Question

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

Answer 1

`f'(x)=cos(e^((lnx-2)^2)) xxe^((lnx-2)^2) xx 2(lnx-2) xx 1/x`

Explanation:

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

`f(g(h(r(x))))=f'(g(h(r(x))))xxg'(h(r(x)))^` `h'(r(x))xxr'(x)`

`f'(x)=cos(e^((lnx-2)^2)) xxe^((lnx-2)^2) xx 2(lnx-2) xx 1/x`