Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

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

$- 2 \frac{d}{d x} [\frac{x}{{(x^{2} - 1)}^{2}}]$

Step 1.2

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 $f (x) = x$ and $g (x) = {(x^{2} - 1)}^{2}$ .

$- 2 \frac{{(x^{2} - 1)}^{2} \frac{d}{d x} [x] - x \frac{d}{d x} [{(x^{2} - 1)}^{2}]}{{({(x^{2} - 1)}^{2})}^{2}}$

Step 1.3

Differentiate using the Power Rule.

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Step 1.3.1

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

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Step 1.3.1.1

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

$- 2 \frac{{(x^{2} - 1)}^{2} \frac{d}{d x} [x] - x \frac{d}{d x} [{(x^{2} - 1)}^{2}]}{{(x^{2} - 1)}^{2 \cdot 2}}$

Step 1.3.1.2

Multiply $2$ by $2$ .

$- 2 \frac{{(x^{2} - 1)}^{2} \frac{d}{d x} [x] - x \frac{d}{d x} [{(x^{2} - 1)}^{2}]}{{(x^{2} - 1)}^{4}}$

Step 1.3.2

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

$- 2 \frac{{(x^{2} - 1)}^{2} \cdot 1 - x \frac{d}{d x} [{(x^{2} - 1)}^{2}]}{{(x^{2} - 1)}^{4}}$

Step 1.3.3

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

$- 2 \frac{{(x^{2} - 1)}^{2} - x \frac{d}{d x} [{(x^{2} - 1)}^{2}]}{{(x^{2} - 1)}^{4}}$

Step 1.4

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

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Step 1.4.1

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

$- 2 \frac{{(x^{2} - 1)}^{2} - x (\frac{d}{d u} [u^{2}] \frac{d}{d x} [x^{2} - 1])}{{(x^{2} - 1)}^{4}}$

Step 1.4.2

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

$- 2 \frac{{(x^{2} - 1)}^{2} - x (2 u \frac{d}{d x} [x^{2} - 1])}{{(x^{2} - 1)}^{4}}$

Step 1.4.3

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

$- 2 \frac{{(x^{2} - 1)}^{2} - x (2 (x^{2} - 1) \frac{d}{d x} [x^{2} - 1])}{{(x^{2} - 1)}^{4}}$

Step 1.5

Simplify with factoring out.

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Step 1.5.1

Multiply $2$ by $- 1$ .

$- 2 \frac{{(x^{2} - 1)}^{2} - 2 x ((x^{2} - 1) \frac{d}{d x} [x^{2} - 1])}{{(x^{2} - 1)}^{4}}$

Step 1.5.2

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

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Step 1.5.2.1

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

$- 2 \frac{(x^{2} - 1) (x^{2} - 1) - 2 x ((x^{2} - 1) \frac{d}{d x} [x^{2} - 1])}{{(x^{2} - 1)}^{4}}$

Step 1.5.2.2

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

$- 2 \frac{(x^{2} - 1) (x^{2} - 1) + (x^{2} - 1) (- 2 x (\frac{d}{d x} [x^{2} - 1]))}{{(x^{2} - 1)}^{4}}$

Step 1.5.2.3

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

$- 2 \frac{(x^{2} - 1) (x^{2} - 1 - 2 x (\frac{d}{d x} [x^{2} - 1]))}{{(x^{2} - 1)}^{4}}$

Step 1.6

Cancel the common factors.

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Step 1.6.1

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

$- 2 \frac{(x^{2} - 1) (x^{2} - 1 - 2 x (\frac{d}{d x} [x^{2} - 1]))}{(x^{2} - 1) {(x^{2} - 1)}^{3}}$

Step 1.6.2

Cancel the common factor.

$- 2 \frac{(x^{2} - 1) (x^{2} - 1 - 2 x (\frac{d}{d x} [x^{2} - 1]))}{(x^{2} - 1) {(x^{2} - 1)}^{3}}$

Step 1.6.3

Rewrite the expression.

$- 2 \frac{x^{2} - 1 - 2 x (\frac{d}{d x} [x^{2} - 1])}{{(x^{2} - 1)}^{3}}$

Step 1.7

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

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

Step 1.8

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

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

Step 1.9

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

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

Step 1.10

Simplify the expression.

