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

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
  1. $\frac{2}{3}$
  2. $\frac{1}{2}$
  3. $\frac{3}{2}$
  4. 1

Solution

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

Asked in: MHT CET 2024 (15 May Shift 2)

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