Question

How do you find the derivative of `Y= (e^x-e^-x)/(e^x+e^-x)`?

Answer 1

`(dY)/dx=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 :- `d/dx(u/v)={v(du)/dx-u(dv)/dx}/v^2`.

By the Chain Rule, `d/dxe^(ax)=e^(ax)*d/dx(ax)=a*e^(ax)`.

As a particular case of this, we have, `d/dxe^-x=-e^-x`.

Hence,

`(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`

`=4/(e^x+e^-x)^2`.

In fact, if we use hyperbolic funs., then, since `Y=tanhx`, we can directly say that `(dY)/dx=sech^2x=4/(e^x+e^-x)^2`.