Question

derivative of g(x) = (2ra^(rx)+n)^p ?

Answer 1

pretty confused here

Answer 2

haha it's been a while since derivatives for me, but i'll try o.O

Answer 3

are r, a, n, and p constants?

Answer 4

the textbook answer is g'(x) = 2r^(2) p (ln a) (2ra^(rx) + n)^(p-1) a^(rx)

Answer 5

guess so

Answer 6

i mean i got as far as p(2ra^(rx) + n)^(p-1) d/dx (2ra^(rx) + n) you know... but at that point i'm sort of at a loss as to what's next

Answer 7

you could use the exponential rule for 2ra^(rx) dx . . .

so that would be 2ra^(rx)ln(a) + a^rx, i think.

Answer 8

after simplifying that's pretty close to the textbook answer, probably i messed up a little on my derivatives. i'm just missing the 2r^2 part :P

Answer 9

yo me too what the hell!!