What will be the derivative of $sin^2x$ with respect to $(\log x)^2$
$\dfrac{x \sin 2x}{\log x}$
$\dfrac{x \sin 2x}{2 \log x}$
$\dfrac{\sin 2x}{2 \log x}$
Solution
Let $u = \sin^2 x, v = (\log x)^2$
$\dfrac{dy}{dx} = 2 \sin x \cos x, \dfrac{dy}{dx} = \dfrac{2 log x}{x}$
$\dfrac{du}{dv} = \dfrac{\frac{du}{dx}}{\dfrac{dv}{dx}} = \dfrac{2 sin x cos x}{2 \frac{log x}{x}}$
$\dfrac{du}{dv} \Rightarrow \dfrac{x \sin x}{2 \log x}$