If y = frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x | vedclass

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(C) Given $y = \frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x}}$.Using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$,where $u = e^{

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$\frac{-8}{(e^{2x} - e^{-2x})^2}$

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(C) Given $y = \frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x}}$.Using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$,where $u = e^{2x} + e^{-2x}$ and $v = e^{2x} - e^{-2x}$.Then $u' = 2e^{2x} - 2e^{-2x} = 2(e^{2x} - e^{-2x})$ and $v' = 2e^{2x} + 2e^{-2x} = 2(e^{2x} + e^{-2x})$.$\frac{dy}{dx} = \frac{(e^{2x} - e^{-2x}) \cdot 2(e^{2x} - e^{-2x}) - (e^{2x} + e^{-2x}) \cdot 2(e^{2x} + e^{-2x})}{(e^{2x} - e^{-2x})^2}$$= \frac{2(e^{2x} - e^{-2x})^2 - 2(e^{2x} + e^{-2x})^2}{(e^{2x} - e^{-2x})^2}$$= \frac{2[(e^{4x} - 2 + e^{-4x}) - (e^{4x} + 2 + e^{-4x})]}{(e^{2x} - e^{-2x})^2}$$= \frac{2[-4]}{(e^{2x} - e^{-2x})^2} = \frac{-8}{(e^{2x} - e^{-2x})^2}$.

If $y = \frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x}}$,then $\frac{dy}{dx} = $

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