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 \(\frac{1}{2}\)
- B \(\frac{1}{\sqrt{2}}\)
- C \(\sqrt{2}\)
- D 1
Answer & Solution
Correct Answer
(B) \(\frac{1}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \begin{aligned} & y=\sec \left(\tan ^{-1} x\right) \\ & \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\sec \left(\tan ^{-1} x\right) \tan \left(\tan ^{-1} x\right) \cdot \frac{1}{1+x^2} \\ &=\sqrt{1+x^2} \cdot \frac{x}{1+x^2} \\ & \ldots\left[\because \tan ^{-1} x=\sec ^{-1} \sqrt{1+x^2}\right] \\ &=\frac{x}{\sqrt{1+x^2}} \\ & \therefore \quad\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)_{x=1}=\frac{1}{\sqrt{1+1^2}}=\frac{1}{\sqrt{2}}\end{aligned} .\end{aligned}\)
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