Question

find g'(x)
g(x) = (1x^2−3x−1)e^x

Answer 1

(2x-3)e^x+(x^2-3x-1)e^x im not sure if its right

Answer 2

Looks fine to me. Can simplify it slightly though.

Answer 3

how would i kinda have trouble factorin out with exponentials

Answer 4

So e^x is a common factor in each term, so you could yank that out:

\[(2x-3)e^{x} + (x^{2}-3x -1)e^{x} = e^{x}(2x-3 + x^{2} - 3x - 1) \implies e^{x}(x^{2}-x-4)\]

Answer 5

Thanks. i knew it was a simple procedure i just dont know why when i see exponential i go crazy -.-

Answer 6

Not used to it is all xD