Q221 / 270 · NDA 2019 (Phase 1) · Mathematics

What is the derivative of $\sec ^2\left(\tan ^{-1} x\right)$ with respect to $x$ ?

  • Correct answer
    A
    2 x
  • B
    $x^2+1$
  • C
    $\mathrm{x}+1$
  • D
    $x^2$

Solution

Let $\mathrm{y}=\sec ^2\left(\tan ^{-1} \mathrm{x}\right)$ $\begin{aligned} & \text { Let } \tan ^{-1} x=\theta \Rightarrow x=\tan \theta \\ & y=\sec ^2 \theta=1+\tan ^2 \theta=1+x^2 \\ & \therefore \frac{d y}{d x}=\frac{d}{d x}\left(1+x^2\right)=2 x\end{aligned}$

Hence, the correct answer is option: 'A'.

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