Q91 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Mathematics

$\frac{d}{d x}\left\{\left(1+x^2\right) \tan ^{-1}(x)\right\}=$

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

Solution

We have, $$ \begin{gathered} \frac{d}{d x}(u \cdot v)=u \cdot \frac{d v}{d x}+v \cdot \frac{d v}{d x} \\ \frac{d}{d x}\left[(1+x)^2 \cdot \tan ^{-1}(x)\right]=\left(1+x^2\right) \\ \frac{d}{d x} \tan ^{-1} x+\frac{d}{d x}\left(1+x^2\right) \cdot \tan ^{-1} x \\ =\left(1+x^2\right) \cdot \frac{1}{1+x^2}+(2 x) \cdot \tan ^{-1} x \\ =1+2 x \tan ^{-1} x=2 x \tan ^{-1} x+1 \end{gathered} $$ Hence, option (3) is correct.

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

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