Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer

up vote 3 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.