Question

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

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

Answer 1

Answer: (A)

We have, $y=\frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}}$
$\Rightarrow y=\frac{x^2+1}{x^2-1}\dots \left(i\right)$
Differentiating $\left(i\right)$ w.r.t. $x$, we get
$\frac{dy}{dx}=\frac{\left(x^2-1\right)\frac{d}{dx}\left(x^2+1\right)-\left(x^2+1\right)\frac{d}{dx}\left(x^2-1\right)}{{\left(x^2-1\right)}^2}$
$=\frac{\left(x^2-1\right)\left(2x\right)-\left(x^2+1\right)\left(2x\right)}{{\left(x^2-1\right)}^2}$
$=\frac{2x^3-2x-2x^3-2x}{{}^{{\left(x^2-1\right)}^2}}$
$=\frac{-4x}{{\left(x^2-1\right)}^2}$