<s>[INST] Differentiate the Function: $f(x)=\ln (\sin^2x)$ $$\begin{align}f(x)&=\ln (\sin^2x)\\
f'(x)&=\frac{1}{\sin^2x}\cdot 2(\sin x)(\cos x)\\
&=\frac{2(\sin x)(\cos x)}{\sin^2x}\\
&=\frac{2\ \ (\cos x)\ }{\sin x}\\
&=2\cot x
\end{align}$$
= $2\ cot\ x$
Is this answer right?
 [/INST] Not quite; you have misapplied the Chain Rule. If $f(x) = g(h(x))$, where here $g(x) = \ln x$ and $h(x) = \sin^2 x$, then $f'(x) = g'(h(x))h'(x)$, so we get
$$
  \frac{1}{\sin^2 x}\cdot\frac{d}{dx}(\sin^2 x) = \frac{2\sin x\cos x}{\sin^2 x}
     = \frac{2\cos x}{\sin x} = 2\cot x.
$$
</s>