How do you differentiate #y= (2e^x) / (1+e^x) #?

1 Answer
Jun 10, 2018

Answer:

#dy/dx=(2e^x)/(1+e^x)^2#

Explanation:

First of all, we see that the function is written in the format of #y=u/v#. This is equivalent to the quotient rule where:

#y=u/v#, #dy/dx=(((du)/dx)v-((dv)/dx)u)/v^2#

Now, separate the numerator and the denominator to differentiate. The easiest thing to do when differentiating is to take the constant term out (#2#) as shown below.

#dy/dx=(2e^x)/(1+e^x)#
#2(dy/dx)(e^x)/(1+e^x)#

#u=e^x#
#(du)/dx=e^x#

#v=1+e^x#
#(dv)/dx=e^x# (as the derivative of #1# is #0#)

Apply quotient rule:

#dy/dx=(((du)/dx)v-((dv)/dx)u)/v^2#
#dy/dx= ((e^x)(1+e^x)-(e^x)(e^x))/(1+e^x)^2#
#dy/dx=(e^x+e^2x-e^2x)/(1+e^x)^2#
#dy/dx=e^x/(1+e^x)^2#

Don't forget to multiply by the constant!

#dy/dx=2(e^x/(1+e^x)^2)#
#dy/dx=(2e^x)/(1+e^x)^2#