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Step 1.10.1

Add $2 x$ and $0$ .

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

Step 1.10.2

Multiply $2$ by $- 2$ .

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

Step 1.11

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

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

Step 1.12

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

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

Step 1.13

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

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

Step 1.14

Add $1$ and $1$ .

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

Step 1.15

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

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

Step 1.16

Combine $- 2$ and $\frac{- 3 x^{2} - 1}{{(x^{2} - 1)}^{3}}$ .

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

Step 1.17

Move the negative in front of the fraction.

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

Step 1.18

Simplify.

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Step 1.18.1

Apply the distributive property.

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

Step 1.18.2

Simplify each term.

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Step 1.18.2.1

Multiply $- 3$ by $2$ .

$- \frac{- 6 x^{2} + 2 \cdot - 1}{{(x^{2} - 1)}^{3}}$

Step 1.18.2.2

Multiply $2$ by $- 1$ .

$- \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}}$

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 $g (x) = \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}}$ .

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

Step 2.2

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

Step 2.3

Differentiate.

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Step 2.3.1

Multiply the exponents in ${({(x^{2} - 1)}^{3})}^{2}$ .

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Step 2.3.1.1

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

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

Step 2.3.1.2

Multiply $3$ by $2$ .

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

Step 2.3.2

By the Sum Rule, the derivative of $- 6 x^{2} - 2$ with respect to $x$ is $\frac{d}{d x} [- 6 x^{2}] + \frac{d}{d x} [- 2]$ .

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

Step 2.3.3

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 x^{2}$ with respect to $x$ is $- 6 \frac{d}{d x} [x^{2}]$ .

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

Step 2.3.4

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

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

Step 2.3.5

Multiply $2$ by $- 6$ .

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x + \frac{d}{d x} (- 2)) - (- 6 x^{2} - 2) \frac{d}{d x} {(x^{2} - 1)}^{3}}{{(x^{2} - 1)}^{6}} + \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}} \cdot \frac{d}{d x} (- 1)$

Step 2.3.6

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x + 0) - (- 6 x^{2} - 2) \frac{d}{d x} {(x^{2} - 1)}^{3}}{{(x^{2} - 1)}^{6}} + \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}} \cdot \frac{d}{d x} (- 1)$

Step 2.3.7

Add $- 12 x$ and $0$ .

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x) - (- 6 x^{2} - 2) \frac{d}{d x} {(x^{2} - 1)}^{3}}{{(x^{2} - 1)}^{6}} + \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}} \cdot \frac{d}{d x} (- 1)$

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 $f (x) = x^{3}$ and $g (x) = x^{2} - 1$ .

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Step 2.4.1

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

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x) - (- 6 x^{2} - 2) (\frac{d}{d u} (u^{3}) \frac{d}{d x} (x^{2} - 1))}{{(x^{2} - 1)}^{6}} + \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}} \cdot \frac{d}{d x} (- 1)$

Step 2.4.2

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

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x) - (- 6 x^{2} - 2) (3 u^{2} \frac{d}{d x} (x^{2} - 1))}{{(x^{2} - 1)}^{6}} + \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}} \cdot \frac{d}{d x} (- 1)$

Step 2.4.3

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

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x) - (- 6 x^{2} - 2) (3 {(x^{2} - 1)}^{2} \frac{d}{d x} (x^{2} - 1))}{{(x^{2} - 1)}^{6}} + \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}} \cdot \frac{d}{d x} (- 1)$

Step 2.5

Differentiate.

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Step 2.5.1

Multiply $3$ by $- 1$ .

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x) - 3 (- 6 x^{2} - 2) ({(x^{2} - 1)}^{2} \frac{d}{d x} (x^{2} - 1))}{{(x^{2} - 1)}^{6}} + \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}} \cdot \frac{d}{d x} (- 1)$

Step 2.5.2

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

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x) - 3 (- 6 x^{2} - 2) ({(x^{2} - 1)}^{2} (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 1)))}{{(x^{2} - 1)}^{6}} + \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}} \cdot \frac{d}{d x} (- 1)$

Step 2.5.3

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

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x) - 3 (- 6 x^{2} - 2) ({(x^{2} - 1)}^{2} (2 x + \frac{d}{d x} (- 1)))}{{(x^{2} - 1)}^{6}} + \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}} \cdot \frac{d}{d x} (- 1)$

Step 2.5.4

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

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x) - 3 (- 6 x^{2} - 2) ({(x^{2} - 1)}^{2} (2 x + 0))}{{(x^{2} - 1)}^{6}} + \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}} \cdot \frac{d}{d x} (- 1)$

Step 2.5.5

Simplify the expression.

