Calculus Examples

${(\tan^{-1} (8 x))}^{2}$

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 $g (x) = \tan^{-1} (8 x)$ .

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To apply the Chain Rule, set $u_{1}$ as $\tan^{-1} (8 x)$ .

$\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [\tan^{-1} (8 x)]$

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

$2 u_{1} \frac{d}{d x} [\tan^{-1} (8 x)]$

Replace all occurrences of $u_{1}$ with $\tan^{-1} (8 x)$ .

$2 \tan^{-1} (8 x) \frac{d}{d x} [\tan^{-1} (8 x)]$

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 $g (x) = 8 x$ .

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To apply the Chain Rule, set $u_{2}$ as $8 x$ .

$2 \tan^{-1} (8 x) (\frac{d}{d u_{2}} [\tan^{-1} (u_{2})] \frac{d}{d x} [8 x])$

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

$2 \tan^{-1} (8 x) (\frac{1}{1 + {u_{2}}^{2}} \frac{d}{d x} [8 x])$

Replace all occurrences of $u_{2}$ with $8 x$ .

$2 \tan^{-1} (8 x) (\frac{1}{1 + {(8 x)}^{2}} \frac{d}{d x} [8 x])$

Step 3

Differentiate.

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Factor $8$ out of $8 x$ .

$2 \tan^{-1} (8 x) (\frac{1}{1 + {(8 (x))}^{2}} \frac{d}{d x} [8 x])$

Simplify the expression.

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Apply the product rule to $8 (x)$ .

$2 \tan^{-1} (8 x) (\frac{1}{1 + 8^{2} x^{2}} \frac{d}{d x} [8 x])$

Raise $8$ to the power of $2$ .

$2 \tan^{-1} (8 x) (\frac{1}{1 + 64 x^{2}} \frac{d}{d x} [8 x])$

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$2 \tan^{-1} (8 x) (\frac{1}{1 + 64 x^{2}} (8 \frac{d}{d x} [x]))$

Combine $8$ and $\frac{1}{1 + 64 x^{2}}$ .

$2 \tan^{-1} (8 x) (\frac{8}{1 + 64 x^{2}} \frac{d}{d x} [x])$

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

$2 \tan^{-1} (8 x) (\frac{8}{1 + 64 x^{2}} \cdot 1)$

Simplify the expression.

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Multiply $\frac{8}{1 + 64 x^{2}}$ by $1$ .

$2 \tan^{-1} (8 x) \frac{8}{1 + 64 x^{2}}$

Reorder terms.

$2 \tan^{-1} (8 x) \frac{8}{64 x^{2} + 1}$

Step 4

Combine terms.

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Combine $\frac{8}{64 x^{2} + 1}$ and $2$ .

$\frac{8 \cdot 2}{64 x^{2} + 1} \tan^{-1} (8 x)$

Multiply $8$ by $2$ .

$\frac{16}{64 x^{2} + 1} \tan^{-1} (8 x)$

Combine $\frac{16}{64 x^{2} + 1}$ and $\tan^{-1} (8 x)$ .

$\frac{16 \tan^{-1} (8 x)}{64 x^{2} + 1}$