Calculus Examples

$y = \sqrt{x^{2} + 9}$

Step 1

Find the first derivative.

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Step 1.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} + 9}$ as ${(x^{2} + 9)}^{\frac{1}{2}}$ .

$f' (x) = \frac{d}{d x} ({(x^{2} + 9)}^{\frac{1}{2}})$

Step 1.2

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 $g (x) = x^{2} + 9$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $x^{2} + 9$ .

$f' (x) = \frac{d}{d u} (u^{\frac{1}{2}}) \frac{d}{d x} (x^{2} + 9)$

Step 1.2.2

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

$f' (x) = \frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} (x^{2} + 9)$

Step 1.2.3

Replace all occurrences of $u$ with $x^{2} + 9$ .

$f' (x) = \frac{1}{2} \cdot {(x^{2} + 9)}^{\frac{1}{2} - 1} \frac{d}{d x} (x^{2} + 9)$

Step 1.3

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

$f' (x) = \frac{1}{2} \cdot {(x^{2} + 9)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} (x^{2} + 9)$

Step 1.4

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

$f' (x) = \frac{1}{2} \cdot {(x^{2} + 9)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} (x^{2} + 9)$

Step 1.5

Combine the numerators over the common denominator.

$f' (x) = \frac{1}{2} \cdot {(x^{2} + 9)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} (x^{2} + 9)$

Step 1.6

Simplify the numerator.

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Step 1.6.1

Multiply $- 1$ by $2$ .

$f' (x) = \frac{1}{2} \cdot {(x^{2} + 9)}^{\frac{1 - 2}{2}} \frac{d}{d x} (x^{2} + 9)$

Step 1.6.2

Subtract $2$ from $1$ .

$f' (x) = \frac{1}{2} \cdot {(x^{2} + 9)}^{\frac{- 1}{2}} \frac{d}{d x} (x^{2} + 9)$

Step 1.7

Combine fractions.

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Step 1.7.1

Move the negative in front of the fraction.

$f' (x) = \frac{1}{2} \cdot {(x^{2} + 9)}^{- \frac{1}{2}} \frac{d}{d x} (x^{2} + 9)$

Step 1.7.2

Combine $\frac{1}{2}$ and ${(x^{2} + 9)}^{- \frac{1}{2}}$ .

$f' (x) = \frac{{(x^{2} + 9)}^{- \frac{1}{2}}}{2} \cdot \frac{d}{d x} (x^{2} + 9)$

Step 1.7.3

Move ${(x^{2} + 9)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (x) = \frac{1}{2 {(x^{2} + 9)}^{\frac{1}{2}}} \cdot \frac{d}{d x} (x^{2} + 9)$

Step 1.8

By the Sum Rule, the derivative of $x^{2} + 9$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [9]$ .

$f' (x) = \frac{1}{2 {(x^{2} + 9)}^{\frac{1}{2}}} \cdot (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (9))$

Step 1.9

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

$f' (x) = \frac{1}{2 {(x^{2} + 9)}^{\frac{1}{2}}} \cdot (2 x + \frac{d}{d x} (9))$

Step 1.10

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$f' (x) = \frac{1}{2 {(x^{2} + 9)}^{\frac{1}{2}}} \cdot (2 x + 0)$

Step 1.11

Simplify terms.

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Step 1.11.1

Add $2 x$ and $0$ .

$f' (x) = \frac{1}{2 {(x^{2} + 9)}^{\frac{1}{2}}} \cdot (2 x)$

Step 1.11.2

Combine $2$ and $\frac{1}{2 {(x^{2} + 9)}^{\frac{1}{2}}}$ .

$f' (x) = \frac{2}{2 {(x^{2} + 9)}^{\frac{1}{2}}} \cdot x$

Step 1.11.3

Combine $\frac{2}{2 {(x^{2} + 9)}^{\frac{1}{2}}}$ and $x$ .

$f' (x) = \frac{2 x}{2 {(x^{2} + 9)}^{\frac{1}{2}}}$

Step 1.11.4

Cancel the common factor.

$f' (x) = \frac{2 x}{2 {(x^{2} + 9)}^{\frac{1}{2}}}$

Step 1.11.5

Rewrite the expression.

$f' (x) = \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}}$

Step 2

Find the second derivative.

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Step 2.1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x$ and $g (x) = {(x^{2} + 9)}^{\frac{1}{2}}$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} \frac{d}{d x} (x) - x \frac{d}{d x} {(x^{2} + 9)}^{\frac{1}{2}}}{{({(x^{2} + 9)}^{\frac{1}{2}})}^{2}}$

Step 2.2

Multiply the exponents in ${({(x^{2} + 9)}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.1

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

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} \frac{d}{d x} (x) - x \frac{d}{d x} {(x^{2} + 9)}^{\frac{1}{2}}}{{(x^{2} + 9)}^{\frac{1}{2} \cdot 2}}$

Step 2.2.2

Cancel the common factor of $2$ .

