Calculus Examples

${(\tan^{-1} (4 x))}^{2}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \tan^{-1} (4 x)$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $\tan^{-1} (4 x)$ .

$\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [\tan^{-1} (4 x)]$

Step 1.2

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

$2 u_{1} \frac{d}{d x} [\tan^{-1} (4 x)]$

Step 1.3

Replace all occurrences of $u_{1}$ with $\tan^{-1} (4 x)$ .

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

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tan^{-1} (x)$ and $g (x) = 4 x$ .

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Step 2.1

To apply the Chain Rule, set $u_{2}$ as $4 x$ .

$2 \tan^{-1} (4 x) (\frac{d}{d u_{2}} [\tan^{-1} (u_{2})] \frac{d}{d x} [4 x])$

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u_{2}$ with $4 x$ .

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

Step 3

Differentiate.

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Step 3.1

Factor $4$ out of $4 x$ .

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

Step 3.2

Simplify the expression.

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

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

$2 \tan^{-1} (4 x) (\frac{4}{1 + 16 x^{2}} \cdot 1)$

Step 3.6

Simplify the expression.

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Step 3.6.1

Multiply $\frac{4}{1 + 16 x^{2}}$ by $1$ .

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

Step 3.6.2

Reorder terms.

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

Step 4

Combine terms.

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Step 4.1

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

$\frac{4 \cdot 2}{16 x^{2} + 1} \tan^{-1} (4 x)$

Step 4.2

Multiply $4$ by $2$ .

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

Step 4.3

Combine $\frac{8}{16 x^{2} + 1}$ and $\tan^{-1} (4 x)$ .

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