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Differentiate E 2 X + E − 2 X E 2 X − E − 2 X ? - Mathematics

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प्रश्न

Differentiate \[\frac{e^{2x} + e^{- 2x}}{e^{2x} - e^{- 2x}}\] ?

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उत्तर

\[\text{ Let } y = \frac{e^{2x} + e^{- 2x}}{e^{2x} - e^{- 2x}}\]

Differentiate with respect to x we get,

\[\frac{d y}{d x} = \frac{d}{dx}\left[ \frac{e^{2x} + e^{- 2x}}{e^{2x} - e^{- 2x}} \right]\]

\[ = \left[ \frac{\left( e^{2x} - e^{- 2x} \right)\frac{d}{dx}\left( e^{2x} + e^{- 2x} \right) - \left( e^{2x} + e^{- 2x} \right)\frac{d}{dx}\left( e^{2x} - e^{- 2x} \right)}{\left( e^{2x} - e^{- 2x} \right)^2} \right] \left[ \text{ Using quotient rule and chain rule } \right]\]

\[ = \frac{\left( e^{2x} - e^{- 2x} \right)\left[ e^{2x} \frac{d}{dx}\left( 2x \right) + e^{- 2x} \frac{d}{dx}\left( - 2x \right) \right] - \left( e^{2x} + e^{- 2x} \right)\left[ e^{2x} \frac{d}{dx}\left( 2x \right) - e^{- 2x} \frac{d}{dx}\left( - 2x \right) \right]}{\left( e^{2x} - e^{- 2x} \right)^2}\]

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

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

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

\[ = \frac{- 8}{\left( e^{2x} - e^{- 2x} \right)^2}\]

\[So, \frac{d}{dx}\left( \frac{e^{2x} + e^{- 2x}}{e^{2x} - e^{- 2x}} \right) = \frac{- 8}{\left( e^{2x} - e^{- 2x} \right)^2}\]

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पाठ 11: Differentiation - Exercise 11.02 [पृष्ठ ३७]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.02 | Q 31 | पृष्ठ ३७

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