Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} + 4}$ as ${(x^{2} + 4)}^{\frac{1}{2}}$ .

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

Step 1.2

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

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

Step 1.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x^{2} + 4$ .

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.4

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 1.5

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

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

Step 1.6

Combine the numerators over the common denominator.

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

Step 1.7

Simplify the numerator.

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Step 1.7.1

Multiply $- 1$ by $2$ .

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

Step 1.7.2

Subtract $2$ from $1$ .

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

Step 1.8

Combine fractions.

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Step 1.8.1

Move the negative in front of the fraction.

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

Step 1.8.2

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

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

Step 1.8.3

Move ${(x^{2} + 4)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 1.8.4

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

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

Step 1.9

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

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

Step 1.10

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

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

Step 1.11

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

$\frac{x}{2 {(x^{2} + 4)}^{\frac{1}{2}}} (2 x + 0) + {(x^{2} + 4)}^{\frac{1}{2}} \frac{d}{d x} [x]$

Step 1.12

Combine fractions.

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Step 1.12.1

Add $2 x$ and $0$ .

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

Step 1.12.2

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

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

Step 1.12.3

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

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

Step 1.13

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

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

Step 1.14

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

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

Step 1.15

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

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

Step 1.16

Reduce the expression by cancelling the common factors.

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Step 1.16.1

Add $1$ and $1$ .

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

Step 1.16.2

Cancel the common factor.

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

Step 1.16.3

Rewrite the expression.

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

Step 1.17

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

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

Step 1.18

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

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

Step 1.19

To write ${(x^{2} + 4)}^{\frac{1}{2}}$ as a fraction with a common denominator, multiply by $\frac{{(x^{2} + 4)}^{\frac{1}{2}}}{{(x^{2} + 4)}^{\frac{1}{2}}}$ .

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

Step 1.20

Combine the numerators over the common denominator.

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

Step 1.21

Multiply ${(x^{2} + 4)}^{\frac{1}{2}}$ by ${(x^{2} + 4)}^{\frac{1}{2}}$ by adding the exponents.

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Step 1.21.1

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

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

Step 1.21.2

Combine the numerators over the common denominator.

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

Step 1.21.3

Add $1$ and $1$ .

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

Step 1.21.4

Divide $2$ by $2$ .

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

Step 1.22

Simplify ${(x^{2} + 4)}^{1}$ .

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

Step 1.23

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

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

Step 2

Find the second 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 $f (x) = 2 x^{2} + 4$ and $g (x) = {(x^{2} + 4)}^{\frac{1}{2}}$ .

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

Step 2.2

Multiply the exponents in ${({(x^{2} + 4)}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.1

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

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

Step 2.2.2

Cancel the common factor of $2$ .

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Step 2.2.2.1

Cancel the common factor.

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

Step 2.2.2.2

Rewrite the expression.

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

Step 2.3

Simplify.

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

Step 2.4

Differentiate.

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.4.3

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

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

Step 2.4.4

Multiply $2$ by $2$ .

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

Step 2.4.5

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

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

Step 2.4.6

Simplify the expression.

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Step 2.4.6.1

Add $4 x$ and $0$ .

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

Step 2.4.6.2

Move $4$ to the left of ${(x^{2} + 4)}^{\frac{1}{2}}$ .

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

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

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Step 2.5.1

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

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

Step 2.5.2

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

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

Step 2.5.3

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

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

Step 2.6

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 2.7

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

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

Step 2.8

Combine the numerators over the common denominator.

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

Step 2.9

Simplify the numerator.

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Step 2.9.1

Multiply $- 1$ by $2$ .

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

Step 2.9.2

Subtract $2$ from $1$ .

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

Step 2.10

Combine fractions.

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Step 2.10.1

Move the negative in front of the fraction.

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

Step 2.10.2

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

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

Step 2.10.3

Move ${(x^{2} + 4)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 2.11

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

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

Step 2.12

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

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

Step 2.13

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

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

Step 2.14

Simplify terms.

