Calculus Examples

$y = \arctan (4 x) + \ln (16 x^{2} + 1)$

Step 1

By the Sum Rule, the derivative of $\arctan (4 x) + \ln (16 x^{2} + 1)$ with respect to $x$ is $\frac{d}{d x} [\arctan (4 x)] + \frac{d}{d x} [\ln (16 x^{2} + 1)]$ .

$\frac{d}{d x} [\arctan (4 x)] + \frac{d}{d x} [\ln (16 x^{2} + 1)]$

Step 2

Evaluate $\frac{d}{d x} [\arctan (4 x)]$ .

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Step 2.1

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

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Step 2.1.1

To apply the Chain Rule, set $u_{1}$ as $4 x$ .

$\frac{d}{d u_{1}} [\arctan (u_{1})] \frac{d}{d x} [4 x] + \frac{d}{d x} [\ln (16 x^{2} + 1)]$

Step 2.1.2

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

$\frac{1}{1 + {u_{1}}^{2}} \frac{d}{d x} [4 x] + \frac{d}{d x} [\ln (16 x^{2} + 1)]$

Step 2.1.3

Replace all occurrences of $u_{1}$ with $4 x$ .

$\frac{1}{1 + {(4 x)}^{2}} \frac{d}{d x} [4 x] + \frac{d}{d x} [\ln (16 x^{2} + 1)]$

Step 2.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$\frac{1}{1 + {(4 x)}^{2}} (4 \frac{d}{d x} [x]) + \frac{d}{d x} [\ln (16 x^{2} + 1)]$

Step 2.3

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

$\frac{1}{1 + {(4 x)}^{2}} (4 \cdot 1) + \frac{d}{d x} [\ln (16 x^{2} + 1)]$

Step 2.4

Factor $4$ out of $4 x$ .

$\frac{1}{1 + {(4 (x))}^{2}} (4 \cdot 1) + \frac{d}{d x} [\ln (16 x^{2} + 1)]$

Step 2.5

Apply the product rule to $4 (x)$ .

$\frac{1}{1 + 4^{2} x^{2}} (4 \cdot 1) + \frac{d}{d x} [\ln (16 x^{2} + 1)]$

Step 2.6

Raise $4$ to the power of $2$ .

$\frac{1}{1 + 16 x^{2}} (4 \cdot 1) + \frac{d}{d x} [\ln (16 x^{2} + 1)]$

Step 2.7

Multiply $4$ by $1$ .

$\frac{1}{1 + 16 x^{2}} \cdot 4 + \frac{d}{d x} [\ln (16 x^{2} + 1)]$

Step 2.8

Combine $\frac{1}{1 + 16 x^{2}}$ and $4$ .

$\frac{4}{1 + 16 x^{2}} + \frac{d}{d x} [\ln (16 x^{2} + 1)]$

Step 3

Evaluate $\frac{d}{d x} [\ln (16 x^{2} + 1)]$ .

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Step 3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 16 x^{2} + 1$ .

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Step 3.1.1

To apply the Chain Rule, set $u_{2}$ as $16 x^{2} + 1$ .

$\frac{4}{1 + 16 x^{2}} + \frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d x} [16 x^{2} + 1]$

Step 3.1.2

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

$\frac{4}{1 + 16 x^{2}} + \frac{1}{u_{2}} \frac{d}{d x} [16 x^{2} + 1]$

Step 3.1.3

Replace all occurrences of $u_{2}$ with $16 x^{2} + 1$ .

$\frac{4}{1 + 16 x^{2}} + \frac{1}{16 x^{2} + 1} \frac{d}{d x} [16 x^{2} + 1]$

Step 3.2

By the Sum Rule, the derivative of $16 x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [16 x^{2}] + \frac{d}{d x} [1]$ .

$\frac{4}{1 + 16 x^{2}} + \frac{1}{16 x^{2} + 1} (\frac{d}{d x} [16 x^{2}] + \frac{d}{d x} [1])$

Step 3.3

Since $16$ is constant with respect to $x$ , the derivative of $16 x^{2}$ with respect to $x$ is $16 \frac{d}{d x} [x^{2}]$ .

$\frac{4}{1 + 16 x^{2}} + \frac{1}{16 x^{2} + 1} (16 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [1])$

Step 3.4

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

$\frac{4}{1 + 16 x^{2}} + \frac{1}{16 x^{2} + 1} (16 (2 x) + \frac{d}{d x} [1])$

Step 3.5

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

$\frac{4}{1 + 16 x^{2}} + \frac{1}{16 x^{2} + 1} (16 (2 x) + 0)$

Step 3.6

Multiply $2$ by $16$ .

$\frac{4}{1 + 16 x^{2}} + \frac{1}{16 x^{2} + 1} (32 x + 0)$

Step 3.7

Add $32 x$ and $0$ .

$\frac{4}{1 + 16 x^{2}} + \frac{1}{16 x^{2} + 1} (32 x)$

Step 3.8

Combine $32$ and $\frac{1}{16 x^{2} + 1}$ .

$\frac{4}{1 + 16 x^{2}} + \frac{32}{16 x^{2} + 1} x$

Step 3.9

Combine $\frac{32}{16 x^{2} + 1}$ and $x$ .

$\frac{4}{1 + 16 x^{2}} + \frac{32 x}{16 x^{2} + 1}$

Step 4

Simplify.

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Step 4.1

Combine terms.

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Step 4.1.1

Reorder terms.

$\frac{4}{16 x^{2} + 1} + \frac{32 x}{16 x^{2} + 1}$

Step 4.1.2

Combine the numerators over the common denominator.

$\frac{4 + 32 x}{16 x^{2} + 1}$

Step 4.2

Reorder terms.

$\frac{32 x + 4}{16 x^{2} + 1}$

Step 4.3

Factor $4$ out of $32 x + 4$ .

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Step 4.3.1

Factor $4$ out of $32 x$ .

$\frac{4 (8 x) + 4}{16 x^{2} + 1}$

Step 4.3.2

Factor $4$ out of $4$ .

$\frac{4 (8 x) + 4 (1)}{16 x^{2} + 1}$

Step 4.3.3

Factor $4$ out of $4 (8 x) + 4 (1)$ .

$\frac{4 (8 x + 1)}{16 x^{2} + 1}$