Q91 / 180 · KCET 2022 · Mathematics

If $y=\left(1+x^2\right) \tan ^{-1} x-x$, then $\frac{d y}{d x}$ is

  • Correct answer
    A
    $2 x \tan ^{-1} x$
  • B
    $\frac{\tan ^{-1} x}{x}$
  • C
    $x^2 \tan ^{-1} x$
  • D
    $x \tan ^{-1} x$

Solution

Given, $$ y=\left(1+x^2\right) \tan ^{-1} x-x $$ Differentiating the given function w.r.t. $x$ $$ \begin{aligned} \frac{d y}{d x} & =\left(1+x^2\right) \frac{d}{d x}\left(\tan ^{-1} x\right)+\tan ^{-1} x \frac{d}{d x}\left(1+x^2\right)-\frac{d}{d x}(x) \\ & =\left(1+x^2\right) \frac{1}{\left(1+x^2\right)}+\tan ^{-1} x(2 x)-1 \\ & =1+2 x \tan ^{-1} x-1=2 x \tan ^{-1} x \end{aligned} $$

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

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