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

1 Answer
Aug 29, 2016

Answer:

#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)#