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

## (2 cos x) / sin x##

Use the .

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!