Question

How do you differentiate `e^((x^2-x)^2)` using the chain rule?

Answer 1

`(df)/(dx)=2x(x-1)(2x-1)e^((x^2-x)^2)`

Explanation:

Here `f(x)=e^(g(x))`, where `g(x)=(h(x))^2` and `h(x)=x^2-x`

According to chain rule `(df)/(dx)=(df)/(dg)xx(dg)/(dh)xx(dh)/(dx)`

Hence `(df)/(dx)=e^((x^2-x)^2)xx2(x^2-x)xx(2x-1)`

= `e^((x^2-x)^2)xx2x(x-1)xx(2x-1)`

= `2x(x-1)(2x-1)e^((x^2-x)^2)`