Question

Mark the correct alternative in each of the following:

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

(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

$$\begin{gathered} y=\frac{1+{\displaystyle \frac{1}{x^2}}}{1-{\displaystyle \frac{1}{x^2}}} \\ =\frac{x^2+1}{x^2-1} \end{gathered}$$

Differentiating both sides with respect to x, we get

$$\begin{gathered} \frac{dy}{dx}=\frac{\left(x^2-1\right)\times {\displaystyle \frac{d}{dx}}\left(x^2+1\right)-\left(x^2+1\right)\times {\displaystyle \frac{d}{dx}}\left(x^2-1\right)}{{\left(x^2-1\right)}^2}\left(\mathrm{Quotient}\mathrm{rule}\right) \\ =\frac{\left(x^2-1\right)\times \left(2x+0\right)-\left(x^2+1\right)\times \left(2x-0\right)}{{\left(x^2-1\right)}^2} \\ =\frac{2x^3-2x-2x^3-2x}{{\left(x^2-1\right)}^2} \\ =\frac{-4x}{{\left(x^2-1\right)}^2} \end{gathered}$$

Hence, the correct answer is option (a).