Derivative of ${\sin }^2\mathrm{x} \mathrm{w}.\mathrm{r}.\mathrm{t} {\left(\mathrm{logx}\right)}^2 \mathrm{is}$

Let $y={\sin }^{2}x$ and $z={\left(\log x\right)}^2$

Differentiate both functions with respect to $x$

     $\frac{dy}{dx}=2\sin x\cos x$

     $\frac{dz}{dx}=2\log x\left(\frac{1}{x}\right)$

The derivative of $y$ with respect to $z$ is 

   $\begin{array}{rcl}\frac{dy}{dz}& =& \frac{dy}{dx}·\frac{dx}{dz}\\ & =& 2\sin x\cos x·\frac{1}{2\log x\left({\displaystyle \frac{1}{x}}\right)}\\ & =& \frac{x\sin x\cos x}{\log x}\end{array}$ 

Therefore, the derivative of ${\sin }^2\mathrm{x}$ with respect to ${\left(\mathrm{logx}\right)}^2 \mathrm{is}$ $\boxed{\frac{x\sin x\cos x}{\log x}}$