Finding the Derivative of y=xe-kx using the Chain Rule

[physics604]

Homework Statement

Find the derivative of y=xe^-kx

Homework Equations

Chain rule

The Attempt at a Solution

y = xe^u

\frac{dy}{du} = xe^u+e^u

u = -kx

\frac{du}{dx} = -k


\frac{dy}{dx} = (xe^-kx+e^-kx)(-k)

= e^-kx(x+1)(-k)

= e^-kx(-kx-k)

The answer is e^-kx(-kx+1). What did I do wrong?

 

[LCKurtz]

Homework Statement

Find the derivative of y=xe^-kx

Homework Equations

Chain rule

The Attempt at a Solution

y = xe^u

\frac{dy}{du} = xe^u+e^u

There is a subtle error here. The problem is that if $u=-kx$ then $x=-\frac u k$ and you have to include $\frac{dx}{du} = -\frac 1 k$ in your second term. But it's best to avoid this error completely by noting that you have a product and should use the product rule before the chain rule. Try$$
\frac d {dx} xe^u = e^u\frac d {dx} x + x \frac d {dx}e^u$$ and use your method on that last term.

 

[physics604]

I'm sorry, but I don't get what you mean.

\frac{d}{dx} xe^u = e^u\frac{d}{dx}x+x\frac{d}{dx}e^u

is still equal to e^u + xe^u.

 

[physics604]

Okay, I get it. Thanks!

 

[LCKurtz]

But it's best to avoid this error completely by noting that you have a product and should use the product rule before the chain rule. Try$$
\frac d {dx} xe^u = e^u\frac d {dx} x + x \frac d {dx}e^u$$ and use your method on that last term.

I'm sorry, but I don't get what you mean.

\frac{d}{dx} xe^u = e^u\frac{d}{dx}x+x\frac{d}{dx}e^u

is still equal to e^u + xe^u.

No it isn't. By your own steps in your original post you showed$$
\frac d {dx} e^u = -ke^{-kx}$$and if you put that in you get the correct answer.

 

[physics604]

I get it now. Thanks!

 

[physics604]

Would you mind taking a look at another question of mine as well?

https://www.physicsforums.com/showthread.php?t=721825