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

1 Answer
Nov 10, 2015

# (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!