MHT CET · Maths · Differentiation
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
- A \(\frac{3}{2 \sqrt{x^2-4}}\)
- B \(\frac{3}{\sqrt{4-x^2}}\)
- C \(\frac{3}{2 \sqrt{1-x^2}}\)
- D \(\frac{4}{\sqrt{4-x^2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{\sqrt{4-x^2}}\)
Step-by-step Solution
Detailed explanation
\(y=\sin ^{-1}\left(\frac{3 x}{2}-\frac{x^3}{2}\right) \)
\( =\sin ^{-1}\left(\frac{3 x}{2}-4\left(\frac{x}{2}\right)^3\right) \)
\( \text { Put } \frac{x}{2}=\sin \theta \Rightarrow \theta=\sin ^{-1}\left(\frac{x}{2}\right) \)
\( \therefore y =\sin ^{-1}\left(3 \sin \theta-4 \sin ^3 \theta\right) \)
\( =\sin ^{-1}(\sin 3 \theta) \)
\( =3 \theta \)
\( \therefore y =3 \sin ^{-1}\left(\frac{x}{2}\right) \)
\( \therefore \frac{\mathrm{d} y}{\mathrm{~d} x} =3 \cdot \frac{1}{\sqrt{1-\left(\frac{x}{2}\right)^2}} \cdot \frac{1}{2}=\frac{3}{\sqrt{4-x^2}}\)
\( =\sin ^{-1}\left(\frac{3 x}{2}-4\left(\frac{x}{2}\right)^3\right) \)
\( \text { Put } \frac{x}{2}=\sin \theta \Rightarrow \theta=\sin ^{-1}\left(\frac{x}{2}\right) \)
\( \therefore y =\sin ^{-1}\left(3 \sin \theta-4 \sin ^3 \theta\right) \)
\( =\sin ^{-1}(\sin 3 \theta) \)
\( =3 \theta \)
\( \therefore y =3 \sin ^{-1}\left(\frac{x}{2}\right) \)
\( \therefore \frac{\mathrm{d} y}{\mathrm{~d} x} =3 \cdot \frac{1}{\sqrt{1-\left(\frac{x}{2}\right)^2}} \cdot \frac{1}{2}=\frac{3}{\sqrt{4-x^2}}\)
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