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Question
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).
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