Derivative of y=axe^-kx Chain rule?

Sep 2009
37
1
Hi,

I'm trying to find the derivative of y=axe^(-kx) (a and k are constants), I have found it using the product rule but it looks to me like the chain rule would be correct for this. I'm having trouble working it out using the chain rule. Can the chain rule be used here, if not why?

Thanks
David
 

Soroban

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May 2006
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Hello, David!

Differentiate: .\(\displaystyle y\:=\:axe^{-kx}\)

We have: .\(\displaystyle y \;=\;a\cdot\bigg[x\cdot e^{-kx}\bigg] \)

The leading \(\displaystyle a\) is a coefficient and can be "left out in front."

Inside the brackets, we have a product . . . Apply the Product Rule.

When differentiating \(\displaystyle e^{-kx}\), we need the Chain Rule.


We will have:

. . \(\displaystyle \frac{dy}{dx} \;=\;a\bigg[x \cdot \left(\text{der. of }e^{-kx}\right) \;+\;e^{-kx} \cdot \text{(der. of }x) \bigg]\)

. . . . .\(\displaystyle =\;a\bigg[x\cdot e^{-kx}\cdot(-k)\;+\;e^{-kx}\cdot 1\bigg] \)

. . . . .\(\displaystyle =\;a\bigg[-kxe^{-kx} + e^{-kx}\bigg]\)

. . . . .\(\displaystyle = \; ae^{-kx}\bigg[1 - kx\bigg] \)

 
Sep 2009
37
1
Thanks for the reply,

So if I understand correctly, when there is e^-kx with or without a constant at the front the chain rule is used. When there is e^-kx with a variable at the front the product rule is used.

Thanks
David
 
Mar 2010
715
381
So if I understand correctly, when there is e^-kx with or without a constant at the front the chain rule is used. When there is e^-kx with a variable at the front the product rule is used.
Precisely. You use the product rule when you have a product of functions, and the chain rule when you have a composition functions. If there was a variable in front of e^-kx, it would mean we have a product of two functions, so the product rule would be used, but if it's just a constant then we can't use the product rule because we don't have a product of two functions - merely a product of a constant and a (composite) function. You wanted to differentiate \(\displaystyle axe^{-kx}\). You can see it's a product of two functions: \(\displaystyle ax \) and \(\displaystyle e^{-kx}\), so we need to use the product rule. The product rule states \(\displaystyle \dfrac{d}{dx}\left\{f(x)\cdot{g(x)}\right\} = f'(x)g(x)+g'(x)f(x). \) If we now take \(\displaystyle ax \) as \(\displaystyle f(x) \) and \(\displaystyle e^{-kx}\) as \(\displaystyle g(x)\), we need to find both \(\displaystyle f'(x)\) and \(\displaystyle g'(x)\). It's easy to find \(\displaystyle f'(x)\) because \(\displaystyle \dfrac{d}{dx}\left(ax\right)= a\), but what about \(\displaystyle g'(x)\)? Well, \(\displaystyle g'(x)\) is the derivative of \(\displaystyle e^{-kx}\), but that's not a product so we cannot use the product rule. It's a composition of two functions, so we have to then use the chain rule, which states \(\displaystyle (f \circ g)'(x) = f'(g(x))g'(x)\). We have \(\displaystyle f'(g(x)) = \dfrac{d}{dx}\left(e^{-kx}\right) = e^{-kx}[/Math], and \(\displaystyle g'(x) = \dfrac{d}{dx}\left(-kx\right) = -k\)\)\(\displaystyle , which means \(\displaystyle f'(g(x))g'(x) = -ke^{-kx}\), then we have found \(\displaystyle g'(x)\). Therefore \(\displaystyle \dfrac{d}{dx}\left\{axe^{-kx}\right\} = ae^{-kx}-ke^{-kx}ax\), which you can of course simply to Soroban's answer. Basically, whenever you have functions of the form \(\displaystyle (f\circ g)(x)\), you use the chain rule, and whenever you have a function of the form \(\displaystyle f(x)\cdot{g(x)}\), you use the product rule.\)
 
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