Question

How do you find the derivative of `ln(sin^2x)`?

Answer 1

`(2 cos x) / sin x`

Explanation:

Use the chain rule.

You can break down your function into the logarithm, the square and the sinus function like follows:

`f(u) = ln(u)`
`u(v) = v^2(x)`
`v(x) = sin(x)`

Now, you need to compute the three derivatives of those three functions (and afterwards plug the respective values of `u` and `v`):

`f'(u) = 1/ u = 1 / v^2 = 1 / sin^2 x`
`u'(v) = 2v = 2 sin x`
`v'(x) = cos(x)`

Now, the only thing left to do is multiplying those three derivatives:

`f'(x) = f'(u) * u'(v) * v'(x) = 1 / sin^2 x * 2 sin x * cos x`
`= (2 cos x) / sin x`

Hope that this helped!