Question

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

• A. $-\frac{4x}{(x^2-1{)}^2}$
• B. $-\frac{4x}{x^2-1}$
• C. $\frac{1-x^2}{4x}$
• D. $\frac{4x}{x^2-1}$

Answer 1

The correct option is A $-\frac{4x}{(x^2-1{)}^2}$

$y=\frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}}$
$=\frac{x^2+1}{x^2-1}$
Differentiating both sides with respect to x, we get
$\frac{dy}{dx}=\frac{(x^2-1)\times \frac{d}{dx}(x^2+1)-(x^2+1)\times \frac{d}{dx}(x^2-1)}{(x^1-1{)}^2}$
$=\frac{(x^2-1)\times (2x+0)-(x^2+1)\times (2x-0)}{(x^2-1{)}^2}$
$=\frac{2x^2-2x-2x^3-2x}{(x^2-1{)}^2}$
$=\frac{-4x}{(x^2-1{)}^2}$
Hence, the correct answer is option (a).