Algebra Examples

$4 \tan^{-1} (e^{u} + \sin (v))$

Step 1

Since $4$ is constant with respect to $v$ , the derivative of $4 \tan^{-1} (e^{u} + \sin (v))$ with respect to $v$ is $4 \frac{d}{d v} [\tan^{-1} (e^{u} + \sin (v))]$ .

$4 \frac{d}{d v} [\tan^{-1} (e^{u} + \sin (v))]$

Step 2

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

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $e^{u} + \sin (v)$ .

$4 (\frac{d}{d u_{1}} [\tan^{-1} (u_{1})] \frac{d}{d v} [e^{u} + \sin (v)])$

Step 2.2

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

$4 (\frac{1}{1 + {u_{1}}^{2}} \frac{d}{d v} [e^{u} + \sin (v)])$

Step 2.3

Replace all occurrences of $u_{1}$ with $e^{u} + \sin (v)$ .

$4 (\frac{1}{1 + {(e^{u} + \sin (v))}^{2}} \frac{d}{d v} [e^{u} + \sin (v)])$

Step 3

Differentiate.

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Step 3.1

Combine $\frac{1}{1 + {(e^{u} + \sin (v))}^{2}}$ and $4$ .

$\frac{4}{1 + {(e^{u} + \sin (v))}^{2}} \frac{d}{d v} [e^{u} + \sin (v)]$

Step 3.2

By the Sum Rule, the derivative of $e^{u} + \sin (v)$ with respect to $v$ is $\frac{d}{d v} [e^{u}] + \frac{d}{d v} [\sin (v)]$ .

$\frac{4}{1 + {(e^{u} + \sin (v))}^{2}} (\frac{d}{d v} [e^{u}] + \frac{d}{d v} [\sin (v)])$

Step 3.3

Since $e^{u}$ is constant with respect to $v$ , the derivative of $e^{u}$ with respect to $v$ is $0$ .

$\frac{4}{1 + {(e^{u} + \sin (v))}^{2}} (0 + \frac{d}{d v} [\sin (v)])$

Step 3.4

Add $0$ and $\frac{d}{d v} [\sin (v)]$ .

$\frac{4}{1 + {(e^{u} + \sin (v))}^{2}} \frac{d}{d v} [\sin (v)]$

Step 4

The derivative of $\sin (v)$ with respect to $v$ is $\cos (v)$ .

$\frac{4}{1 + {(e^{u} + \sin (v))}^{2}} \cos (v)$

Step 5

Combine fractions.

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Step 5.1

Combine $\frac{4}{1 + {(e^{u} + \sin (v))}^{2}}$ and $\cos (v)$ .

$\frac{4 \cos (v)}{1 + {(e^{u} + \sin (v))}^{2}}$

Step 5.2

Reorder terms.

$\frac{4 \cos (v)}{{(e^{u} + \sin (v))}^{2} + 1}$