Question

Find the derivative using the "Chain Rule".

Answer 1

Right, post!

Answer 2

\[y=xe^-kx\]

Answer 3

I don't think that's the question...

Answer 4

thats xe raised to the -kx

Answer 5

Right, gotcha.\[{d\over dx}f(g(x)) = f'(g(x)) g'(x) \]

Answer 6

Raising to \(-kx\) is \(f\), and multiplying by \(e\) is \(g\).

Answer 7

hmm how do I identify f and g?

Answer 8

Let's go with the easy way, okay? There is always an "outside function" and "inside function".

Answer 9

right, x i would assume is outside?

Answer 10

e^-kx is the inside

Answer 11

\[(\sin(x))^7\]Here, raising to 7 is outside and sin is inside.

Answer 12

ohhh, so 7 would go out front, and 6 would be left over 7sinx^6

Answer 13

You always do it like this:
Derivative of outside function Inside function * Derivative of inside function

Answer 14

7cosx^6?

Answer 15

Not really.

Answer 16

7sinx^6(cosx) ?

Answer 17

yep

Answer 18

ahhh i see

Answer 19

so for my problem, it would be
-kx(xe)^-2kx(e^-kx)

Answer 20

that doesent look right

Answer 21

@dfresenius I gotta go... exam tomorrow.
You can check this out and see steps. http://www.wolframalpha.com/input/?i=differentiate+xe%5E%28%E2%88%92kx%29#socialtoggle

Answer 22

you there math?

Answer 23

oh ,yes,now i m

Answer 24

so
x'e^(-kx) + x {e^(-kx)}' = ??

Answer 25

e^-kx+x(e^-kx)

Answer 26

thats the product rule.

Answer 27

oh u didn't do {e^(-kx)}' correctly....

Answer 28

i thought e^x = e^x

Answer 29

\(\huge \frac{d}{dx}e^{ax}=ae^{ax}\)

Answer 30

e^-kx +x(-ke^-kx)

Answer 31

yep.
\(\huge(1-kx) e^{-kx}\)

Answer 32

wow, thats it?

Answer 33

yup.

Answer 34

thanks that was very helpful!

Answer 35

welcome and
\(\huge \frac{d}{dx}e^{ax}=e^{ax}\frac{d}{dx}ax=ae^{ax}<-chain rule\)

Answer 36

u got where the chain rule came to play ?

Answer 37

well, sort of, I just thought i used product rule

Answer 38

because e^ax = ae^ax pretty much solved it

Answer 39

yes, but that formula is because of chain rule, as i have shown there...

Answer 40

ok sorry i couldnt medal you , gave one out in the beginning to quickly

Answer 41

no problem at all ! i m not here for medals ....