How do you differentiate #f(x)=8e^(x^2)/(e^x+1)# using the chain rule?

Question

2326 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-13T18:25:10

The only trick here is that #(e^(x^2))'=e^(x^2)*(x^2)'=e^(x^2)*2x#
Final derivative is:

#f'(x)=8e^(x^2)(2x*(e^x+1)-e^x)/(e^x+1)^2#
or
#f'(x)=8e^(x^2)(e^x*(2x-1)+2x+1)/(e^x+1)^2#

#f(x)=8(e^(x^2))/(e^x+1)#

#f'(x)=8((e^(x^2))'(e^x+1)-e^(x^2)(e^x+1)')/(e^x+1)^2#

#f'(x)=8(e^(x^2)*(x^2)'(e^x+1)-e^(x^2)*e^x)/(e^x+1)^2#

#f'(x)=8(e^(x^2)2x*(e^x+1)-e^(x^2)*e^x)/(e^x+1)^2#

#f'(x)=8(e^(x^2)(2x*(e^x+1)-e^x))/(e^x+1)^2#

#f'(x)=8e^(x^2)(2x*(e^x+1)-e^x)/(e^x+1)^2#

or (if you want to factor #e^x# in the nominator)

#f'(x)=8e^(x^2)(e^x*(2x-1)+2x+1)/(e^x+1)^2#

Note: if you want to study the sign, you are gonna have a bad time. Just look at the graph:

graph{8(e^(x^2))/(e^x+1) [-50.25, 53.75, -2.3, 49.76]}