MHT CET · Maths · Differentiation
If \(\mathrm{y}=\tan ^{-1}\left[\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}}\right]\), then \(\left(\frac{d y}{d x}\right)=\)
- A \(\frac{-1}{\sqrt{1-x^{2}}}\)
- B \(\frac{-x}{\sqrt{1-x^{2}}}\)
- C \(\frac{1}{\sqrt{1-x^{2}}}\)
- D \(\frac{x}{\sqrt{1-x^{2}}}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{\sqrt{1-x^{2}}}\)
Step-by-step Solution
Detailed explanation
Given \(y=\tan ^{-1}\left[\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}}\right]\)
Put \(x=\cos \theta \Rightarrow \theta=\cos ^{-1} x\)
\(y=\tan ^{-1}\left[\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\right]=\tan ^{-1}\left[\frac{1-\tan \theta}{1+\tan \theta}\right]\)
\(\quad=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\theta\right)\right]=\frac{\pi}{4}-\theta\)
\(\therefore \frac{d y}{d x}=\frac{\pi}{4}-\cos ^{-1} x\)
\(\begin{aligned} y &=\frac{1}{\sqrt{1-x^{2}}} \\ y &=\frac{1}{\sqrt{1-x^{2}}} \end{aligned}\)
Put \(x=\cos \theta \Rightarrow \theta=\cos ^{-1} x\)
\(y=\tan ^{-1}\left[\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\right]=\tan ^{-1}\left[\frac{1-\tan \theta}{1+\tan \theta}\right]\)
\(\quad=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\theta\right)\right]=\frac{\pi}{4}-\theta\)
\(\therefore \frac{d y}{d x}=\frac{\pi}{4}-\cos ^{-1} x\)
\(\begin{aligned} y &=\frac{1}{\sqrt{1-x^{2}}} \\ y &=\frac{1}{\sqrt{1-x^{2}}} \end{aligned}\)
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