Question

I need help with differentiating the function:

f(x)=ln(sin^2x)

Answer 1

\[f(x)=\ln (\sin ^2x)\]

Answer 2

Start off by letting y = f(x) and rearranging to
\[y= 2\ln{\sin(x)}\]
Now, let \[u = \sin{x}\]
\[\frac{du}{dx} = \cos{x}\]
And y can now be written as:
\[y = 2\ln{u}\]
\[\frac{dy}{du} = \frac{2}{u}\]
And by the chain rule:
\[\frac{dy}{dx} = \frac{dy}{du}*\frac{du}{dx}\]
So
\[\frac{dy}{dx} = \frac{2}{u} * \cos{x}\]
Replacing u with sin(x):
\[\frac{dy}{dx} = \frac{2}{\sin{x}} * \cos{x}\]
And tidying up:
\[\frac{dy}{dx} = 2\cot{x}\]

Answer 3

Thank you! I couldn't figure this prob out all the way :)