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