MHT CET · Maths · Differentiation
If \(y=\sec \left(\tan ^{-1} x\right)\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) at \(x=1\) is equal to
- A \(\frac{-1}{\sqrt{2}}\)
- B \(\frac{1}{2}\)
- C \(\frac{1}{\sqrt{2}}\)
- D \(\sqrt{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & y=\sec \left(\tan ^{-1} x\right) \\ \therefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=\sec \left(\tan ^{-1} x\right) \tan \left(\tan ^{-1} x\right) \cdot \frac{1}{1+x^2} \\ = & \sqrt{1+x^2} \cdot \frac{x}{1+x^2} \quad \ldots\left[\because \tan ^{-1} x=\sec ^{-1} \sqrt{1+x^2}\right] \\ = & \frac{x}{\sqrt{1+x^2}} \\ \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x_{\mathrm{at} x=1}}=\frac{1}{\sqrt{2}}\end{aligned}\)
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