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If Y = E X − E − X E X + E − X .Prove that D Y D X = 1 − Y 2 ? - Mathematics

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प्रश्न

If \[y = \frac{e^x - e^{- x}}{e^x + e^{- x}}\] .prove that \[\frac{dy}{dx} = 1 - y^2\] ?

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उत्तर

\[\text{We have, y }= \frac{e^x - e^{- x}}{e^x + e^{- x}}\]

Differentiating with respect to x, 

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

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

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

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

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

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

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

\[ = 1 - y^2 \]

\[So, \frac{d y}{d x} = 1 - y^2 \]

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पाठ 11: Differentiation - Exercise 11.02 [पृष्ठ ३८]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.02 | Q 65 | पृष्ठ ३८

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