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Mark the Correct Alternative in of the Following: If Y = 1 + 1 X 2 1 − 1 X 2 Then D Y D X = - Mathematics

MCQ

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

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

  • \[- \frac{4x}{x^2 - 1}\]

  • \[\frac{1 - x^2}{4x}\]

  • \[\frac{4x}{x^2 - 1}\] 

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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).

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Q 5 | Page 48
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