Q91 / 200 · TS EAMCET 2001 · Mathematics

$\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^3\right)$ is equal to

  • A
    $\frac{3}{\sqrt{4-x^2}}$
  • Correct answer
    B
    $\frac{3}{\sqrt{1-x^2}}$
  • C
    $\frac{1}{\sqrt{4-x^2}}$
  • D
    $-\frac{1}{\sqrt{4-x^2}}$

Solution

$\begin{aligned} & \frac{d}{d x}\left\{\sin ^{-1}\left(3 x-4 x^3\right)\right\} \\ & \text { Put } x=\sin \theta \\ & \Rightarrow \quad \theta=\sin ^{-1} x \\ & =\frac{d}{d x}\left\{\sin ^{-1}\left(3 \sin \theta-4 \sin ^3 \theta\right)\right\} \\ & =\frac{d}{d x}\left\{\sin ^{-1} \sin 3 \theta\right\} \\ & =\frac{d}{d x}\{3 \theta\}=3 \frac{d}{d x} \sin ^{-1} x=\frac{3}{\sqrt{1-x^2}} \\ & \end{aligned}$

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

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