Calculus Examples

$f (x) = \arctan (\sqrt{1 + 3 x})$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{1 + 3 x}$ as ${(1 + 3 x)}^{\frac{1}{2}}$ .

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

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \arctan (x)$ and $g (x) = {(1 + 3 x)}^{\frac{1}{2}}$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as ${(1 + 3 x)}^{\frac{1}{2}}$ .

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

Step 2.2

The derivative of $\arctan (u_{1})$ with respect to $u_{1}$ is $\frac{1}{1 + {u_{1}}^{2}}$ .

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

Step 2.3

Replace all occurrences of $u_{1}$ with ${(1 + 3 x)}^{\frac{1}{2}}$ .

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

Step 3

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

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Step 3.1

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

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

Step 3.2

Cancel the common factor of $2$ .

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Step 3.2.1

Cancel the common factor.

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

Step 3.2.2

Rewrite the expression.

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

Step 4

Simplify.

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

Step 5

Add $1$ and $1$ .

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

Step 6

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 $g (x) = 1 + 3 x$ .

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Step 6.1

To apply the Chain Rule, set $u_{2}$ as $1 + 3 x$ .

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

Step 6.2

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

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

Step 6.3

Replace all occurrences of $u_{2}$ with $1 + 3 x$ .

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

Step 7

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

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

Step 8

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

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

Step 9

Combine the numerators over the common denominator.

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

Step 10

Simplify the numerator.

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Step 10.1

Multiply $- 1$ by $2$ .

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

Step 10.2

Subtract $2$ from $1$ .

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

Step 11

Combine fractions.

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Step 11.1

Move the negative in front of the fraction.

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

Step 11.2

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

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

Step 11.3

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

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

Step 11.4

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

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

Step 12

By the Sum Rule, the derivative of $1 + 3 x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [3 x]$ .

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

Step 13

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

$\frac{1}{2 {(1 + 3 x)}^{\frac{1}{2}} (2 + 3 x)} (0 + \frac{d}{d x} [3 x])$

Step 14

Add $0$ and $\frac{d}{d x} [3 x]$ .

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

Step 15

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

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

Step 16

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

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

Step 17

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

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

Step 18

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

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