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Step 2.5.5.1

Add $2 x$ and $0$ .

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x) - 3 (- 6 x^{2} - 2) ({(x^{2} - 1)}^{2} (2 x))}{{(x^{2} - 1)}^{6}} + \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}} \cdot \frac{d}{d x} (- 1)$

Step 2.5.5.2

Move $2$ to the left of ${(x^{2} - 1)}^{2}$ .

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x) - 3 (- 6 x^{2} - 2) (2 \cdot ({(x^{2} - 1)}^{2} x))}{{(x^{2} - 1)}^{6}} + \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}} \cdot \frac{d}{d x} (- 1)$

Step 2.5.5.3

Multiply $2$ by $- 3$ .

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x) - 6 (- 6 x^{2} - 2) ({(x^{2} - 1)}^{2} x)}{{(x^{2} - 1)}^{6}} + \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}} \cdot \frac{d}{d x} (- 1)$

Step 2.5.6

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

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x) - 6 (- 6 x^{2} - 2) ({(x^{2} - 1)}^{2} x)}{{(x^{2} - 1)}^{6}} + \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}} \cdot 0$

Step 2.5.7

Simplify the expression.

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Step 2.5.7.1

Multiply $\frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}}$ by $0$ .

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x) - 6 (- 6 x^{2} - 2) ({(x^{2} - 1)}^{2} x)}{{(x^{2} - 1)}^{6}} + 0$

Step 2.5.7.2

Add $- \frac{{(x^{2} - 1)}^{3} (- 12 x) - 6 (- 6 x^{2} - 2) ({(x^{2} - 1)}^{2} x)}{{(x^{2} - 1)}^{6}}$ and $0$ .

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x) - 6 (- 6 x^{2} - 2) ({(x^{2} - 1)}^{2} x)}{{(x^{2} - 1)}^{6}}$

Step 2.6

Simplify.

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Step 2.6.1

Apply the distributive property.

$f'' (x) = - \frac{{(x^{2} - 1)}^{3} (- 12 x) + (- 6 (- 6 x^{2}) - 6 \cdot - 2) ({(x^{2} - 1)}^{2} x)}{{(x^{2} - 1)}^{6}}$

Step 2.6.2

Simplify the numerator.

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Step 2.6.2.1

Factor ${(x^{2} - 1)}^{2} x$ out of ${(x^{2} - 1)}^{3} (- 12 x) + (- 6 (- 6 x^{2}) - 6 \cdot - 2) {(x^{2} - 1)}^{2} x$ .

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Step 2.6.2.1.1

Factor ${(x^{2} - 1)}^{2} x$ out of ${(x^{2} - 1)}^{3} (- 12 x)$ .

$f'' (x) = - \frac{{(x^{2} - 1)}^{2} x ((x^{2} - 1) \cdot (- 12)) + (- 6 (- 6 x^{2}) - 6 \cdot - 2) {(x^{2} - 1)}^{2} x}{{(x^{2} - 1)}^{6}}$

Step 2.6.2.1.2

Factor ${(x^{2} - 1)}^{2} x$ out of $(- 6 (- 6 x^{2}) - 6 \cdot - 2) {(x^{2} - 1)}^{2} x$ .

$f'' (x) = - \frac{{(x^{2} - 1)}^{2} x ((x^{2} - 1) \cdot (- 12)) + {(x^{2} - 1)}^{2} x (- 6 (- 6 x^{2}) - 6 \cdot - 2)}{{(x^{2} - 1)}^{6}}$

Step 2.6.2.1.3

Factor ${(x^{2} - 1)}^{2} x$ out of ${(x^{2} - 1)}^{2} x ((x^{2} - 1) \cdot (- 12)) + {(x^{2} - 1)}^{2} x (- 6 (- 6 x^{2}) - 6 \cdot - 2)$ .

