Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

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

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

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

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

Step 1.3

Differentiate.

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

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

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

Step 1.3.5

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

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

Step 1.3.6

Simplify the expression.

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Step 1.3.6.1

Add $2 x$ and $0$ .

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

Step 1.3.6.2

Multiply $2$ by $- 1$ .

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

Step 1.4

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

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

Step 1.5

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

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

Step 1.6

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

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

Step 1.7

Add $1$ and $1$ .

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

Step 1.8

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

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

Step 1.9

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

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

Step 1.10

Simplify.

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Step 1.10.1

Apply the distributive property.

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

Step 1.10.2

Simplify each term.

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Step 1.10.2.1

Multiply $- 1$ by $4$ .

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

Step 1.10.2.2

Multiply $4$ by $1$ .

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

Step 2

Find the second derivative of the function.

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Step 2.1

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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Step 2.2.1.1

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

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

Step 2.2.1.2

Multiply $2$ by $2$ .

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Multiply $2$ by $- 4$ .

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

Step 2.2.6

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

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

Step 2.2.7

Add $- 8 x$ and $0$ .

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

Step 2.3

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 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.4

Differentiate.

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Step 2.4.1

Multiply $2$ by $- 1$ .

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

Step 2.4.2

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

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

Step 2.4.3

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

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

Step 2.4.4

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

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

Step 2.4.5

Simplify the expression.

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Step 2.4.5.1

Add $2 x$ and $0$ .

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

Step 2.4.5.2

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

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

Step 2.4.5.3

Multiply $2$ by $- 2$ .

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

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

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

Step 2.5.2

Apply the distributive property.

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

Step 2.5.3

Simplify the numerator.

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Step 2.5.3.1

Simplify each term.

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Step 2.5.3.1.1

Rewrite using the commutative property of multiplication.

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

Step 2.5.3.1.2

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

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

Step 2.5.3.1.3

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

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Step 2.5.3.1.3.1

Apply the distributive property.

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

Step 2.5.3.1.3.2

Apply the distributive property.

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

Step 2.5.3.1.3.3

Apply the distributive property.

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

Step 2.5.3.1.4

Simplify and combine like terms.

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Step 2.5.3.1.4.1

Simplify each term.

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Step 2.5.3.1.4.1.1

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

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Step 2.5.3.1.4.1.1.1

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

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

Step 2.5.3.1.4.1.1.2

Add $2$ and $2$ .

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

Step 2.5.3.1.4.1.2

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

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

Step 2.5.3.1.4.1.3

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

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

Step 2.5.3.1.4.1.4

Multiply $1$ by $1$ .

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

Step 2.5.3.1.4.2

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

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

Step 2.5.3.1.5

Apply the distributive property.

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

Step 2.5.3.1.6

Simplify.

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Step 2.5.3.1.6.1

Multiply $2$ by $- 8$ .

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

Step 2.5.3.1.6.2

Multiply $- 8$ by $1$ .

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

Step 2.5.3.1.7

Apply the distributive property.

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

Step 2.5.3.1.8

Simplify.

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Step 2.5.3.1.8.1

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

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Step 2.5.3.1.8.1.1

Move $x$ .

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

Step 2.5.3.1.8.1.2

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

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Step 2.5.3.1.8.1.2.1

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

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

Step 2.5.3.1.8.1.2.2

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

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

Step 2.5.3.1.8.1.3

Add $1$ and $4$ .

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

Step 2.5.3.1.8.2

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

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Step 2.5.3.1.8.2.1

Move $x$ .

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

Step 2.5.3.1.8.2.2

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

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Step 2.5.3.1.8.2.2.1

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

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

Step 2.5.3.1.8.2.2.2

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

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

Step 2.5.3.1.8.2.3

Add $1$ and $2$ .

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

Step 2.5.3.1.9

Simplify each term.

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Step 2.5.3.1.9.1

Multiply $- 4$ by $- 4$ .

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

Step 2.5.3.1.9.2

Multiply $- 4$ by $4$ .

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

Step 2.5.3.1.10

Simplify each term.

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Step 2.5.3.1.10.1

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

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Step 2.5.3.1.10.1.1

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

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Step 2.5.3.1.10.1.1.1

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

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

Step 2.5.3.1.10.1.1.2

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

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

Step 2.5.3.1.10.1.2

Add $2$ and $1$ .

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

Step 2.5.3.1.10.2

Multiply $x$ by $1$ .

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

Step 2.5.3.1.11

Expand $(16 x^{2} - 16) (x^{3} + x)$ using the FOIL Method.

