Question

How do you find the derivative of `e^(x(3x^2 + 2x-1)^2`?

Answer 1

The answer is:

`y'=e^(x(3x^2 + 2x-1)^2)[1*(3x^2 + 2x-1)^2+x*2(3x^2 + 2x-1)*(6x+2)]=`

`=e^(x(3x^2 + 2x-1)^2)(3x^2 + 2x-1)(3x^2 + 2x-1+12x^2+4x)=`

`=e^(x(3x^2 + 2x-1)^2)(3x^2 + 2x-1)(15x^2+6x-1)`.