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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
MCQ
If $\text{y}=\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ then $\frac{\text{dy}}{\text{dx}}=$
  • ✓
    $-\frac{2}{1+\text{x}^2}$
  • B
    $\frac{2}{1+\text{x}^2}$
  • C
    $\frac{1}{2-\text{x}^2}$
  • D
    $\frac{2}{2-\text{x}^2}$
✓

Answer

Correct option: A.
$-\frac{2}{1+\text{x}^2}$
Let $\text{y}=\sin^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
Differentiating with respect to $x$ using chain rule, we get,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)^2}}\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{x}^2)}{\sqrt{(1+\text{x}^2)^2-(1-\text{x}^2)^2}}\bigg[\frac{(1+\text{x}^2)\frac{\text{d}}{\text{dx}}(1-\text{x}^2)-(1-\text{x}^2)\frac{\text{d}}{\text{dx}}(1+\text{x}^2)}{(1+\text{x}^2)^2}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{x}^2)}{\sqrt{(1+\text{x}^2)^2-(1-\text{x}^2)^2}}\bigg[\frac{(1+\text{x}^2)(-2\text{x})-(1-\text{x}^2)(2\text{x})}{(1+\text{x}^2)^2}\bigg]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(1+\text{x}^2)}{2\text{x}}\Big[\frac{-2\text{x}-2\text{x}^3-2\text{x}+2\text{x}^3}{(1+\text{x}^2)^2}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-4\text{x}}{2\text{x}(1+\text{x}^2)}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-2}{1+\text{x}^2}$

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