Question

How do you differentiate `f(x)=xe^((lnx-2)^2` using the chain rule.?

Answer 1

`f'(x)=2(lnx-2)e^((lnx-2)^2)+e^((lnx-2)^2)`

Explanation:

`f(x)=xe^((lnx-2)^2)`

Use product rule and chain rule

`f=x, g=e^((lnx-2)^2)`

`f'=1, g'=e^((lnx-2)^2)*2(lnx-2) 1/x`

`f'(x)=fg'+gf'`

`f'(x)=2(lnx-2)e^((lnx-2)^2)+e^((lnx-2)^2)`