If y= 1+1 x^ 2 1-1 x^ 2 then dy dx is equal to: - Vidyadip

MCQ

If $\text{y}=\frac{1+\frac{1}{\text{x}^{2}}}{1-\frac{1}{\text{x}^{2}}}$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:

✓
$\frac{-4\text{x}}{(\text{x}^{2}-1)^{2}}$
B
$\frac{-4\text{x}}{(\text{x}^{2}-1)^{2}}$
C
$\frac{1-\text{x}^{2}}{4\text{x}}$
D
$\frac{4\text{x}}{\text{x}^{2}-1}$

✓

Answer

Correct option: A.

$\frac{-4\text{x}}{(\text{x}^{2}-1)^{2}}$

Given, $\text{y}=\frac{1+\frac{1}{\text{x}^{2}}}{1-\frac{1}{\text{x}^{2}}}$
$\Rightarrow\text{y}=\frac{\text{x}^{2}+1}{\text{x}^{2}-1}$
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{(\text{x}^{2}-1).2\text{x}-(\text{x}^{2}+1).2\text{x}}{(\text{x}^{2}-1)^{2}}$
$=\frac{2\text{x}(\text{x}^{2}-1-\text{x}^{2}-1)}{(\text{x}^{2}-1)^{2}}$
$=\frac{2\text{x}(-2)}{(\text{x}^{2}-1)^{2}}$
$=\frac{-4\text{x}}{(\text{x}^{2}-1)^2{}}$

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