$y=\log _{\sin ^{-1} x} \cos ^{-1} x$, then $\frac{d y}{d x}$ is

none of these

$y=\log _{\sin ^{-1} x} \cos ^{-1} x,=\frac{\log \cos ^{-1} x}{\log \sin ^{-1} x}$

$\frac{d y}{d x}=\frac{1}{\log \sin ^{-1} x} \times \frac{1}{\cos ^{-1} x} \times \frac{-1}{\sqrt{1-x^2}}+\log \left(\cos ^{-1} x\right) \times-1 \times \frac{1}{\left(\log \sin ^{-1} x\right)^2} \times \frac{1}{\sin ^{-1} x} \times \frac{1}{\sqrt{1-x^2}}$

hence none of these

Hence (4) is correct answer.