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