Question

derivative of u^8e(-u)

Answer 1

This one is tricky, and the first step is probably the hardest part. First, recognize that 8*e is a constant. From there, can you think of a way you can rewrite the expression as an exponential with "e" as the base?

Answer 2

I am sorry I missed typed the equation. the 8 is the exponent of u
and the -u is the exponent of e

Answer 3

Oh, so is the question supposed to be like this? \[u ^{8}e ^{-u}\]

Answer 4

yes

Answer 5

Ok, whew, that's much easier. The key here is to remember the chain rule. You basically have two expressions that are multiplied together, so apply the chain rule like this: \[\frac{ d }{ dx } (f(x)*g(x)) = f'(x)*g(x) + g'(x)*f(x)\]Do you understand how to apply that formula to this problem?

Answer 6

I think so. There are examples I can use to help

Answer 7

Knowing the rule helps alot.

Answer 8

Ok, I can carry it out a little bit further by plugging in your example into the formula, if that would help.

Answer 9

That would be aswesome

Answer 10

Thank you

Answer 11

Ok, so with your example,\[f(x) = u ^{8}\] and \[g(x) = e^{-u}\]and since they are multiplied together, \[\frac{ d }{ dx }(f(x)*g(x)) = \frac{d}{dx}(u ^{8})*e ^{-u} + \frac{d}{dx}(e^{-u})*u^8\] Do you think you can take those derivatives?

Answer 12

?

Answer 13

Sorry, I just realized that I wrote that in a very poor way. I'm so used to writing derivatives with respect to x that I forgot that yours was with respect to u.

Answer 14

Sorry I had an answer it did not take

Answer 15

With your numbers plugged in, it should be \[\frac{d}{du}(u^8)*e^{-u} + \frac{d}{du}(e^{-u})*u^8\]

Answer 16

and the derivative of e^(-u) is e^(-u) right?

Answer 17

So it would be 8u^7*e^(-u)+e^(-u)*u^8

Answer 18

I used to love math

Answer 19

The derivative of e^(something) is (derivative of something)*e^(something), so \[\frac{d}{dx}(e^{-u}) = -e^{-u}\]Note the negative sign.

Answer 20

Because of the -1

Answer 21

right, so your previous answer was almost right, but it was missing that negative sign.

Answer 22

Awesome

Answer 23

I think I can do it from here. I really appreciate your help

Answer 24

No problem!