Question

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

Answer 1

You can find this derivative by applying the Chain Rule, with `cosx` as the inner function, and `lnx` as the outer function.

Process:

To apply the chain rule, we first find the derivative of the outer function, `lnu`, with `u = cosx`. Remember that the derivative of `lnu = 1/u = 1/cosx`.

Now we just need to find the derivative of the inner function, `cosx`, and multiply it by the derivative of the outer function we just found.

Since the derivative of `cosx` is (`-sinx`), we end up with:

`dy/dx = (1/cosx) * (-sinx) = (-sinx/cosx) = -tanx`.

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

Answer 2

`dy/dx=-tanx`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`• d/dx(ln(f(x)))=(f'(x))/(f(x))`

`rArrdy/dx=(-sinx)/(cosx)=-tanx`