Calculus Examples

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

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 $g (x) = \sin (x) + \cos (x)$ .

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

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

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

Step 2

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

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

Step 3

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

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

Step 4

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

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

Step 5

Simplify.

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Step 5.1

Reorder the factors of $\frac{1}{\sin (x) + \cos (x)} (\cos (x) - \sin (x))$ .

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

Step 5.2

Apply the distributive property.

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

Combine the numerators over the common denominator.

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