$f'' (x) = - \frac{{(x^{2} - 1)}^{2} x ((x^{2} - 1) \cdot (- 12) - 6 (- 6 x^{2}) - 6 \cdot - 2)}{{(x^{2} - 1)}^{6}}$

Step 2.6.2.2

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

$f'' (x) = - \frac{{(x^{2} - 1^{2})}^{2} x ((x^{2} - 1) \cdot (- 12) - 6 (- 6 x^{2}) - 6 \cdot - 2)}{{(x^{2} - 1)}^{6}}$

Step 2.6.2.3

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

$f'' (x) = - \frac{{((x + 1) (x - 1))}^{2} x ((x^{2} - 1) \cdot (- 12) - 6 (- 6 x^{2}) - 6 \cdot - 2)}{{(x^{2} - 1)}^{6}}$

Step 2.6.2.4

Apply the product rule to $(x + 1) (x - 1)$ .

$f'' (x) = - \frac{{(x + 1)}^{2} {(x - 1)}^{2} x ((x^{2} - 1) \cdot (- 12) - 6 (- 6 x^{2}) - 6 \cdot - 2)}{{(x^{2} - 1)}^{6}}$

Step 2.6.2.5

Combine exponents.

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Step 2.6.2.5.1

Multiply $- 6$ by $- 6$ .

$f'' (x) = - \frac{{(x + 1)}^{2} {(x - 1)}^{2} x ((x^{2} - 1) \cdot (- 12) + 36 x^{2} - 6 \cdot - 2)}{{(x^{2} - 1)}^{6}}$

Step 2.6.2.5.2

Multiply $- 6$ by $- 2$ .

$f'' (x) = - \frac{{(x + 1)}^{2} {(x - 1)}^{2} x ((x^{2} - 1) \cdot (- 12) + 36 x^{2} + 12)}{{(x^{2} - 1)}^{6}}$

Step 2.6.2.6

Simplify each term.

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Step 2.6.2.6.1

Apply the distributive property.

$f'' (x) = - \frac{{(x + 1)}^{2} {(x - 1)}^{2} x (x^{2} \cdot - 12 - 1 \cdot - 12 + 36 x^{2} + 12)}{{(x^{2} - 1)}^{6}}$

Step 2.6.2.6.2

Move $- 12$ to the left of $x^{2}$ .

$f'' (x) = - \frac{{(x + 1)}^{2} {(x - 1)}^{2} x (- 12 \cdot x^{2} - 1 \cdot - 12 + 36 x^{2} + 12)}{{(x^{2} - 1)}^{6}}$

Step 2.6.2.6.3

Multiply $- 1$ by $- 12$ .

$f'' (x) = - \frac{{(x + 1)}^{2} {(x - 1)}^{2} x (- 12 x^{2} + 12 + 36 x^{2} + 12)}{{(x^{2} - 1)}^{6}}$

Step 2.6.2.7

Add $- 12 x^{2}$ and $36 x^{2}$ .

$f'' (x) = - \frac{{(x + 1)}^{2} {(x - 1)}^{2} x (24 x^{2} + 12 + 12)}{{(x^{2} - 1)}^{6}}$

Step 2.6.2.8

Add $12$ and $12$ .

$f'' (x) = - \frac{{(x + 1)}^{2} {(x - 1)}^{2} x (24 x^{2} + 24)}{{(x^{2} - 1)}^{6}}$

Step 2.6.2.9

Factor $24$ out of $24 x^{2} + 24$ .

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Step 2.6.2.9.1

Factor $24$ out of $24 x^{2}$ .

$f'' (x) = - \frac{{(x + 1)}^{2} {(x - 1)}^{2} x (24 (x^{2}) + 24)}{{(x^{2} - 1)}^{6}}$

Step 2.6.2.9.2

Factor $24$ out of $24$ .

$f'' (x) = - \frac{{(x + 1)}^{2} {(x - 1)}^{2} x (24 (x^{2}) + 24 (1))}{{(x^{2} - 1)}^{6}}$

Step 2.6.2.9.3

Factor $24$ out of $24 (x^{2}) + 24 (1)$ .

$f'' (x) = - \frac{{(x + 1)}^{2} {(x - 1)}^{2} x (24 (x^{2} + 1))}{{(x^{2} - 1)}^{6}}$

$f'' (x) = - \frac{{(x + 1)}^{2} {(x - 1)}^{2} x \cdot (24 (x^{2} + 1))}{{(x^{2} - 1)}^{6}}$

Step 2.6.3

Move $24$ to the left of ${(x + 1)}^{2} {(x - 1)}^{2} x$ .

