# How do you find the derivative of y=ln(cos(x)) ?

Aug 22, 2014

You can find this derivative by applying the Chain Rule, with $\cos x$ as the inner function, and $\ln x$ as the outer function.

Process:

To apply the chain rule, we first find the derivative of the outer function, $\ln u$, with $u = \cos x$. Remember that the derivative of $\ln u = \frac{1}{u} = \frac{1}{\cos} x$.

Now we just need to find the derivative of the inner function, $\cos x$, and multiply it by the derivative of the outer function we just found.

Since the derivative of $\cos x$ is ($- \sin x$), we end up with:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{\cos} x\right) \cdot \left(- \sin x\right) = \left(- \sin \frac{x}{\cos} x\right) = - \tan x$.

A shorter way to do these is to just know that the derivative of a $\ln \left(u\right)$-type function is the derivative of the inside over the original of what's inside.

Jun 4, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \tan x$

#### Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

• d/dx(ln(f(x)))=(f'(x))/(f(x))

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \sin x}{\cos x} = - \tan x$