Question

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

Answer 1

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`

Explanation:

`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]}