Calculus Examples

$4 {(x + 2)}^{\frac{5}{3}}$

Step 1

Find the first derivative.

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Step 1.1

Since $4$ is constant with respect to $x$ , the derivative of $4 {(x + 2)}^{\frac{5}{3}}$ with respect to $x$ is $4 \frac{d}{d x} [{(x + 2)}^{\frac{5}{3}}]$ .

$4 \frac{d}{d x} [{(x + 2)}^{\frac{5}{3}}]$

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 $f (x) = x^{\frac{5}{3}}$ and $g (x) = x + 2$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $x + 2$ .

$4 (\frac{d}{d u} [u^{\frac{5}{3}}] \frac{d}{d x} [x + 2])$

Step 1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{5}{3}$ .

$4 (\frac{5}{3} u^{\frac{5}{3} - 1} \frac{d}{d x} [x + 2])$

Step 1.2.3

Replace all occurrences of $u$ with $x + 2$ .

$4 (\frac{5}{3} {(x + 2)}^{\frac{5}{3} - 1} \frac{d}{d x} [x + 2])$

Step 1.3

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

$4 (\frac{5}{3} {(x + 2)}^{\frac{5}{3} - 1 \cdot \frac{3}{3}} \frac{d}{d x} [x + 2])$

Step 1.4

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

$4 (\frac{5}{3} {(x + 2)}^{\frac{5}{3} + \frac{- 1 \cdot 3}{3}} \frac{d}{d x} [x + 2])$

Step 1.5

Combine the numerators over the common denominator.

$4 (\frac{5}{3} {(x + 2)}^{\frac{5 - 1 \cdot 3}{3}} \frac{d}{d x} [x + 2])$

Step 1.6

Simplify the numerator.

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Step 1.6.1

Multiply $- 1$ by $3$ .

$4 (\frac{5}{3} {(x + 2)}^{\frac{5 - 3}{3}} \frac{d}{d x} [x + 2])$

Step 1.6.2

Subtract $3$ from $5$ .

$4 (\frac{5}{3} {(x + 2)}^{\frac{2}{3}} \frac{d}{d x} [x + 2])$

Step 1.7

Combine fractions.

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Step 1.7.1

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

$4 (\frac{5 {(x + 2)}^{\frac{2}{3}}}{3} \frac{d}{d x} [x + 2])$

Step 1.7.2

Combine $\frac{5 {(x + 2)}^{\frac{2}{3}}}{3}$ and $4$ .

$\frac{5 {(x + 2)}^{\frac{2}{3}} \cdot 4}{3} \frac{d}{d x} [x + 2]$

Step 1.7.3

Multiply $4$ by $5$ .

$\frac{20 {(x + 2)}^{\frac{2}{3}}}{3} \frac{d}{d x} [x + 2]$

Step 1.8

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

$\frac{20 {(x + 2)}^{\frac{2}{3}}}{3} (\frac{d}{d x} [x] + \frac{d}{d x} [2])$

Step 1.9

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

$\frac{20 {(x + 2)}^{\frac{2}{3}}}{3} (1 + \frac{d}{d x} [2])$

Step 1.10

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$\frac{20 {(x + 2)}^{\frac{2}{3}}}{3} (1 + 0)$

Step 1.11

Simplify the expression.

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Step 1.11.1

Add $1$ and $0$ .

$\frac{20 {(x + 2)}^{\frac{2}{3}}}{3} \cdot 1$

Step 1.11.2

Multiply $\frac{20 {(x + 2)}^{\frac{2}{3}}}{3}$ by $1$ .

$f' (x) = \frac{20 {(x + 2)}^{\frac{2}{3}}}{3}$

Step 2

Find the second derivative.

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Step 2.1

Since $\frac{20}{3}$ is constant with respect to $x$ , the derivative of $\frac{20 {(x + 2)}^{\frac{2}{3}}}{3}$ with respect to $x$ is $\frac{20}{3} \frac{d}{d x} [{(x + 2)}^{\frac{2}{3}}]$ .