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Step 2.2.2.1

Cancel the common factor.

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} \frac{d}{d x} (x) - x \frac{d}{d x} {(x^{2} + 9)}^{\frac{1}{2}}}{{(x^{2} + 9)}^{\frac{1}{2} \cdot 2}}$

Step 2.2.2.2

Rewrite the expression.

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} \frac{d}{d x} (x) - x \frac{d}{d x} {(x^{2} + 9)}^{\frac{1}{2}}}{x^{2} + 9}$

Step 2.3

Simplify.

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} \frac{d}{d x} (x) - x \frac{d}{d x} {(x^{2} + 9)}^{\frac{1}{2}}}{x^{2} + 9}$

Step 2.4

Differentiate using the Power Rule.

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Step 2.4.1

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

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} \cdot 1 - x \frac{d}{d x} {(x^{2} + 9)}^{\frac{1}{2}}}{x^{2} + 9}$

Step 2.4.2

Multiply ${(x^{2} + 9)}^{\frac{1}{2}}$ by $1$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - x \frac{d}{d x} {(x^{2} + 9)}^{\frac{1}{2}}}{x^{2} + 9}$

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 $g (x) = x^{2} + 9$ .

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Step 2.5.1

To apply the Chain Rule, set $u$ as $x^{2} + 9$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - x (\frac{d}{d u} (u^{\frac{1}{2}}) \frac{d}{d x} (x^{2} + 9))}{x^{2} + 9}$

Step 2.5.2

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

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - x (\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} (x^{2} + 9))}{x^{2} + 9}$

Step 2.5.3

Replace all occurrences of $u$ with $x^{2} + 9$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - x (\frac{1}{2} \cdot {(x^{2} + 9)}^{\frac{1}{2} - 1} \frac{d}{d x} (x^{2} + 9))}{x^{2} + 9}$

Step 2.6

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

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - x (\frac{1}{2} \cdot {(x^{2} + 9)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} (x^{2} + 9))}{x^{2} + 9}$

Step 2.7

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

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - x (\frac{1}{2} \cdot {(x^{2} + 9)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} (x^{2} + 9))}{x^{2} + 9}$

Step 2.8

Combine the numerators over the common denominator.

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - x (\frac{1}{2} \cdot {(x^{2} + 9)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} (x^{2} + 9))}{x^{2} + 9}$

Step 2.9

Simplify the numerator.

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Step 2.9.1

Multiply $- 1$ by $2$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - x (\frac{1}{2} \cdot {(x^{2} + 9)}^{\frac{1 - 2}{2}} \frac{d}{d x} (x^{2} + 9))}{x^{2} + 9}$

Step 2.9.2

Subtract $2$ from $1$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - x (\frac{1}{2} \cdot {(x^{2} + 9)}^{\frac{- 1}{2}} \frac{d}{d x} (x^{2} + 9))}{x^{2} + 9}$

Step 2.10

Combine fractions.

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Step 2.10.1

Move the negative in front of the fraction.

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - x (\frac{1}{2} \cdot {(x^{2} + 9)}^{- \frac{1}{2}} \frac{d}{d x} (x^{2} + 9))}{x^{2} + 9}$

Step 2.10.2

Combine $\frac{1}{2}$ and ${(x^{2} + 9)}^{- \frac{1}{2}}$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - x (\frac{{(x^{2} + 9)}^{- \frac{1}{2}}}{2} \cdot \frac{d}{d x} (x^{2} + 9))}{x^{2} + 9}$

Step 2.10.3

Move ${(x^{2} + 9)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - x (\frac{1}{2 {(x^{2} + 9)}^{\frac{1}{2}}} \cdot \frac{d}{d x} (x^{2} + 9))}{x^{2} + 9}$

Step 2.10.4

Combine $\frac{1}{2 {(x^{2} + 9)}^{\frac{1}{2}}}$ and $x$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - \frac{x}{2 {(x^{2} + 9)}^{\frac{1}{2}}} \cdot \frac{d}{d x} (x^{2} + 9)}{x^{2} + 9}$

Step 2.11

By the Sum Rule, the derivative of $x^{2} + 9$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [9]$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - \frac{x}{2 {(x^{2} + 9)}^{\frac{1}{2}}} \cdot (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (9))}{x^{2} + 9}$

Step 2.12

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

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - \frac{x}{2 {(x^{2} + 9)}^{\frac{1}{2}}} \cdot (2 x + \frac{d}{d x} (9))}{x^{2} + 9}$

Step 2.13

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - \frac{x}{2 {(x^{2} + 9)}^{\frac{1}{2}}} \cdot (2 x + 0)}{x^{2} + 9}$

Step 2.14

Combine fractions.

