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

MCQ

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

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

✓

Answer

Correct option: A.

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

$\text{y}=\frac{1+\frac{1}{\text{x}^2}}{1-\frac{1}{\text{x}^2}}$
$=\frac{\text{x}^2+1}{\text{x}^2-1}$
Differentiate both the sides with respect to $x,$ we get
$\frac{\text{dy}}{\text{dx}}=\frac{(\text{x}^2-1)\frac{\text{d}}{\text{dx}}(\text{x}^2+1)-(\text{x}^2+1)\frac{\text{d}}{\text{dx}}(\text{x}^2-1)}{(\text{x}^2-1)^2} ($Quotient rule$)$
$=\frac{(\text{x}^2-1)(\text{2x}+0)-(\text{x}^2+1)(\text{2x}-0)}{(\text{x}^2-1)^2}$
$=\frac{\text{2x}^3-\text{2x}-\text{2x}^3-\text{2x}}{(\text{x}^2-1)^2}$
$=\frac{-\text{4x}}{(\text{x}^2-1)^2}$

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