}{1}$

Step 10.3

Simplify the expression.

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Step 10.3.1

Multiply $0$ by $3$ .

$\frac{0}{1}$

Step 10.3.2

Divide $0$ by $1$ .

$0$

Step 11

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

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Step 11.1

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

$(- \infty, - 1) \cup (- 1, 0) \cup (0, 1) \cup (1, \infty)$

Step 11.2

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

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Step 11.2.1

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

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

Step 11.2.2

Simplify the result.

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Step 11.2.2.1

Remove parentheses.

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

Step 11.2.2.2

Simplify the numerator.

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Step 11.2.2.2.1

Multiply $- 1$ by $- 4$ .

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

Step 11.2.2.2.2

Subtract $4$ from $1$ .

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

Step 11.2.2.2.3

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

$f' (- 4) = \frac{- 64 \cdot (- 3 (1 + 4))}{{(- 2 {(- 4)}^{2} + 1)}^{2}}$

Step 11.2.2.2.4

Add $1$ and $4$ .

$f' (- 4) = \frac{- 64 \cdot - 3 \cdot 5}{{(- 2 {(- 4)}^{2} + 1)}^{2}}$

Step 11.2.2.2.5

Combine exponents.

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Step 11.2.2.2.5.1

Multiply $- 64$ by $- 3$ .

$f' (- 4) = \frac{192 \cdot 5}{{(- 2 {(- 4)}^{2} + 1)}^{2}}$

Step 11.2.2.2.5.2

Multiply $192$ by $5$ .

$f' (- 4) = \frac{960}{{(- 2 {(- 4)}^{2} + 1)}^{2}}$

Step 11.2.2.3

Simplify the denominator.

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Step 11.2.2.3.1

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

$f' (- 4) = \frac{960}{{(- 2 \cdot 16 + 1)}^{2}}$

Step 11.2.2.3.2

Multiply $- 2$ by $16$ .

$f' (- 4) = \frac{960}{{(- 32 + 1)}^{2}}$

Step 11.2.2.3.3

Add $- 32$ and $1$ .

$f' (- 4) = \frac{960}{{(- 31)}^{2}}$

Step 11.2.2.3.4

Raise $- 31$ to the power of $2$ .

$f' (- 4) = \frac{960}{961}$

Step 11.2.2.4

The final answer is $\frac{960}{961}$ .

$\frac{960}{961}$

Step 11.3

Substitute any number, such as $- 0.85$ , from the interval $(- 1, 0)$ in the first derivative $\frac{x^{3} (1 + x) (1 - x)}{{(- 2 x^{2} + 1)}^{2}}$ to check if the result is negative or positive.

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Step 11.3.1

Replace the variable $x$ with $- 0.85$ in the expression.

$f' (- 0.85) = \frac{{(- 0.85)}^{3} (1 - 0.85) (1 - (- 0.85))}{{(- 2 {(- 0.85)}^{2} + 1)}^{2}}$

Step 11.3.2

Simplify the result.

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Step 11.3.2.1

Remove parentheses.

$f' (- 0.85) = \frac{{(- 0.85)}^{3} (1 - 0.85) (1 - (- 0.85))}{{(- 2 {(- 0.85)}^{2} + 1)}^{2}}$

Step 11.3.2.2

Simplify the numerator.

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Step 11.3.2.2.1

Multiply $- 1$ by $- 0.85$ .

$f' (- 0.85) = \frac{{(- 0.85)}^{3} (1 - 0.85) (1 + 0.85)}{{(- 2 {(- 0.85)}^{2} + 1)}^{2}}$

Step 11.3.2.2.2

Subtract $0.85$ from $1$ .

$f' (- 0.85) = \frac{{(- 0.85)}^{3} \cdot (0.15 (1 + 0.85))}{{(- 2 {(- 0.85)}^{2} + 1)}^{2}}$

Step 11.3.2.2.3

Combine exponents.

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Step 11.3.2.2.3.1

Multiply ${(- 0.85)}^{3}$ by $0.15$ .

$f' (- 0.85) = \frac{- 0.09211875 (1 + 0.85)}{{(- 2 {(- 0.85)}^{2} + 1)}^{2}}$

Step 11.3.2.2.3.2

Multiply $- 0.09211875$ by $1 + 0.85$ .

$f' (- 0.85) = \frac{- 0.17041968}{{(- 2 {(- 0.85)}^{2} + 1)}^{2}}$

Step 11.3.2.3

Simplify the denominator.

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Step 11.3.2.3.1

Raise $- 0.85$ to the power of $2$ .

$f' (- 0.85) = \frac{- 0.17041968}{{(- 2 \cdot 0.7225 + 1)}^{2}}$

Step 11.3.2.3.2

Multiply $- 2$ by $0.7225$ .

$f' (- 0.85) = \frac{- 0.17041968}{{(- 1.445 + 1)}^{2}}$

Step 11.3.2.3.3

Add $- 1.445$ and $1$ .

$f' (- 0.85) = \frac{- 0.17041968}{{(- 0.445)}^{2}}$

Step 11.3.2.3.4

Raise $- 0.445$ to the power of $2$ .

$f' (- 0.85) = \frac{- 0.17041968}{0.198025}$

Step 11.3.2.4

Divide $- 0.17041968$ by $0.198025$ .

$f' (- 0.85) = - 0.86059683$

Step 11.3.2.5

The final answer is $- 0.86059683$ .

$- 0.86059683$

Step 11.4

Substitute any number, such as $0.35$ , from the interval $(0, 1)$ in the first derivative $\frac{x^{3} (1 + x) (1 - x)}{{(- 2 x^{2} + 1)}^{2}}$ to check if the result is negative or positive.

