Calculus Examples

$\frac{x^{2} + 1}{x^{2} - 1}$

Step 1

Find the first derivative.

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Step 1.1

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

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.2.4

Simplify the expression.

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Step 1.2.4.1

Add $2 x$ and $0$ .

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

Step 1.2.4.2

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

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

Step 1.2.5

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

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

Step 1.2.6

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

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

Step 1.2.7

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

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

Step 1.2.8

Simplify the expression.

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Step 1.2.8.1

Add $2 x$ and $0$ .

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

Step 1.2.8.2

Multiply $2$ by $- 1$ .

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

Step 1.3

Simplify.

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Step 1.3.1

Apply the distributive property.

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

Step 1.3.2

Apply the distributive property.

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

Step 1.3.3

Apply the distributive property.

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

Step 1.3.4

Apply the distributive property.

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

Step 1.3.5

Simplify the numerator.

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Step 1.3.5.1

Combine the opposite terms in $2 x^{2} x + 2 \cdot - 1 x - 2 x^{2} x - 2 \cdot 1 x$ .

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Step 1.3.5.1.1

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

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

Step 1.3.5.1.2

Add $2 \cdot - 1 x$ and $0$ .

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

Step 1.3.5.2

Simplify each term.

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Step 1.3.5.2.1

Multiply $2$ by $- 1$ .

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

Step 1.3.5.2.2

Multiply $- 2$ by $1$ .

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

Step 1.3.5.3

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

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

Step 1.3.6

Move the negative in front of the fraction.

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

Step 1.3.7

Simplify the denominator.

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Step 1.3.7.1

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

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

Step 1.3.7.2

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 $b = 1$ .

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

Step 1.3.7.3

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

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

Step 2

Find the second derivative.

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Step 2.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- \frac{4 x}{{(x + 1)}^{2} {(x - 1)}^{2}}$ with respect to $x$ is $- 4 \frac{d}{d x} [\frac{x}{{(x + 1)}^{2} {(x - 1)}^{2}}]$ .

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

Step 2.2

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

Step 2.3

Differentiate using the Power Rule.

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.4

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

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

Step 2.5

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 $g (x) = x - 1$ .

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Step 2.5.1

To apply the Chain Rule, set $u_{1}$ as $x - 1$ .

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

Step 2.5.2

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

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

Step 2.5.3

Replace all occurrences of $u_{1}$ with $x - 1$ .

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

Step 2.6

Differentiate.

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Step 2.6.1

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

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

Step 2.6.2

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

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

Step 2.6.3

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

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

Step 2.6.4

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

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

Step 2.6.5

Simplify the expression.

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Step 2.6.5.1

Add $1$ and $0$ .

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

Step 2.6.5.2

Multiply $2$ by $1$ .

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

Step 2.7

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 $g (x) = x + 1$ .

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Step 2.7.1

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

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

Step 2.7.2

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

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

Step 2.7.3

Replace all occurrences of $u_{2}$ with $x + 1$ .

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

Step 2.8

Differentiate.

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Step 2.8.1

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

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

Step 2.8.2

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

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

Step 2.8.3

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

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

Step 2.8.4

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

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

Step 2.8.5

Combine fractions.

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Step 2.8.5.1

Add $1$ and $0$ .

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

Step 2.8.5.2

Multiply $2$ by $1$ .

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

Step 2.8.5.3

Combine $- 4$ and $\frac{{(x + 1)}^{2} {(x - 1)}^{2} - x (2 {(x + 1)}^{2} (x - 1) + 2 {(x - 1)}^{2} (x + 1))}{{({(x + 1)}^{2} {(x - 1)}^{2})}^{2}}$ .

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

Step 2.8.5.4

Move the negative in front of the fraction.

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

Step 2.9

Simplify.

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Step 2.9.1

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

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

Step 2.9.2

Apply the distributive property.

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

Step 2.9.3

Apply the distributive property.

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

Step 2.9.4

Simplify the numerator.

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Step 2.9.4.1

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

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Step 2.9.4.1.1

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

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

Step 2.9.4.1.2

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

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

Step 2.9.4.1.3

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

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

Step 2.9.4.1.4

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

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

Step 2.9.4.1.5

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

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

Step 2.9.4.2

Combine exponents.

