Calculus Examples

$f (x) = \ln (1 + \sin (x))$

Step 1

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

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Step 1.1

To apply the Chain Rule, set $u$ as $1 + \sin (x)$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [1 + \sin (x)]$

Step 1.2

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

$\frac{1}{u} \frac{d}{d x} [1 + \sin (x)]$

Step 1.3

Replace all occurrences of $u$ with $1 + \sin (x)$ .

$\frac{1}{1 + \sin (x)} \frac{d}{d x} [1 + \sin (x)]$

Step 2

Differentiate.

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Step 2.1

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

$\frac{1}{1 + \sin (x)} (\frac{d}{d x} [1] + \frac{d}{d x} [\sin (x)])$

Step 2.2

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

$\frac{1}{1 + \sin (x)} (0 + \frac{d}{d x} [\sin (x)])$

Step 2.3

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

$\frac{1}{1 + \sin (x)} \frac{d}{d x} [\sin (x)]$

Step 3

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

$\frac{1}{1 + \sin (x)} \cos (x)$

Step 4

Combine $\frac{1}{1 + \sin (x)}$ and $\cos (x)$ .

$\frac{\cos (x)}{1 + \sin (x)}$