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Step 2.14.1

Add $2 x$ and $0$ .

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

Step 2.14.2

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

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

Step 2.14.3

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

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

Step 2.14.4

Cancel the common factor.

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

Step 2.14.5

Rewrite the expression.

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

Step 2.15

Simplify.

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Step 2.15.1

Apply the distributive property.

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

Step 2.15.2

Simplify the numerator.

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Step 2.15.2.1

Simplify each term.

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Step 2.15.2.1.1

Multiply $2$ by $- 1$ .

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

Step 2.15.2.1.2

Multiply $- 1$ by $4$ .

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

Step 2.15.2.2

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

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

Step 2.15.2.3

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

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Step 2.15.2.3.1

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

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

Step 2.15.2.3.2

Factor $2$ out of $- 4$ .

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

Step 2.15.2.3.3

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

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

Step 2.15.2.4

To write $4 {(x^{2} + 4)}^{\frac{1}{2}} x$ as a fraction with a common denominator, multiply by $\frac{{(x^{2} + 4)}^{\frac{1}{2}}}{{(x^{2} + 4)}^{\frac{1}{2}}}$ .

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

Step 2.15.2.5

Combine the numerators over the common denominator.

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

Step 2.15.2.6

Rewrite $\frac{4 {(x^{2} + 4)}^{\frac{1}{2}} x {(x^{2} + 4)}^{\frac{1}{2}} + 2 (- x^{2} - 2) x}{{(x^{2} + 4)}^{\frac{1}{2}}}$ in a factored form.

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Step 2.15.2.6.1

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

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Step 2.15.2.6.1.1

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

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

Step 2.15.2.6.1.2

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

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

Step 2.15.2.6.1.3

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

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

Step 2.15.2.6.2

Combine exponents.

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Step 2.15.2.6.2.1

Multiply ${(x^{2} + 4)}^{\frac{1}{2}}$ by ${(x^{2} + 4)}^{\frac{1}{2}}$ by adding the exponents.

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Step 2.15.2.6.2.1.1

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

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

Step 2.15.2.6.2.1.2

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

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

Step 2.15.2.6.2.1.3

Combine the numerators over the common denominator.

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

Step 2.15.2.6.2.1.4

Add $1$ and $1$ .

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

Step 2.15.2.6.2.1.5

Divide $2$ by $2$ .

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

Step 2.15.2.6.2.2

Simplify $2 {(x^{2} + 4)}^{1}$ .

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

Step 2.15.2.7

Simplify the numerator.

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Step 2.15.2.7.1

Apply the distributive property.

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

Step 2.15.2.7.2

Multiply $2$ by $4$ .

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

Step 2.15.2.7.3

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

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

Step 2.15.2.7.4

Subtract $2$ from $8$ .

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

Step 2.15.3

Combine terms.

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Step 2.15.3.1

Rewrite $\frac{\frac{2 x (x^{2} + 6)}{{(x^{2} + 4)}^{\frac{1}{2}}}}{x^{2} + 4}$ as a product.

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

Step 2.15.3.2

Multiply $\frac{2 x (x^{2} + 6)}{{(x^{2} + 4)}^{\frac{1}{2}}}$ by $\frac{1}{x^{2} + 4}$ .

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

Step 2.15.3.3

Multiply ${(x^{2} + 4)}^{\frac{1}{2}}$ by $x^{2} + 4$ by adding the exponents.

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Step 2.15.3.3.1

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

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Step 2.15.3.3.1.1

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

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

Step 2.15.3.3.1.2

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

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

Step 2.15.3.3.2

Write $1$ as a fraction with a common denominator.

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

Step 2.15.3.3.3

Combine the numerators over the common denominator.

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

Step 2.15.3.3.4

Add $1$ and $2$ .

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

Step 3

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

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

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