$f'' (x) = - \frac{24 {(x + 1)}^{2} {(x - 1)}^{2} x (x^{2} + 1)}{{(x^{2} - 1)}^{6}}$

Step 2.6.4

Simplify the denominator.

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Step 2.6.4.1

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

$f'' (x) = - \frac{24 {(x + 1)}^{2} {(x - 1)}^{2} x (x^{2} + 1)}{{(x^{2} - 1^{2})}^{6}}$

Step 2.6.4.2

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

$f'' (x) = - \frac{24 {(x + 1)}^{2} {(x - 1)}^{2} x (x^{2} + 1)}{{((x + 1) (x - 1))}^{6}}$

Step 2.6.4.3

Apply the product rule to $(x + 1) (x - 1)$ .

$f'' (x) = - \frac{24 {(x + 1)}^{2} {(x - 1)}^{2} x (x^{2} + 1)}{{(x + 1)}^{6} {(x - 1)}^{6}}$

Step 2.6.5

Cancel the common factor of ${(x + 1)}^{2}$ and ${(x + 1)}^{6}$ .

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Step 2.6.5.1

Factor ${(x + 1)}^{2}$ out of $24 {(x + 1)}^{2} {(x - 1)}^{2} x (x^{2} + 1)$ .

$f'' (x) = - \frac{{(x + 1)}^{2} (24 {(x - 1)}^{2} x (x^{2} + 1))}{{(x + 1)}^{6} {(x - 1)}^{6}}$

Step 2.6.5.2

Cancel the common factors.

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Step 2.6.5.2.1

Factor ${(x + 1)}^{2}$ out of ${(x + 1)}^{6} {(x - 1)}^{6}$ .

$f'' (x) = - \frac{{(x + 1)}^{2} (24 {(x - 1)}^{2} x (x^{2} + 1))}{{(x + 1)}^{2} ({(x + 1)}^{4} {(x - 1)}^{6})}$

Step 2.6.5.2.2

Cancel the common factor.

$f'' (x) = - \frac{{(x + 1)}^{2} (24 {(x - 1)}^{2} x (x^{2} + 1))}{{(x + 1)}^{2} ({(x + 1)}^{4} {(x - 1)}^{6})}$

Step 2.6.5.2.3

Rewrite the expression.

$f'' (x) = - \frac{24 {(x - 1)}^{2} x (x^{2} + 1)}{{(x + 1)}^{4} {(x - 1)}^{6}}$

Step 2.6.6

Cancel the common factor of ${(x - 1)}^{2}$ and ${(x - 1)}^{6}$ .

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Step 2.6.6.1

Factor ${(x - 1)}^{2}$ out of $24 {(x - 1)}^{2} x (x^{2} + 1)$ .

$f'' (x) = - \frac{{(x - 1)}^{2} (24 x (x^{2} + 1))}{{(x + 1)}^{4} {(x - 1)}^{6}}$

Step 2.6.6.2

Cancel the common factors.

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Step 2.6.6.2.1

Factor ${(x - 1)}^{2}$ out of ${(x + 1)}^{4} {(x - 1)}^{6}$ .

$f'' (x) = - \frac{{(x - 1)}^{2} (24 x (x^{2} + 1))}{{(x - 1)}^{2} ({(x + 1)}^{4} {(x - 1)}^{4})}$

Step 2.6.6.2.2

Cancel the common factor.

$f'' (x) = - \frac{{(x - 1)}^{2} (24 x (x^{2} + 1))}{{(x - 1)}^{2} ({(x + 1)}^{4} {(x - 1)}^{4})}$

Step 2.6.6.2.3

Rewrite the expression.

$f'' (x) = - \frac{24 x (x^{2} + 1)}{{(x + 1)}^{4} {(x - 1)}^{4}}$

Step 3

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

$- \frac{- 6 x^{2} - 2}{{(x^{2} - 1)}^{3}} = 0$

Step 4

Since there is no value of $x$ that makes the first derivative equal to $0$ , there are no local extrema.

No Local Extrema

Step 5

No Local Extrema

Step 6