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Step 2.14.1

Add $2 x$ and $0$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - \frac{x}{2 {(x^{2} + 9)}^{\frac{1}{2}}} \cdot (2 x)}{x^{2} + 9}$

Step 2.14.2

Multiply $2$ by $- 1$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - 2 \frac{x}{2 {(x^{2} + 9)}^{\frac{1}{2}}} \cdot x}{x^{2} + 9}$

Step 2.14.3

Combine $- 2$ and $\frac{x}{2 {(x^{2} + 9)}^{\frac{1}{2}}}$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} + \frac{- 2 x}{2 {(x^{2} + 9)}^{\frac{1}{2}}} \cdot x}{x^{2} + 9}$

Step 2.14.4

Combine $\frac{- 2 x}{2 {(x^{2} + 9)}^{\frac{1}{2}}}$ and $x$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} + \frac{- 2 x \cdot x}{2 {(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.15

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

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} + \frac{- 2 (x \cdot x)}{2 {(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.16

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

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} + \frac{- 2 (x \cdot x)}{2 {(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.17

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

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} + \frac{- 2 x^{1 + 1}}{2 {(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.18

Add $1$ and $1$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} + \frac{- 2 x^{2}}{2 {(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.19

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

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} + \frac{2 (- x^{2})}{2 {(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.20

Cancel the common factors.

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Step 2.20.1

Factor $2$ out of $2 {(x^{2} + 9)}^{\frac{1}{2}}$ .

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} + \frac{2 (- x^{2})}{2 ({(x^{2} + 9)}^{\frac{1}{2}})}}{x^{2} + 9}$

Step 2.20.2

Cancel the common factor.

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} + \frac{2 (- x^{2})}{2 {(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.20.3

Rewrite the expression.

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} + \frac{- x^{2}}{{(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.21

Move the negative in front of the fraction.

$f'' (x) = \frac{{(x^{2} + 9)}^{\frac{1}{2}} - \frac{x^{2}}{{(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.22

To write ${(x^{2} + 9)}^{\frac{1}{2}}$ as a fraction with a common denominator, multiply by $\frac{{(x^{2} + 9)}^{\frac{1}{2}}}{{(x^{2} + 9)}^{\frac{1}{2}}}$ .

$f'' (x) = \frac{\frac{{(x^{2} + 9)}^{\frac{1}{2}} {(x^{2} + 9)}^{\frac{1}{2}}}{{(x^{2} + 9)}^{\frac{1}{2}}} - \frac{x^{2}}{{(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.23

Combine the numerators over the common denominator.

$f'' (x) = \frac{\frac{{(x^{2} + 9)}^{\frac{1}{2}} {(x^{2} + 9)}^{\frac{1}{2}} - x^{2}}{{(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.24

Multiply ${(x^{2} + 9)}^{\frac{1}{2}}$ by ${(x^{2} + 9)}^{\frac{1}{2}}$ by adding the exponents.

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Step 2.24.1

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

$f'' (x) = \frac{\frac{{(x^{2} + 9)}^{\frac{1}{2} + \frac{1}{2}} - x^{2}}{{(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.24.2

Combine the numerators over the common denominator.

$f'' (x) = \frac{\frac{{(x^{2} + 9)}^{\frac{1 + 1}{2}} - x^{2}}{{(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.24.3

Add $1$ and $1$ .

$f'' (x) = \frac{\frac{{(x^{2} + 9)}^{\frac{2}{2}} - x^{2}}{{(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.24.4

Divide $2$ by $2$ .

$f'' (x) = \frac{\frac{(x^{2} + 9) - x^{2}}{{(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.25

Simplify ${(x^{2} + 9)}^{1}$ .

$f'' (x) = \frac{\frac{x^{2} + 9 - x^{2}}{{(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.26

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

$f'' (x) = \frac{\frac{0 + 9}{{(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.27

Add $0$ and $9$ .

$f'' (x) = \frac{\frac{9}{{(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$

Step 2.28

Rewrite $\frac{\frac{9}{{(x^{2} + 9)}^{\frac{1}{2}}}}{x^{2} + 9}$ as a product.

$f'' (x) = \frac{9}{{(x^{2} + 9)}^{\frac{1}{2}}} \cdot \frac{1}{x^{2} + 9}$

Step 2.29

Multiply $\frac{9}{{(x^{2} + 9)}^{\frac{1}{2}}}$ by $\frac{1}{x^{2} + 9}$ .

$f'' (x) = \frac{9}{{(x^{2} + 9)}^{\frac{1}{2}} (x^{2} + 9)}$

Step 2.30

Multiply ${(x^{2} + 9)}^{\frac{1}{2}}$ by $x^{2} + 9$ by adding the exponents.

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Step 2.30.1

Multiply ${(x^{2} + 9)}^{\frac{1}{2}}$ by $x^{2} + 9$ .

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Step 2.30.1.1

Raise $x^{2} + 9$ to the power of $1$ .

$f'' (x) = \frac{9}{{(x^{2} + 9)}^{\frac{1}{2}} (x^{2} + 9)}$

Step 2.30.1.2

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

$f'' (x) = \frac{9}{{(x^{2} + 9)}^{\frac{1}{2} + 1}}$

Step 2.30.2

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

$f'' (x) = \frac{9}{{(x^{2} + 9)}^{\frac{1}{2} + \frac{2}{2}}}$

Step 2.30.3

Combine the numerators over the common denominator.

