Differentiate `(e^x+e^(-x))/(e^x-e^(-x))` with respect to `x` .

Explanation

Let `y=(e^x+e^(-x))/(e^x-e^(-x))`

By quotient rule,

`(d)/(dx)((u)/(v))=(v(du)/(dx)-u(dv)/(dx))/(v^(2))`


We know that, `y=e^(ax)`

`y'=ay=ae^(ax)`


`frac{dy}{dx}=((e^(x)-e^(-x))(d)/(dx)(e^(x)+e^(-x))-(e^(x)+e^(-x))(d)/(dx)(e^(x)-e^(-x)))/((e^(x)-e^(-x))^(2))`

`=((e^(x)-e^(-x))(e^(x)-e^(-x))-(e^(x)+e^(-x))(e^(x)+e^(-x)))/((e^(x)+e^(-x))^(2))`

`=((e^(x)-e^(-x))^2-(e^(x)+e^(-x))^2)/((e^(x)+e^(-x))^(2))`

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