MHT CET · Maths · Differentiation
If \(y=\sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right)\), then \(\frac{d y}{d x}\) is equal to
- A \(\frac{-1}{\sqrt{1-x^2}}\)
- B \(\frac{-x}{\sqrt{1-x^2}}\)
- C \(\frac{1}{\sqrt{1-x^2}}\)
- D \(\frac{-2 x}{\sqrt{1-x^2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{-x}{\sqrt{1-x^2}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & y=\sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right) \\ & \Rightarrow y=\sin \left(2 \tan ^{-1}\left(\cot \frac{\theta}{2}\right)\right)[\text { let } x=\cos \theta] \\ & \Rightarrow y=\sin \left(2 \tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right\}\right) \\ & \Rightarrow y=\sin \left(2\left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right) \\ & \Rightarrow y=\sin (\pi-\theta)=\sin \theta=\sqrt{1-x^2}\end{aligned}\)
\(\Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{1-x^2}} \times(-2 x)=\frac{-x}{\sqrt{1-x^2}}\)
\(\Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{1-x^2}} \times(-2 x)=\frac{-x}{\sqrt{1-x^2}}\)
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