$f'' (x) = \frac{9}{{(x^{2} + 9)}^{\frac{1 + 2}{2}}}$

Step 2.30.4

Add $1$ and $2$ .

$f'' (x) = \frac{9}{{(x^{2} + 9)}^{\frac{3}{2}}}$

Step 3

Find the third derivative.

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Step 3.1

Differentiate using the Constant Multiple Rule.

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Step 3.1.1

Since $9$ is constant with respect to $x$ , the derivative of $\frac{9}{{(x^{2} + 9)}^{\frac{3}{2}}}$ with respect to $x$ is $9 \frac{d}{d x} [\frac{1}{{(x^{2} + 9)}^{\frac{3}{2}}}]$ .

$f''' (x) = 9 \frac{d}{d x} (\frac{1}{{(x^{2} + 9)}^{\frac{3}{2}}})$

Step 3.1.2

Apply basic rules of exponents.

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Step 3.1.2.1

Rewrite $\frac{1}{{(x^{2} + 9)}^{\frac{3}{2}}}$ as ${({(x^{2} + 9)}^{\frac{3}{2}})}^{- 1}$ .

$f''' (x) = 9 \frac{d}{d x} ({({(x^{2} + 9)}^{\frac{3}{2}})}^{- 1})$

Step 3.1.2.2

Multiply the exponents in ${({(x^{2} + 9)}^{\frac{3}{2}})}^{- 1}$ .

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Step 3.1.2.2.1

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

$f''' (x) = 9 \frac{d}{d x} ({(x^{2} + 9)}^{\frac{3}{2} \cdot - 1})$

Step 3.1.2.2.2

Multiply $\frac{3}{2} \cdot - 1$ .

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Step 3.1.2.2.2.1

Combine $\frac{3}{2}$ and $- 1$ .

$f''' (x) = 9 \frac{d}{d x} ({(x^{2} + 9)}^{\frac{3 \cdot - 1}{2}})$

Step 3.1.2.2.2.2

Multiply $3$ by $- 1$ .

$f''' (x) = 9 \frac{d}{d x} ({(x^{2} + 9)}^{\frac{- 3}{2}})$

Step 3.1.2.2.3

Move the negative in front of the fraction.

$f''' (x) = 9 \frac{d}{d x} ({(x^{2} + 9)}^{- \frac{3}{2}})$

Step 3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- \frac{3}{2}}$ and $g (x) = x^{2} + 9$ .

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Step 3.2.1

To apply the Chain Rule, set $u$ as $x^{2} + 9$ .

$f''' (x) = 9 (\frac{d}{d u} (u^{- \frac{3}{2}}) \frac{d}{d x} (x^{2} + 9))$

Step 3.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = - \frac{3}{2}$ .

$f''' (x) = 9 (- \frac{3}{2} u^{- \frac{3}{2} - 1} \frac{d}{d x} (x^{2} + 9))$

Step 3.2.3

Replace all occurrences of $u$ with $x^{2} + 9$ .

$f''' (x) = 9 (- \frac{3}{2} \cdot {(x^{2} + 9)}^{- \frac{3}{2} - 1} \frac{d}{d x} (x^{2} + 9))$

Step 3.3

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

$f''' (x) = 9 (- \frac{3}{2} \cdot {(x^{2} + 9)}^{- \frac{3}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} (x^{2} + 9))$

Step 3.4

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

$f''' (x) = 9 (- \frac{3}{2} \cdot {(x^{2} + 9)}^{- \frac{3}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} (x^{2} + 9))$

Step 3.5

Combine the numerators over the common denominator.

$f''' (x) = 9 (- \frac{3}{2} \cdot {(x^{2} + 9)}^{\frac{- 3 - 1 \cdot 2}{2}} \frac{d}{d x} (x^{2} + 9))$

Step 3.6

Simplify the numerator.

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Step 3.6.1

Multiply $- 1$ by $2$ .

$f''' (x) = 9 (- \frac{3}{2} \cdot {(x^{2} + 9)}^{\frac{- 3 - 2}{2}} \frac{d}{d x} (x^{2} + 9))$

Step 3.6.2

Subtract $2$ from $- 3$ .

$f''' (x) = 9 (- \frac{3}{2} \cdot {(x^{2} + 9)}^{\frac{- 5}{2}} \frac{d}{d x} (x^{2} + 9))$

Step 3.7

Combine fractions.

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Step 3.7.1

Move the negative in front of the fraction.

$f''' (x) = 9 (- \frac{3}{2} \cdot {(x^{2} + 9)}^{- \frac{5}{2}} \frac{d}{d x} (x^{2} + 9))$

Step 3.7.2

Combine ${(x^{2} + 9)}^{- \frac{5}{2}}$ and $\frac{3}{2}$ .

