The derivative of \(\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)\) w.r.t. \(\sin ^{-1}\left(3 x-4 x^3\right)\) is

Let \(y=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)\)
and \(\mathrm{z}=\sin ^{-1}\left(3 x-4 x^3\right)\)
Put \(x=\sin \theta \Rightarrow \theta=\sin ^{-1} x\)
\(\therefore \quad y=\sin ^{-1}\left(2 \sin \theta \sqrt{1-\sin ^2 \theta}\right)\) and
\(\begin{aligned} & \mathrm{z}=\sin ^{-1}\left(3 \sin \theta-4 \sin ^3 \theta\right) \\ & \Rightarrow y=\sin ^{-1}(\sin 2 \theta) \text { and } z=\sin ^{-1}(\sin 3 \theta) \\ & \Rightarrow y=2 \theta=2 \sin ^{-1} x \text { and } z=3 \theta=3 \sin ^{-1} x\end{aligned}\)
\(\begin{aligned} & \therefore \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2}{\sqrt{1-x^2}} \text { and } \frac{\mathrm{d} z}{\mathrm{~d} x}=\frac{3}{\sqrt{1-x^2}} \\ & \therefore \frac{\mathrm{~d} y}{\mathrm{~d} z}=\frac{\frac{\mathrm{d} y}{\mathrm{~d} x}}{\frac{\mathrm{dz}}{\mathrm{d} x}}=\frac{2}{3}\end{aligned}\)