Find the Derivative - d/dt natural log of cos(t)

$\ln (\cos (t))$

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \ln (t)$ and $g (t) = \cos (t)$ .

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

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

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

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

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

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

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

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

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

$\sec (t) (- \sin (t))$

Simplify.

Reorder the factors of $\sec (t) (- \sin (t))$ .

$- \sec (t) \sin (t)$

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

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

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

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

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

$- \tan (t)$

Find the Derivative - d/dt natural log of cos(t)

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