$f''' (x) = 9 (- \frac{{(x^{2} + 9)}^{- \frac{5}{2}} \cdot 3}{2} \cdot \frac{d}{d x} (x^{2} + 9))$

Step 3.7.3

Simplify the expression.

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Step 3.7.3.1

Move $3$ to the left of ${(x^{2} + 9)}^{- \frac{5}{2}}$ .

$f''' (x) = 9 (- \frac{3 \cdot {(x^{2} + 9)}^{- \frac{5}{2}}}{2} \cdot \frac{d}{d x} (x^{2} + 9))$

Step 3.7.3.2

Move ${(x^{2} + 9)}^{- \frac{5}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f''' (x) = 9 (- \frac{3}{2 {(x^{2} + 9)}^{\frac{5}{2}}} \cdot \frac{d}{d x} (x^{2} + 9))$

Step 3.7.3.3

Multiply $- 1$ by $9$ .

$f''' (x) = - 9 (\frac{3}{2 {(x^{2} + 9)}^{\frac{5}{2}}} \cdot \frac{d}{d x} (x^{2} + 9))$

Step 3.7.4

Combine $\frac{3}{2 {(x^{2} + 9)}^{\frac{5}{2}}}$ and $- 9$ .

$f''' (x) = \frac{3 \cdot - 9}{2 {(x^{2} + 9)}^{\frac{5}{2}}} \cdot \frac{d}{d x} (x^{2} + 9)$

Step 3.7.5

Simplify the expression.

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Step 3.7.5.1

Multiply $3$ by $- 9$ .

$f''' (x) = \frac{- 27}{2 {(x^{2} + 9)}^{\frac{5}{2}}} \cdot \frac{d}{d x} (x^{2} + 9)$

Step 3.7.5.2

Move the negative in front of the fraction.

$f''' (x) = - \frac{27}{2 {(x^{2} + 9)}^{\frac{5}{2}}} \cdot \frac{d}{d x} (x^{2} + 9)$

Step 3.8

By the Sum Rule, the derivative of $x^{2} + 9$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [9]$ .

$f''' (x) = - \frac{27}{2 {(x^{2} + 9)}^{\frac{5}{2}}} \cdot (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (9))$

Step 3.9

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

$f''' (x) = - \frac{27}{2 {(x^{2} + 9)}^{\frac{5}{2}}} \cdot (2 x + \frac{d}{d x} (9))$

Step 3.10

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$f''' (x) = - \frac{27}{2 {(x^{2} + 9)}^{\frac{5}{2}}} \cdot (2 x + 0)$

Step 3.11

Simplify terms.

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Step 3.11.1

Add $2 x$ and $0$ .

$f''' (x) = - \frac{27}{2 {(x^{2} + 9)}^{\frac{5}{2}}} \cdot (2 x)$

Step 3.11.2

Multiply $2$ by $- 1$ .

$f''' (x) = - 2 \frac{27}{2 {(x^{2} + 9)}^{\frac{5}{2}}} \cdot x$

Step 3.11.3

Combine $- 2$ and $\frac{27}{2 {(x^{2} + 9)}^{\frac{5}{2}}}$ .

$f''' (x) = \frac{- 2 \cdot 27}{2 {(x^{2} + 9)}^{\frac{5}{2}}} \cdot x$

Step 3.11.4

Multiply $- 2$ by $27$ .

$f''' (x) = \frac{- 54}{2 {(x^{2} + 9)}^{\frac{5}{2}}} \cdot x$

Step 3.11.5

Combine $\frac{- 54}{2 {(x^{2} + 9)}^{\frac{5}{2}}}$ and $x$ .

$f''' (x) = \frac{- 54 x}{2 {(x^{2} + 9)}^{\frac{5}{2}}}$

Step 3.11.6

Factor $2$ out of $- 54 x$ .

$f''' (x) = \frac{2 (- 27 x)}{2 {(x^{2} + 9)}^{\frac{5}{2}}}$

Step 3.12

Cancel the common factors.

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Step 3.12.1

Factor $2$ out of $2 {(x^{2} + 9)}^{\frac{5}{2}}$ .

$f''' (x) = \frac{2 (- 27 x)}{2 ({(x^{2} + 9)}^{\frac{5}{2}})}$

Step 3.12.2

Cancel the common factor.

$f''' (x) = \frac{2 (- 27 x)}{2 {(x^{2} + 9)}^{\frac{5}{2}}}$

Step 3.12.3

Rewrite the expression.

$f''' (x) = \frac{- 27 x}{{(x^{2} + 9)}^{\frac{5}{2}}}$

Step 3.13

Move the negative in front of the fraction.

$f''' (x) = - \frac{27 x}{{(x^{2} + 9)}^{\frac{5}{2}}}$

Step 4

Find the fourth derivative.

