Question

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

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

Answer 1

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

Given: $y=\frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}}$

To find: $\frac{dy}{dx}$

$\Rightarrow y=\frac{x^2+1}{x^2-1}$

$\Rightarrow \frac{dy}{dx}=\frac{2x(x^2-1)-2x(x^2+1)}{(x^2-1{)}^2}$

$(\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{\frac{vdu}{dx}-\frac{udv}{dx}}{v^2})$

$\Rightarrow \frac{dy}{dx}=\frac{-4x}{(x^2-1{)}^2}$

Hence, the correct option is (a)