Calculus Examples

\ln (\cos (x))

$\ln (\cos (x))$

Step 1

Differentiate using the chain rule, which states that

\frac{d}{d x} [f (g (x))]

$\frac{d}{d x} [f (g (x))]$ is

f' (g (x)) g' (x)

$f' (g (x)) g' (x)$ where

f (x) = \ln (x)

$f (x) = \ln (x)$ and

g (x) = \cos (x)

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

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Step 1.1

To apply the Chain Rule, set

u

$u$ as

\cos (x)

$\cos (x)$ .

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

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

Step 1.2

The derivative of

\ln (u)

$\ln (u)$ with respect to

u

$u$ is

\frac{1}{u}

$\frac{1}{u}$ .

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

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

Step 1.3

Replace all occurrences of

u

$u$ with

\cos (x)

$\cos (x)$ .

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

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

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

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

Step 2

Convert from

\frac{1}{\cos (x)}

$\frac{1}{\cos (x)}$ to

\sec (x)

$\sec (x)$ .

\sec (x) \frac{d}{d x} [\cos (x)]

$\sec (x) \frac{d}{d x} [\cos (x)]$

Step 3

The derivative of

\cos (x)

$\cos (x)$ with respect to

x

$x$ is

- \sin (x)

$- \sin (x)$ .

\sec (x) (- \sin (x))

$\sec (x) (- \sin (x))$

Step 4

Simplify.

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Step 4.1

Reorder the factors of

\sec (x) (- \sin (x))

$\sec (x) (- \sin (x))$ .

- \sec (x) \sin (x)

$- \sec (x) \sin (x)$

Step 4.2

Rewrite

\sec (x)

$\sec (x)$ in terms of sines and cosines.

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

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

Step 4.3

Combine

\sin (x)

$\sin (x)$ and

\frac{1}{\cos (x)}

$\frac{1}{\cos (x)}$ .

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

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

Step 4.4

Convert from

\frac{\sin (x)}{\cos (x)}

$\frac{\sin (x)}{\cos (x)}$ to

\tan (x)

$\tan (x)$ .

- \tan (x)

$- \tan (x)$

- \tan (x)

$- \tan (x)$