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Step 2.5.3.1.11.1

Apply the distributive property.

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

Step 2.5.3.1.11.2

Apply the distributive property.

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

Step 2.5.3.1.11.3

Apply the distributive property.

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

Step 2.5.3.1.12

Simplify and combine like terms.

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Step 2.5.3.1.12.1

Simplify each term.

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Step 2.5.3.1.12.1.1

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

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Step 2.5.3.1.12.1.1.1

Move $x^{3}$ .

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

Step 2.5.3.1.12.1.1.2

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

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

Step 2.5.3.1.12.1.1.3

Add $3$ and $2$ .

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

Step 2.5.3.1.12.1.2

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

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Step 2.5.3.1.12.1.2.1

Move $x$ .

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

Step 2.5.3.1.12.1.2.2

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

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Step 2.5.3.1.12.1.2.2.1

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

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

Step 2.5.3.1.12.1.2.2.2

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

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

Step 2.5.3.1.12.1.2.3

Add $1$ and $2$ .

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

Step 2.5.3.1.12.2

Subtract $16 x^{3}$ from $16 x^{3}$ .

$f'' (x) = \frac{- 8 x^{5} - 16 x^{3} - 8 x + 16 x^{5} + 0 - 16 x}{{(x^{2} + 1)}^{4}}$

Step 2.5.3.1.12.3

Add $16 x^{5}$ and $0$ .

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

Step 2.5.3.2

Add $- 8 x^{5}$ and $16 x^{5}$ .

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

Step 2.5.3.3

Subtract $16 x$ from $- 8 x$ .

$f'' (x) = \frac{8 x^{5} - 16 x^{3} - 24 x}{{(x^{2} + 1)}^{4}}$

Step 2.5.4

Simplify the numerator.

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Step 2.5.4.1

Factor $8 x$ out of $8 x^{5} - 16 x^{3} - 24 x$ .

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Step 2.5.4.1.1

Factor $8 x$ out of $8 x^{5}$ .

$f'' (x) = \frac{8 x (x^{4}) - 16 x^{3} - 24 x}{{(x^{2} + 1)}^{4}}$

Step 2.5.4.1.2

Factor $8 x$ out of $- 16 x^{3}$ .

$f'' (x) = \frac{8 x (x^{4}) + 8 x (- 2 x^{2}) - 24 x}{{(x^{2} + 1)}^{4}}$

Step 2.5.4.1.3

Factor $8 x$ out of $- 24 x$ .

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

Step 2.5.4.1.4

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

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

Step 2.5.4.1.5

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

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

Step 2.5.4.2

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

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

Step 2.5.4.3

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

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

Step 2.5.4.4

Factor $u^{2} - 2 u - 3$ using the AC method.

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Step 2.5.4.4.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Step 2.5.4.4.2

Write the factored form using these integers.

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

Step 2.5.4.5

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

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

Step 2.5.5

Cancel the common factor of $x^{2} + 1$ and ${(x^{2} + 1)}^{4}$ .

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Step 2.5.5.1

Factor $x^{2} + 1$ out of $8 x (x^{2} - 3) (x^{2} + 1)$ .

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

Step 2.5.5.2

Cancel the common factors.

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Step 2.5.5.2.1

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

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

Step 2.5.5.2.2

Cancel the common factor.

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

Step 2.5.5.2.3

Rewrite the expression.

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

Step 3

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

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

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

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

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

Step 4.1.2

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

Step 4.1.3

Differentiate.

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Step 4.1.3.1

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

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

Step 4.1.3.2

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

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

Step 4.1.3.3

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

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

Step 4.1.3.4

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

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

Step 4.1.3.5

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

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

Step 4.1.3.6

Simplify the expression.

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Step 4.1.3.6.1

Add $2 x$ and $0$ .

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

Step 4.1.3.6.2

Multiply $2$ by $- 1$ .

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

Step 4.1.4

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

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

Step 4.1.5

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

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

Step 4.1.6

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

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

Step 4.1.7

Add $1$ and $1$ .

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

Step 4.1.8

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

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

Step 4.1.9

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

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

Step 4.1.10

Simplify.

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Step 4.1.10.1

Apply the distributive property.

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

Step 4.1.10.2

Simplify each term.

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Step 4.1.10.2.1

Multiply $- 1$ by $4$ .

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

Step 4.1.10.2.2

Multiply $4$ by $1$ .

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

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{- 4 x^{2} + 4}{{(x^{2} + 1)}^{2}}$ .