$\frac{20}{3} \frac{d}{d x} [{(x + 2)}^{\frac{2}{3}}]$

Step 2.2

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

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Step 2.2.1

To apply the Chain Rule, set $u$ as $x + 2$ .

$\frac{20}{3} (\frac{d}{d u} [u^{\frac{2}{3}}] \frac{d}{d x} [x + 2])$

Step 2.2.2

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

$\frac{20}{3} (\frac{2}{3} u^{\frac{2}{3} - 1} \frac{d}{d x} [x + 2])$

Step 2.2.3

Replace all occurrences of $u$ with $x + 2$ .

$\frac{20}{3} (\frac{2}{3} {(x + 2)}^{\frac{2}{3} - 1} \frac{d}{d x} [x + 2])$

Step 2.3

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

$\frac{20}{3} (\frac{2}{3} {(x + 2)}^{\frac{2}{3} - 1 \cdot \frac{3}{3}} \frac{d}{d x} [x + 2])$

Step 2.4

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

$\frac{20}{3} (\frac{2}{3} {(x + 2)}^{\frac{2}{3} + \frac{- 1 \cdot 3}{3}} \frac{d}{d x} [x + 2])$

Step 2.5

Combine the numerators over the common denominator.

$\frac{20}{3} (\frac{2}{3} {(x + 2)}^{\frac{2 - 1 \cdot 3}{3}} \frac{d}{d x} [x + 2])$

Step 2.6

Simplify the numerator.

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Step 2.6.1

Multiply $- 1$ by $3$ .

$\frac{20}{3} (\frac{2}{3} {(x + 2)}^{\frac{2 - 3}{3}} \frac{d}{d x} [x + 2])$

Step 2.6.2

Subtract $3$ from $2$ .

$\frac{20}{3} (\frac{2}{3} {(x + 2)}^{\frac{- 1}{3}} \frac{d}{d x} [x + 2])$

Step 2.7

Combine fractions.

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Step 2.7.1

Move the negative in front of the fraction.

$\frac{20}{3} (\frac{2}{3} {(x + 2)}^{- \frac{1}{3}} \frac{d}{d x} [x + 2])$

Step 2.7.2

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

$\frac{20}{3} (\frac{2 {(x + 2)}^{- \frac{1}{3}}}{3} \frac{d}{d x} [x + 2])$

Step 2.7.3

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

$\frac{20}{3} (\frac{2}{3 {(x + 2)}^{\frac{1}{3}}} \frac{d}{d x} [x + 2])$

Step 2.7.4

Multiply $\frac{2}{3 {(x + 2)}^{\frac{1}{3}}}$ by $\frac{20}{3}$ .

$\frac{2 \cdot 20}{3 {(x + 2)}^{\frac{1}{3}} \cdot 3} \frac{d}{d x} [x + 2]$

Step 2.7.5

Multiply.

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Step 2.7.5.1

Multiply $2$ by $20$ .

$\frac{40}{3 {(x + 2)}^{\frac{1}{3}} \cdot 3} \frac{d}{d x} [x + 2]$

Step 2.7.5.2

Multiply $3$ by $3$ .

$\frac{40}{9 {(x + 2)}^{\frac{1}{3}}} \frac{d}{d x} [x + 2]$

Step 2.8

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

$\frac{40}{9 {(x + 2)}^{\frac{1}{3}}} (\frac{d}{d x} [x] + \frac{d}{d x} [2])$

Step 2.9

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

$\frac{40}{9 {(x + 2)}^{\frac{1}{3}}} (1 + \frac{d}{d x} [2])$

Step 2.10

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$\frac{40}{9 {(x + 2)}^{\frac{1}{3}}} (1 + 0)$

Step 2.11

Simplify the expression.

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Step 2.11.1

Add $1$ and $0$ .

$\frac{40}{9 {(x + 2)}^{\frac{1}{3}}} \cdot 1$

Step 2.11.2

Multiply $\frac{40}{9 {(x + 2)}^{\frac{1}{3}}}$ by $1$ .

$f'' (x) = \frac{40}{9 {(x + 2)}^{\frac{1}{3}}}$