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Step 11.4.1

Replace the variable $x$ with $0.35$ in the expression.

$f' (0.35) = \frac{{(0.35)}^{3} (1 + 0.35) (1 - (0.35))}{{(- 2 {(0.35)}^{2} + 1)}^{2}}$

Step 11.4.2

Simplify the result.

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Step 11.4.2.1

Remove parentheses.

$f' (0.35) = \frac{{(0.35)}^{3} (1 + 0.35) (1 - (0.35))}{{(- 2 {(0.35)}^{2} + 1)}^{2}}$

Step 11.4.2.2

Simplify the numerator.

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Step 11.4.2.2.1

Multiply $- 1$ by $0.35$ .

$f' (0.35) = \frac{{0.35}^{3} (1 + 0.35) (1 - 0.35)}{{(- 2 \cdot {0.35}^{2} + 1)}^{2}}$

Step 11.4.2.2.2

Add $1$ and $0.35$ .

$f' (0.35) = \frac{{0.35}^{3} \cdot (1.35 (1 - 0.35))}{{(- 2 \cdot {0.35}^{2} + 1)}^{2}}$

Step 11.4.2.2.3

Combine exponents.

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Step 11.4.2.2.3.1

Multiply ${0.35}^{3}$ by $1.35$ .

$f' (0.35) = \frac{0.05788125 (1 - 0.35)}{{(- 2 \cdot {0.35}^{2} + 1)}^{2}}$

Step 11.4.2.2.3.2

Multiply $0.05788125$ by $1 - 0.35$ .

$f' (0.35) = \frac{0.03762281}{{(- 2 \cdot {0.35}^{2} + 1)}^{2}}$

Step 11.4.2.3

Simplify the denominator.

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Step 11.4.2.3.1

Raise $0.35$ to the power of $2$ .

$f' (0.35) = \frac{0.03762281}{{(- 2 \cdot 0.1225 + 1)}^{2}}$

Step 11.4.2.3.2

Multiply $- 2$ by $0.1225$ .

$f' (0.35) = \frac{0.03762281}{{(- 0.245 + 1)}^{2}}$

Step 11.4.2.3.3

Add $- 0.245$ and $1$ .

$f' (0.35) = \frac{0.03762281}{{0.755}^{2}}$

Step 11.4.2.3.4

Raise $0.755$ to the power of $2$ .

$f' (0.35) = \frac{0.03762281}{0.570025}$

Step 11.4.2.4

Divide $0.03762281$ by $0.570025$ .

$f' (0.35) = 0.06600203$

Step 11.4.2.5

The final answer is $0.06600203$ .

$0.06600203$

Step 11.5

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

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Step 11.5.1

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

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

Step 11.5.2

Simplify the result.

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Step 11.5.2.1

Remove parentheses.

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

Step 11.5.2.2

Simplify the numerator.

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Step 11.5.2.2.1

Multiply $- 1$ by $4$ .

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

Step 11.5.2.2.2

Add $1$ and $4$ .

$f' (4) = \frac{4^{3} \cdot (5 (1 - 4))}{{(- 2 \cdot 4^{2} + 1)}^{2}}$

Step 11.5.2.2.3

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

$f' (4) = \frac{64 \cdot (5 (1 - 4))}{{(- 2 \cdot 4^{2} + 1)}^{2}}$

Step 11.5.2.2.4

Subtract $4$ from $1$ .

$f' (4) = \frac{64 \cdot 5 \cdot - 3}{{(- 2 \cdot 4^{2} + 1)}^{2}}$

Step 11.5.2.2.5

Combine exponents.

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Step 11.5.2.2.5.1

Multiply $64$ by $5$ .

$f' (4) = \frac{320 \cdot - 3}{{(- 2 \cdot 4^{2} + 1)}^{2}}$

Step 11.5.2.2.5.2

Multiply $320$ by $- 3$ .

$f' (4) = \frac{- 960}{{(- 2 \cdot 4^{2} + 1)}^{2}}$

Step 11.5.2.3

Simplify the denominator.

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Step 11.5.2.3.1

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

$f' (4) = \frac{- 960}{{(- 2 \cdot 16 + 1)}^{2}}$

Step 11.5.2.3.2

Multiply $- 2$ by $16$ .

$f' (4) = \frac{- 960}{{(- 32 + 1)}^{2}}$

Step 11.5.2.3.3

Add $- 32$ and $1$ .

$f' (4) = \frac{- 960}{{(- 31)}^{2}}$

Step 11.5.2.3.4

Raise $- 31$ to the power of $2$ .

$f' (4) = \frac{- 960}{961}$

Step 11.5.2.4

Move the negative in front of the fraction.

$f' (4) = - \frac{960}{961}$

Step 11.5.2.5

The final answer is $- \frac{960}{961}$ .

$- \frac{960}{961}$

Step 11.6

Since the first derivative changed signs from positive to negative around $x = - 1$ , then $x = - 1$ is a local maximum.

$x = - 1$ is a local maximum

Step 11.7

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

$x = 0$ is a local minimum

Step 11.8

Since the first derivative changed signs from positive to negative around $x = 1$ , then $x = 1$ is a local maximum.

$x = 1$ is a local maximum

Step 11.9

These are the local extrema for $f (x) = \frac{x^{4}}{4 - 8 x^{2}}$ .

$x = - 1$ is a local maximum

$x = 0$ is a local minimum

$x = 1$ is a local maximum

$x = - 1$ is a local maximum

$x = 0$ is a local minimum

$x = 1$ is a local maximum

Step 12