Question

How do you differentiate `f(x)=e^(x^2-3x-4)` using the chain rule?

Answer 1

`f'(x)=(2x-3)e^(x^2-3x-4)`

Explanation:

Differentiate using the `color(blue)"chain rule"`

`color(orange)"Reminder " color(red)(|bar(ul(color(white)(a/a)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(a/a)|)))....(A)`

let `u=x^2-3x-4rArr(du)/(dx)=2x-3`

then `f(x)=y=e^urArr(dy)/(du)=e^u`

Substitute these values into (A) and change u back into terms of x.

`rArrdy/dx=e^u(2x-3)=(2x-3)e^(x^2-3x-4)`