Calculus Examples

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

Step 1

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

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Step 1.1

To apply the Chain Rule, set $u$ as $x + 9$ .

$\frac{d}{d u} [u^{\frac{2}{3}}] \frac{d}{d x} [x + 9]$

Step 1.2

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

$\frac{2}{3} u^{\frac{2}{3} - 1} \frac{d}{d x} [x + 9]$

Step 1.3

Replace all occurrences of $u$ with $x + 9$ .

$\frac{2}{3} {(x + 9)}^{\frac{2}{3} - 1} \frac{d}{d x} [x + 9]$

Step 2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{2}{3} {(x + 9)}^{\frac{2}{3} - 1 \cdot \frac{3}{3}} \frac{d}{d x} [x + 9]$

Step 3

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

$\frac{2}{3} {(x + 9)}^{\frac{2}{3} + \frac{- 1 \cdot 3}{3}} \frac{d}{d x} [x + 9]$

Step 4

Combine the numerators over the common denominator.

$\frac{2}{3} {(x + 9)}^{\frac{2 - 1 \cdot 3}{3}} \frac{d}{d x} [x + 9]$

Step 5

Simplify the numerator.

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Step 5.1

Multiply $- 1$ by $3$ .

$\frac{2}{3} {(x + 9)}^{\frac{2 - 3}{3}} \frac{d}{d x} [x + 9]$

Step 5.2

Subtract $3$ from $2$ .

$\frac{2}{3} {(x + 9)}^{\frac{- 1}{3}} \frac{d}{d x} [x + 9]$

Step 6

Combine fractions.

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Step 6.1

Move the negative in front of the fraction.

$\frac{2}{3} {(x + 9)}^{- \frac{1}{3}} \frac{d}{d x} [x + 9]$

Step 6.2

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

$\frac{2 {(x + 9)}^{- \frac{1}{3}}}{3} \frac{d}{d x} [x + 9]$

Step 6.3

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

$\frac{2}{3 {(x + 9)}^{\frac{1}{3}}} \frac{d}{d x} [x + 9]$

Step 7

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

$\frac{2}{3 {(x + 9)}^{\frac{1}{3}}} (\frac{d}{d x} [x] + \frac{d}{d x} [9])$

Step 8

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

$\frac{2}{3 {(x + 9)}^{\frac{1}{3}}} (1 + \frac{d}{d x} [9])$

Step 9

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

$\frac{2}{3 {(x + 9)}^{\frac{1}{3}}} (1 + 0)$

Step 10

Simplify the expression.

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Step 10.1

Add $1$ and $0$ .

$\frac{2}{3 {(x + 9)}^{\frac{1}{3}}} \cdot 1$

Step 10.2

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

$\frac{2}{3 {(x + 9)}^{\frac{1}{3}}}$