Question

Differentiate y=xe^(-x/2).

Answer 1

y'=e^(-x/2)+xe^(-x/2)(-1/2)

Answer 2

Thanks, but I'm a bit confused on differentiating e^(-x/2).
why does it give you e^(-x/2)(-1/2) I thought it's -2x(e^(-x/2)).

Answer 3

for \[e^{f(x)}\]
the derivative is

\[e^{f(x)}\] *f'(x)

Answer 4

It is a product rule and a chain rule. So applying the product rule first you have: f'(x)g(x) which is (1)e^(-x/2). Then the second part +f(x)g'(x). This is where the chain rule comes in. You get x(-1/2)e^(-x/2). Add them together. You get: e^(-x/2)+(-x/2)e^(-x/2).

Answer 5

So you use the rule: \[\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)\]. (product)
And the chain rule: \[\frac{d}{dx}[f(g(x))]=g'(x)f'(g(x))\].

Answer 6

@malevolence19, which part of the equation do you use the chain rule?
if im not wrong is it the e^(-x/2)?

Answer 7

yes, right

Answer 8

Exactly :)

Answer 9

And one q, can't we just differentiate it normally? Why is it a chain rule? Sorry if I'm asking too many questions, my brain is not functioning properly today haha.

Answer 10

You know the derivative of e^x is just e^x right? Well, technically, it is a chain rule. But the derivative of the x is just 1 right? But in math, we never right implied 1's. It is only a chain rule because it is not straight e^x Its e^g(x). Where g(x), in this case, is -x/2. Does that help a little?

Answer 11

Hmm, I think I kind of get it now. So for the second part of the product rule, you get x(-1/2)(e^-x/2) because of the chain rule, the -1/2 is the derivative of the power of e?

Answer 12

Exactly :) And the x is just from copying it since you took a derivative of it in the first term :P So for example if you had xe^(x^2) The SECOND term would be: x(2x)e^(x^2). Because 2x is the derivative of x^2.

Answer 13

If you get stuck again just post the question and I'll be able to answer. Calculus is my thing xP

Answer 14

I'll be on tomorrow O:

Answer 15

Hahaha, thanks a lot! :D