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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 =

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प्रश्न

Mark the correct alternative in  of the following: 

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

विकल्प

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

  • \[- \frac{4x}{x^2 - 1}\]

  • \[\frac{1 - x^2}{4x}\]

  • \[\frac{4x}{x^2 - 1}\] 

MCQ
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उत्तर

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

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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अध्याय 30: Derivatives - Exercise 30.7 [पृष्ठ ४८]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 30 Derivatives
Exercise 30.7 | Q 5 | पृष्ठ ४८

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