Question

How do you find the derivative of $Y=\frac{e^x-e^{-x}}{e^x+e^{-x}}$?

Answer 1

$\frac{dy}{dx}=\frac{4}{(e^x+e^{-x}{)}^2}$
Explanation:
To find the derivative, we have to use the Quotient Rule, and, the Chain Rule given below for ready reference:-
The Quotient Rule:- $\frac{d}{dx}=\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$
By the Chain Rule, $\frac{d}{dx}=e^{az}=e^{ax}.\frac{d}{dx}\left(ax\right)=a.e^{ax}$
As a particular case of this, we have, $\frac{d}{dx}{e}^{-x}=-e^{-x}$
Hence,
$\frac{dy}{dx}$
$=\frac{(e^x+e^{-x})\frac{d}{dx}(e^x-e^{-x})-(e^x-e^{-x})\frac{d}{dx}(e^x+e^{-x})}{(e^x+e^{-x}{)}^2}$
$=\frac{(e^x+e^{-x})(e^x-(-e^{-x}\left)\right)-(e^x-e^{-x})(e^x-e^{-x})}{(e^x+e^{-x}{)}^2}$
$=\frac{(e^x+e^{-x}{)}^2-(e^x-e^{-x}{)}^2}{(e^x+e^{-x}{)}^2}$
$=\frac{4}{(e^x+e^{-x}{)}^2}$
In fact, if we use hyperbolic funs., then, since $Y=tanhx$, we can directly say that $\frac{dy}{dx}=sec{h}^{2}x=\frac{4}{(e^x+e^{-x}{)}^2}$