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Step 4.1

Since $- 27$ is constant with respect to $x$ , the derivative of $- \frac{27 x}{{(x^{2} + 9)}^{\frac{5}{2}}}$ with respect to $x$ is $- 27 \frac{d}{d x} [\frac{x}{{(x^{2} + 9)}^{\frac{5}{2}}}]$ .

$f^{4} (x) = - 27 \frac{d}{d x} \frac{x}{{(x^{2} + 9)}^{\frac{5}{2}}}$

Step 4.2

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} \frac{d}{d x} (x) - x \frac{d}{d x} {(x^{2} + 9)}^{\frac{5}{2}}}{{({(x^{2} + 9)}^{\frac{5}{2}})}^{2}}$

Step 4.3

Differentiate using the Power Rule.

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Step 4.3.1

Multiply the exponents in ${({(x^{2} + 9)}^{\frac{5}{2}})}^{2}$ .

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Step 4.3.1.1

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

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} \frac{d}{d x} (x) - x \frac{d}{d x} {(x^{2} + 9)}^{\frac{5}{2}}}{{(x^{2} + 9)}^{\frac{5}{2} \cdot 2}}$

Step 4.3.1.2

Cancel the common factor of $2$ .

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Step 4.3.1.2.1

Cancel the common factor.

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} \frac{d}{d x} (x) - x \frac{d}{d x} {(x^{2} + 9)}^{\frac{5}{2}}}{{(x^{2} + 9)}^{\frac{5}{2} \cdot 2}}$

Step 4.3.1.2.2

Rewrite the expression.

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} \frac{d}{d x} (x) - x \frac{d}{d x} {(x^{2} + 9)}^{\frac{5}{2}}}{{(x^{2} + 9)}^{5}}$

Step 4.3.2

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

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} \cdot 1 - x \frac{d}{d x} {(x^{2} + 9)}^{\frac{5}{2}}}{{(x^{2} + 9)}^{5}}$

Step 4.3.3

Multiply ${(x^{2} + 9)}^{\frac{5}{2}}$ by $1$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - x \frac{d}{d x} {(x^{2} + 9)}^{\frac{5}{2}}}{{(x^{2} + 9)}^{5}}$

Step 4.4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{5}{2}}$ and $g (x) = x^{2} + 9$ .

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Step 4.4.1

To apply the Chain Rule, set $u$ as $x^{2} + 9$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - x (\frac{d}{d u} (u^{\frac{5}{2}}) \frac{d}{d x} (x^{2} + 9))}{{(x^{2} + 9)}^{5}}$

Step 4.4.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{5}{2}$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - x (\frac{5}{2} u^{\frac{5}{2} - 1} \frac{d}{d x} (x^{2} + 9))}{{(x^{2} + 9)}^{5}}$

Step 4.4.3

Replace all occurrences of $u$ with $x^{2} + 9$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - x (\frac{5}{2} \cdot {(x^{2} + 9)}^{\frac{5}{2} - 1} \frac{d}{d x} (x^{2} + 9))}{{(x^{2} + 9)}^{5}}$

Step 4.5

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

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - x (\frac{5}{2} \cdot {(x^{2} + 9)}^{\frac{5}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} (x^{2} + 9))}{{(x^{2} + 9)}^{5}}$

Step 4.6

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

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - x (\frac{5}{2} \cdot {(x^{2} + 9)}^{\frac{5}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} (x^{2} + 9))}{{(x^{2} + 9)}^{5}}$

Step 4.7

Combine the numerators over the common denominator.

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - x (\frac{5}{2} \cdot {(x^{2} + 9)}^{\frac{5 - 1 \cdot 2}{2}} \frac{d}{d x} (x^{2} + 9))}{{(x^{2} + 9)}^{5}}$

Step 4.8

Simplify the numerator.

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Step 4.8.1

Multiply $- 1$ by $2$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - x (\frac{5}{2} \cdot {(x^{2} + 9)}^{\frac{5 - 2}{2}} \frac{d}{d x} (x^{2} + 9))}{{(x^{2} + 9)}^{5}}$

Step 4.8.2

Subtract $2$ from $5$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - x (\frac{5}{2} \cdot {(x^{2} + 9)}^{\frac{3}{2}} \frac{d}{d x} (x^{2} + 9))}{{(x^{2} + 9)}^{5}}$

Step 4.9

Combine fractions.

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Step 4.9.1

Combine $\frac{5}{2}$ and ${(x^{2} + 9)}^{\frac{3}{2}}$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - x (\frac{5 {(x^{2} + 9)}^{\frac{3}{2}}}{2} \cdot \frac{d}{d x} (x^{2} + 9))}{{(x^{2} + 9)}^{5}}$

Step 4.9.2

Combine $\frac{5 {(x^{2} + 9)}^{\frac{3}{2}}}{2}$ and $x$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - \frac{5 {(x^{2} + 9)}^{\frac{3}{2}} x}{2} \cdot \frac{d}{d x} (x^{2} + 9)}{{(x^{2} + 9)}^{5}}$

Step 4.10

By the Sum Rule, the derivative of $x^{2} + 9$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [9]$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - \frac{5 {(x^{2} + 9)}^{\frac{3}{2}} x}{2} \cdot (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (9))}{{(x^{2} + 9)}^{5}}$

Step 4.11

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

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - \frac{5 {(x^{2} + 9)}^{\frac{3}{2}} x}{2} \cdot (2 x + \frac{d}{d x} (9))}{{(x^{2} + 9)}^{5}}$

Step 4.12

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - \frac{5 {(x^{2} + 9)}^{\frac{3}{2}} x}{2} \cdot (2 x + 0)}{{(x^{2} + 9)}^{5}}$

Step 4.13

Combine fractions.

