# derivatives using chain rule

I am completely lost on this one, I have no idea what to do at all on it. I know that the derivative of $e^x$ is $e^x$ but that doesn't seem to help here. I am suppose to find the derivative of $y=xe^{-kx}$ I figured I could use the product rule which would be the derivative of e $e^{-kx} (x) + e^{-kx} (1)$ I need to find the derivative of $e^{-kx}$ and for that I got $-e^{-kx}$ I likely made many mistakes because my answer isn't even close.

-
The derivative of $e^{-kx}$ is not $-e^{-kx}$. You seemed to have pulled the sign out the exponent correctly, but you forget the scaling factor.. – anon Sep 22 '11 at 20:41
I don't know what the scaling factor is, what I did though was make $y=e^u$ and $u=-kx$ is that wrong? It is problaby time I give up for the day, I can't do a single problem on my own. – user138246 Sep 22 '11 at 20:44
What is the derivative of -$kx$ w.r.t $x$? – M.B. Sep 22 '11 at 20:49
Jordan: I was talking about $k$ is all. Your method is completely right, but you didn't do the math correctly: what's $du/dx$? – anon Sep 22 '11 at 20:49
@Jordan: I think that you should know that a meta question has been brought up concerning your questions. – mixedmath Jun 13 '12 at 18:26

This is a combination of product rule and chain rule. To review, the product rule states

$$\frac{d f g}{dx} = \frac{df}{dx} g + f \frac{dg}{dx}$$

or in Newton's notation

$$(fg)' = f' g + g' f.$$

And as you've identified correctly,

$$y' = (x e^{- k x})' = (1) e^{-kx} + x (e^{-kx})',$$

so we just need to determine what $(e^{-kx})'$ is. From previous questions, you know that the chain rule states that

$$\frac{df(g(x))}{dx} = \frac{df(g)}{dg} \frac{dg(x)}{dx},$$

or more succinctly

$$f'(g(x)) = f'(g) g'(x).$$

So, in $(e^{-kx})'$, we can see that $f(g) = e^g$ and $g(x) = -k x$. You can either memorize how to treat the derivative of a constant times your variable, or you could apply the product rule, again. The product rule gives

$$g'(x) = (-k)' x + -k (x)'.$$

So, what are the derivatives of $-k$ and $x$ with respect to $x$? Once you have those values in hand, plug back in.

-