If $y=\sin ^{-1}\left(\frac{3 x}{2}-\frac{x^3}{2}\right)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ is…
If $y=\sin ^{-1}\left(\frac{3 x}{2}-\frac{x^3}{2}\right)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ is equal to
- $\frac{3}{2 \sqrt{x^2-4}}$
- $\frac{3}{\sqrt{4-x^2}}$
- $\frac{3}{2 \sqrt{1-x^2}}$
- $\frac{4}{\sqrt{4-x^2}}$
Solution
$\begin{aligned} & y=\sin ^{-1}\left(\frac{3 x}{2}-\frac{x^3}{2}\right) \\ & =\sin ^{-1}\left(\frac{3 x}{2}-4\left(\frac{x}{2}\right)^3\right) \\ & \text { Put } \frac{x}{2}=\sin \theta \Rightarrow \theta=\sin ^{-1}\left(\frac{x}{2}\right) \\ \therefore \quad y & =\sin ^{-1}\left(3 \sin \theta-4 \sin ^3 \theta\right) \\ & =\sin ^{-1}(\sin 3 \theta) \\ & =3 \theta \\ \therefore \quad y & =3 \sin ^{-1}\left(\frac{x}{2}\right) \\ \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x} & =3 \cdot \frac{1}{\sqrt{1-\left(\frac{x}{2}\right)^2}} \cdot \frac{1}{2}=\frac{3}{\sqrt{4-x^2}}\end{aligned}$
Asked in: MHT CET 2024 (09 May Shift 2)
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