Q108 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Mathematics

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

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

Solution

$y=\tan \left(\cos ^{-1} x\right)$ $\begin{aligned} \Rightarrow \quad \frac{d y}{d x} & =\sec ^2\left(\cos ^{-1} x\right) \cdot\left(\frac{-1}{\sqrt{1-x^2}}\right) \\ & =\sec ^2\left(\sec ^{-1} \frac{1}{x}\right)\left(\frac{-1}{\sqrt{1-x^2}}\right) \\ & =\left(\frac{1}{x}\right)^2 \times\left(\frac{-1}{\sqrt{1-x^2}}\right)=\frac{-1}{x^2 \sqrt{1-x^2}}\end{aligned}$

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

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