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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
MCQ
Choose the correct answers from the given four options:
If $\text{y}=\log\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:
  • A
    $\frac{4\text{x}^3}{1-\text{x}^4}$
  • ✓
    $\frac{-4\text{x}}{1-\text{x}^4}$
  • C
    $\frac{1}{4-\text{x}^4}$
  • D
    $\frac{-4\text{x}^3}{1-\text{x}^4}$
✓

Answer

Correct option: B.
$\frac{-4\text{x}}{1-\text{x}^4}$
We have, $\text{y}=\log\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\frac{1-\text{x}^2}{1+\text{x}^2}}\cdot\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$

$=\frac{(1+\text{x}^2)}{(1-\text{x}^2)}\cdot\frac{(1+\text{x}^2)\cdot(-2\text{x})-(1-\text{x}^2)\cdot2\text{x}}{(1+\text{x}^2)^2}$

$=\frac{-2\text{x}[1+\text{x}^2+1-\text{x}^2]}{(1-\text{x}^2)\cdot(1+\text{x}^2)}=\frac{-4\text{x}}{1-\text{x}^4}$

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