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

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{- 4 x^{2} + 4}{{(x^{2} + 1)}^{2}} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

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

Step 5.2

Set the numerator equal to zero.

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

Step 5.3

Solve the equation for $x$ .

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Step 5.3.1

Subtract $4$ from both sides of the equation.

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

Step 5.3.2

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

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Step 5.3.2.1

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

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

Step 5.3.2.2

Simplify the left side.

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Step 5.3.2.2.1

Cancel the common factor of $- 4$ .

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Step 5.3.2.2.1.1

Cancel the common factor.

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

Step 5.3.2.2.1.2

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

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

Step 5.3.2.3

Simplify the right side.

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Step 5.3.2.3.1

Divide $- 4$ by $- 4$ .

$x^{2} = 1$

Step 5.3.3

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

$x = \pm \sqrt{1}$

Step 5.3.4

Any root of $1$ is $1$ .

$x = \pm 1$

Step 5.3.5

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

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Step 5.3.5.1

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

$x = 1$

Step 5.3.5.2

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

$x = - 1$

Step 5.3.5.3

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

$x = 1, - 1$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$x = 1, - 1$

Step 8

Evaluate the second derivative at $x = 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

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

Step 9

Evaluate the second derivative.

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Step 9.1

Multiply $8$ by $1$ .

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

Step 9.2

Simplify the denominator.

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Step 9.2.1

One to any power is one.

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

Step 9.2.2

Add $1$ and $1$ .

$\frac{8 (1^{2} - 3)}{2^{3}}$

Step 9.2.3

Raise $2$ to the power of $3$ .

$\frac{8 (1^{2} - 3)}{8}$

Step 9.3

Simplify the numerator.

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Step 9.3.1

One to any power is one.

$\frac{8 (1 - 3)}{8}$

Step 9.3.2

Subtract $3$ from $1$ .

$\frac{8 \cdot - 2}{8}$

Step 9.4

Simplify the expression.

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Step 9.4.1

Multiply $8$ by $- 2$ .

$\frac{- 16}{8}$

Step 9.4.2

Divide $- 16$ by $8$ .

$- 2$

Step 10

$x = 1$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 1$ is a local maximum

Step 11

Find the y-value when $x = 1$ .

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Step 11.1

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

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

Step 11.2

Simplify the result.

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Step 11.2.1

Multiply $4$ by $1$ .

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

Step 11.2.2

Simplify the denominator.

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Step 11.2.2.1

One to any power is one.

$f (1) = \frac{4}{1 + 1}$

Step 11.2.2.2

Add $1$ and $1$ .

$f (1) = \frac{4}{2}$

Step 11.2.3

Divide $4$ by $2$ .

$f (1) = 2$

Step 11.2.4

The final answer is $2$ .

$y = 2$

Step 12

Evaluate the second derivative at $x = - 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

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

Step 13

Evaluate the second derivative.

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Step 13.1

Multiply $8$ by $- 1$ .

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

Step 13.2

Simplify the denominator.

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Step 13.2.1

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

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

Step 13.2.2

Add $1$ and $1$ .

$\frac{- 8 ({(- 1)}^{2} - 3)}{2^{3}}$

Step 13.2.3

Raise $2$ to the power of $3$ .

$\frac{- 8 ({(- 1)}^{2} - 3)}{8}$

Step 13.3

Simplify the numerator.

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Step 13.3.1

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

$\frac{- 8 (1 - 3)}{8}$

Step 13.3.2

Subtract $3$ from $1$ .

$\frac{- 8 \cdot - 2}{8}$

Step 13.4

Simplify the expression.

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Step 13.4.1

Multiply $- 8$ by $- 2$ .

$\frac{16}{8}$

Step 13.4.2

Divide $16$ by $8$ .

$2$

Step 14

$x = - 1$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = - 1$ is a local minimum

Step 15

Find the y-value when $x = - 1$ .

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Step 15.1

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

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

Step 15.2

Simplify the result.

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Step 15.2.1

Multiply $4$ by $- 1$ .

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

Step 15.2.2

Simplify the denominator.

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Step 15.2.2.1

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

$f (- 1) = \frac{- 4}{1 + 1}$

Step 15.2.2.2

Add $1$ and $1$ .

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

Step 15.2.3

Divide $- 4$ by $2$ .

$f (- 1) = - 2$

Step 15.2.4

The final answer is $- 2$ .

$y = - 2$

Step 16

These are the local extrema for $f (x) = \frac{4 x}{x^{2} + 1}$ .

$(1, 2)$ is a local maxima

$(- 1, - 2)$ is a local minima

Step 17