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Step 4.13.1

Add $2 x$ and $0$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - \frac{5 {(x^{2} + 9)}^{\frac{3}{2}} x}{2} \cdot (2 x)}{{(x^{2} + 9)}^{5}}$

Step 4.13.2

Multiply $2$ by $- 1$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - 2 \frac{5 {(x^{2} + 9)}^{\frac{3}{2}} x}{2} \cdot x}{{(x^{2} + 9)}^{5}}$

Step 4.13.3

Combine $- 2$ and $\frac{5 {(x^{2} + 9)}^{\frac{3}{2}} x}{2}$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} + \frac{- 2 (5 {(x^{2} + 9)}^{\frac{3}{2}} x)}{2} \cdot x}{{(x^{2} + 9)}^{5}}$

Step 4.13.4

Multiply $5$ by $- 2$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} + \frac{- 10 ({(x^{2} + 9)}^{\frac{3}{2}} x)}{2} \cdot x}{{(x^{2} + 9)}^{5}}$

Step 4.13.5

Combine $\frac{- 10 ({(x^{2} + 9)}^{\frac{3}{2}} x)}{2}$ and $x$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} + \frac{- 10 ({(x^{2} + 9)}^{\frac{3}{2}} x) x}{2}}{{(x^{2} + 9)}^{5}}$

Step 4.14

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

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} + \frac{- 10 ({(x^{2} + 9)}^{\frac{3}{2}} (x \cdot x))}{2}}{{(x^{2} + 9)}^{5}}$

Step 4.15

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

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} + \frac{- 10 ({(x^{2} + 9)}^{\frac{3}{2}} (x \cdot x))}{2}}{{(x^{2} + 9)}^{5}}$

Step 4.16

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

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} + \frac{- 10 ({(x^{2} + 9)}^{\frac{3}{2}} x^{1 + 1})}{2}}{{(x^{2} + 9)}^{5}}$

Step 4.17

Add $1$ and $1$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} + \frac{- 10 ({(x^{2} + 9)}^{\frac{3}{2}} x^{2})}{2}}{{(x^{2} + 9)}^{5}}$

Step 4.18

Factor $2$ out of $- 10 {(x^{2} + 9)}^{\frac{3}{2}} x^{2}$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} + \frac{2 (- 5 {(x^{2} + 9)}^{\frac{3}{2}} x^{2})}{2}}{{(x^{2} + 9)}^{5}}$

Step 4.19

Cancel the common factors.

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Step 4.19.1

Factor $2$ out of $2$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} + \frac{2 (- 5 {(x^{2} + 9)}^{\frac{3}{2}} x^{2})}{2 (1)}}{{(x^{2} + 9)}^{5}}$

Step 4.19.2

Cancel the common factor.

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} + \frac{2 (- 5 {(x^{2} + 9)}^{\frac{3}{2}} x^{2})}{2 \cdot 1}}{{(x^{2} + 9)}^{5}}$

Step 4.19.3

Rewrite the expression.

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} + \frac{- 5 {(x^{2} + 9)}^{\frac{3}{2}} x^{2}}{1}}{{(x^{2} + 9)}^{5}}$

Step 4.19.4

Divide $- 5 {(x^{2} + 9)}^{\frac{3}{2}} x^{2}$ by $1$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - 5 {(x^{2} + 9)}^{\frac{3}{2}} x^{2}}{{(x^{2} + 9)}^{5}}$

Step 4.20

Factor ${(x^{2} + 9)}^{\frac{3}{2}}$ out of ${(x^{2} + 9)}^{\frac{5}{2}} - 5 {(x^{2} + 9)}^{\frac{3}{2}} x^{2}$ .

