anonymous
  • anonymous
derivative of g(x) = (2ra^(rx)+n)^p ?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
pretty confused here
anonymous
  • anonymous
haha it's been a while since derivatives for me, but i'll try o.O
anonymous
  • anonymous
are r, a, n, and p constants?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
the textbook answer is g'(x) = 2r^(2) p (ln a) (2ra^(rx) + n)^(p-1) a^(rx)
anonymous
  • anonymous
guess so
anonymous
  • anonymous
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
anonymous
  • anonymous
you could use the exponential rule for 2ra^(rx) dx . . . so that would be 2ra^(rx)ln(a) + a^rx, i think.
anonymous
  • anonymous
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
anonymous
  • anonymous
yo me too what the hell!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.