Question

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

Options

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

• ${\displaystyle \frac{-4x}{x^2-1}}$

• ${\displaystyle \frac{1-x^2}{4x}}$

• ${\displaystyle \frac{4x}{x^2-1}}$

Answer 1

If ${\displaystyle y=\frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}}}$ then ${\displaystyle \frac{dy}{dx}}$ is ${\displaystyle \frac{-4x}{{(x^2-1)}^2}}$.

Explanation:

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

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

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

= ${\displaystyle \frac{2x(x^2-1-x^2-1)}{{(x62-1)}^2}}$

= ${\displaystyle \frac{2x(-2)}{{(x^2-1)}^2}}$

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