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

1 Answer
Jul 31, 2016

(dY)/dx=4/(e^x+e^-x)^2dYdx=4(ex+ex)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^2ddx(uv)=vdudxudvdxv2.

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

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

Hence,

(dY)/dxdYdx

={(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=(ex+ex)ddx(exex)(exex)ddx(ex+ex)(ex+ex)2

={(e^x+e^-x)(e^x-(-e^-x))-(e^x-e^-x)(e^x-e^-x)}/(e^x+e^-x)^2=(ex+ex)(ex(ex))(exex)(exex)(ex+ex)2

={(e^x+e^-x)^2-(e^x-e^-x)^2}/(e^x+e^-x)^2=(ex+ex)2(exex)2(ex+ex)2

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

In fact, if we use hyperbolic funs., then, since Y=tanhxY=tanhx, we can directly say that (dY)/dx=sech^2x=4/(e^x+e^-x)^2dYdx=sech2x=4(ex+ex)2.