Calculus Examples

$- \ln (\cos (x))$

Step 1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \ln (\cos (x))$ with respect to $x$ is $- \frac{d}{d x} [\ln (\cos (x))]$ .

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

Step 2

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

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 3

Convert from $\frac{1}{\cos (x)}$ to $\sec (x)$ .

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

Step 4

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

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

Step 5

Multiply.

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Step 5.1

Multiply $- 1$ by $- 1$ .

$1 \sec (x) \sin (x)$

Step 5.2

Multiply $\sec (x)$ by $1$ .

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

Step 6

Simplify.

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

Convert from $\frac{\sin (x)}{\cos (x)}$ to $\tan (x)$ .

$\tan (x)$