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