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Step 4.20.1

Move ${(x^{2} + 9)}^{\frac{3}{2}}$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{5}{2}} - 5 x^{2} {(x^{2} + 9)}^{\frac{3}{2}}}{{(x^{2} + 9)}^{5}}$

Step 4.20.2

Factor ${(x^{2} + 9)}^{\frac{3}{2}}$ out of ${(x^{2} + 9)}^{\frac{5}{2}}$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{3}{2}} {(x^{2} + 9)}^{\frac{2}{2}} - 5 x^{2} {(x^{2} + 9)}^{\frac{3}{2}}}{{(x^{2} + 9)}^{5}}$

Step 4.20.3

Factor ${(x^{2} + 9)}^{\frac{3}{2}}$ out of $- 5 x^{2} {(x^{2} + 9)}^{\frac{3}{2}}$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{3}{2}} {(x^{2} + 9)}^{\frac{2}{2}} + {(x^{2} + 9)}^{\frac{3}{2}} (- 5 x^{2})}{{(x^{2} + 9)}^{5}}$

Step 4.20.4

Factor ${(x^{2} + 9)}^{\frac{3}{2}}$ out of ${(x^{2} + 9)}^{\frac{3}{2}} {(x^{2} + 9)}^{\frac{2}{2}} + {(x^{2} + 9)}^{\frac{3}{2}} (- 5 x^{2})$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{3}{2}} ({(x^{2} + 9)}^{\frac{2}{2}} - 5 x^{2})}{{(x^{2} + 9)}^{5}}$

Step 4.21

Cancel the common factor of $2$ .

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Step 4.21.1

Cancel the common factor.

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{3}{2}} ({(x^{2} + 9)}^{\frac{2}{2}} - 5 x^{2})}{{(x^{2} + 9)}^{5}}$

Step 4.21.2

Rewrite the expression.

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{3}{2}} ((x^{2} + 9) - 5 x^{2})}{{(x^{2} + 9)}^{5}}$

Step 4.22

Simplify.

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{3}{2}} (x^{2} + 9 - 5 x^{2})}{{(x^{2} + 9)}^{5}}$

Step 4.23

Subtract $5 x^{2}$ from $x^{2}$ .

$f^{4} (x) = - 27 \frac{{(x^{2} + 9)}^{\frac{3}{2}} (- 4 x^{2} + 9)}{{(x^{2} + 9)}^{5}}$

Step 4.24

Move ${(x^{2} + 9)}^{\frac{3}{2}}$ to the denominator using the negative exponent rule $b^{n} = \frac{1}{b^{- n}}$ .

$f^{4} (x) = - 27 \frac{- 4 x^{2} + 9}{{(x^{2} + 9)}^{5} {(x^{2} + 9)}^{- \frac{3}{2}}}$

Step 4.25

Multiply ${(x^{2} + 9)}^{5}$ by ${(x^{2} + 9)}^{- \frac{3}{2}}$ by adding the exponents.

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Step 4.25.1

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

$f^{4} (x) = - 27 \frac{- 4 x^{2} + 9}{{(x^{2} + 9)}^{5 - \frac{3}{2}}}$

Step 4.25.2

To write $5$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f^{4} (x) = - 27 \frac{- 4 x^{2} + 9}{{(x^{2} + 9)}^{5 \cdot \frac{2}{2} - \frac{3}{2}}}$

Step 4.25.3

Combine $5$ and $\frac{2}{2}$ .

$f^{4} (x) = - 27 \frac{- 4 x^{2} + 9}{{(x^{2} + 9)}^{\frac{5 \cdot 2}{2} - \frac{3}{2}}}$

Step 4.25.4

Combine the numerators over the common denominator.

$f^{4} (x) = - 27 \frac{- 4 x^{2} + 9}{{(x^{2} + 9)}^{\frac{5 \cdot 2 - 3}{2}}}$

Step 4.25.5

Simplify the numerator.

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Step 4.25.5.1

Multiply $5$ by $2$ .

$f^{4} (x) = - 27 \frac{- 4 x^{2} + 9}{{(x^{2} + 9)}^{\frac{10 - 3}{2}}}$

Step 4.25.5.2

Subtract $3$ from $10$ .

$f^{4} (x) = - 27 \frac{- 4 x^{2} + 9}{{(x^{2} + 9)}^{\frac{7}{2}}}$

Step 4.26

Combine $- 27$ and $\frac{- 4 x^{2} + 9}{{(x^{2} + 9)}^{\frac{7}{2}}}$ .

$f^{4} (x) = \frac{- 27 (- 4 x^{2} + 9)}{{(x^{2} + 9)}^{\frac{7}{2}}}$

Step 4.27

Move the negative in front of the fraction.

$f^{4} (x) = - \frac{(27) (- 4 x^{2} + 9)}{{(x^{2} + 9)}^{\frac{7}{2}}}$

Step 4.28

Simplify.

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Step 4.28.1

Apply the distributive property.

$f^{4} (x) = - \frac{27 (- 4 x^{2}) + 27 \cdot 9}{{(x^{2} + 9)}^{\frac{7}{2}}}$

Step 4.28.2

Simplify each term.

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Step 4.28.2.1

Multiply $- 4$ by $27$ .

$f^{4} (x) = - \frac{- 108 x^{2} + 27 \cdot 9}{{(x^{2} + 9)}^{\frac{7}{2}}}$

Step 4.28.2.2

Multiply $27$ by $9$ .

$f^{4} (x) = - \frac{- 108 x^{2} + 243}{{(x^{2} + 9)}^{\frac{7}{2}}}$