Mark the correct alternative in of the following:

If \[y = \frac{1 + \frac{1}{x^2}}{1 - \frac{1}{x^2}}\] then \[\frac{dy}{dx} =\]

#### Options

\[- \frac{4x}{\left( x^2 - 1 \right)^2}\]

\[- \frac{4x}{x^2 - 1}\]

\[\frac{1 - x^2}{4x}\]

\[\frac{4x}{x^2 - 1}\]

#### Solution

\[y = \frac{1 + \frac{1}{x^2}}{1 - \frac{1}{x^2}}\]

\[ = \frac{x^2 + 1}{x^2 - 1}\]

Differentiating both sides with respect to *x*, we get

\[\frac{dy}{dx} = \frac{\left( x^2 - 1 \right) \times \frac{d}{dx}\left( x^2 + 1 \right) - \left( x^2 + 1 \right) \times \frac{d}{dx}\left( x^2 - 1 \right)}{\left( x^2 - 1 \right)^2} \left( \text{ Quotient rule } \right)\]

\[ = \frac{\left( x^2 - 1 \right) \times \left( 2x + 0 \right) - \left( x^2 + 1 \right) \times \left( 2x - 0 \right)}{\left( x^2 - 1 \right)^2}\]

\[ = \frac{2 x^3 - 2x - 2 x^3 - 2x}{\left( x^2 - 1 \right)^2}\]

\[ = \frac{- 4x}{\left( x^2 - 1 \right)^2}\]

Hence, the correct answer is option (a).