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Step 2.9.4.2.1

Multiply $2$ by $- 1$ .

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

Step 2.9.4.2.2

Multiply $2$ by $- 1$ .

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

Step 2.9.4.3

Simplify each term.

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Step 2.9.4.3.1

Expand $(x + 1) (x - 1)$ using the FOIL Method.

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Step 2.9.4.3.1.1

Apply the distributive property.

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

Step 2.9.4.3.1.2

Apply the distributive property.

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

Step 2.9.4.3.1.3

Apply the distributive property.

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

Step 2.9.4.3.2

Simplify and combine like terms.

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Step 2.9.4.3.2.1

Simplify each term.

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Step 2.9.4.3.2.1.1

Multiply $x$ by $x$ .

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

Step 2.9.4.3.2.1.2

Move $- 1$ to the left of $x$ .

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

Step 2.9.4.3.2.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 2.9.4.3.2.1.4

Multiply $x$ by $1$ .

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

Step 2.9.4.3.2.1.5

Multiply $- 1$ by $1$ .

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

Step 2.9.4.3.2.2

Add $- x$ and $x$ .

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

Step 2.9.4.3.2.3

Add $x^{2}$ and $0$ .

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

Step 2.9.4.3.3

Apply the distributive property.

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

Step 2.9.4.3.4

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

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Step 2.9.4.3.4.1

Move $x$ .

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

Step 2.9.4.3.4.2

Multiply $x$ by $x$ .

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

Step 2.9.4.3.5

Multiply $- 2$ by $1$ .

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

Step 2.9.4.3.6

Apply the distributive property.

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

Step 2.9.4.3.7

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

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Step 2.9.4.3.7.1

Move $x$ .

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

Step 2.9.4.3.7.2

Multiply $x$ by $x$ .

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

Step 2.9.4.3.8

Multiply $- 1$ by $- 2$ .

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

Step 2.9.4.4

Combine the opposite terms in $x^{2} - 1 - 2 x^{2} - 2 x - 2 x^{2} + 2 x$ .

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Step 2.9.4.4.1

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

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

Step 2.9.4.4.2

Add $x^{2} - 1 - 2 x^{2} - 2 x^{2}$ and $0$ .

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

Step 2.9.4.5

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

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

Step 2.9.4.6

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

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

Step 2.9.5

Combine terms.

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Step 2.9.5.1

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

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Step 2.9.5.1.1

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

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

Step 2.9.5.1.2

Multiply $2$ by $2$ .

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

Step 2.9.5.2

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

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Step 2.9.5.2.1

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

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

Step 2.9.5.2.2

Multiply $2$ by $2$ .

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

Step 2.9.5.3

Cancel the common factor of $x + 1$ and ${(x + 1)}^{4}$ .

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Step 2.9.5.3.1

Factor $x + 1$ out of $4 (x + 1) (x - 1) (- 3 x^{2} - 1)$ .

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

Step 2.9.5.3.2

Cancel the common factors.

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Step 2.9.5.3.2.1

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

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

Step 2.9.5.3.2.2

Cancel the common factor.

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

Step 2.9.5.3.2.3

Rewrite the expression.

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

Step 2.9.5.4

Cancel the common factor of $x - 1$ and ${(x - 1)}^{4}$ .

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Step 2.9.5.4.1

Factor $x - 1$ out of $4 (x - 1) (- 3 x^{2} - 1)$ .

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

Step 2.9.5.4.2

Cancel the common factors.

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Step 2.9.5.4.2.1

Factor $x - 1$ out of ${(x + 1)}^{3} {(x - 1)}^{4}$ .

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

Step 2.9.5.4.2.2

Cancel the common factor.

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

Step 2.9.5.4.2.3

Rewrite the expression.

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

Step 2.9.6

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

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

Step 2.9.7

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

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

Step 2.9.8

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

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

Step 2.9.9

Rewrite $- (3 x^{2} + 1)$ as $- 1 (3 x^{2} + 1)$ .

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

Step 2.9.10

Move the negative in front of the fraction.

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

Step 2.9.11

Multiply $- 1$ by $- 1$ .

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

Step 2.9.12

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

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