If y= ( 1-x^2 1+x^2 ), then dy dx = - Vidyadip

MCQ

If $\text{y}=\log\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big),$ then $\frac{\text{dy}}{\text{dx}}=$

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, $\text{y}=\log\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\frac{1-\text{x}^2}{1+\text{x}^2}}\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1+\text{x}^2}{1-\text{x}^2}\Big[\frac{(1+\text{x}^2)(-2\text{x})-(1-\text{x}^2)(2\text{x})}{(1+\text{x}^2)^2}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1-\text{x}^2}\Big[\frac{-2\text{x}-2\text{x}^3-2\text{x}+2\text{x}^3}{(1+\text{x}^2)}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{-4\text{x}}{1-\text{x}^4}$

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