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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NDA 2017 · Mathematics · Inverse Trigonometric Functions

If y= ^ -1 ((2 x)/(1+x²)), then (d y)/(d x)is equal to

If y=cos⁡−1y=\cos ^{-1}\left(\frac{2 x}{1+x²}\right),thenthen\frac{d y}{d x}is equal to

Options

  1. A−21+x2-\frac{2}{1+\mathrm{x}^2} for all ∣x∣<1|\mathrm{x}| < 1
  2. B−21+x2-\frac{2}{1+\mathrm{x}^2} for all ∣x∣>1|\mathrm{x}| > 1
  3. C21+x2\frac{2}{1+\mathrm{x}^2} for all ∣x∣<1|\mathrm{x}| < 1
  4. DNone of the above

Correct answer

A. −21+x2-\frac{2}{1+\mathrm{x}^2} for all ∣x∣<1|\mathrm{x}| < 1

Step-by-step solution

y=\cos ^{-1} \left(\frac{2 x}{1+x^2}\right) Put \mathrm{x}=\tan \theta . \Rightarrow \theta=\tan ^{-1} \mathrm{x}y=cos⁡−1(2tan⁡θ(1+tan⁡2θ))=cos⁡−1(sin⁡y=\cos ^{-1}\left(\frac{2 \tan \theta}{\left(1+\tan ^2 \theta\right)}\right)=\cos ^{-1}(\sin2 θ)\theta)=cos⁡−1(cos⁡(π2−2θ))=π2−2θ∴dydx=ddx(π2−2θ)\begin{aligned}& =\cos ^{-1}\left(\cos \left(\frac{\pi}{2}-2 \theta\right)\right)\\ & =\frac{\pi}{2}-2 \theta\\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{2}-2 \theta\right)\end{aligned}$\begin{aligned}& =\frac{d}{d x}\left(\frac{\pi}{2}-2 \tan ^{-1} x\right) \\ & \frac{d y}{d x}=\frac{-2}{1+x^2}, \text{ when }|x| \lt 1\end{aligned}
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