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Homework Help: Derivates of e^x

  1. Feb 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the derivative of these e^x functions [paraphrased]

    2. Relevant equations
    [itex]x^{2} e^{x}[/itex]

    3. The attempt at a solution
    [itex]2x e^{x}[/itex]

    This is how I believe it to be correct from my understanding. It does feel wrong, with the answers agreeing with the feeling.
  2. jcsd
  3. Feb 25, 2013 #2


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    Gold Member

    You need to use the product rule to differentiate that.

    $$\frac{d}{dx}(u\cdot v)=u(\frac{dv}{dx})+ v(\frac{du}{dx})$$
  4. Feb 25, 2013 #3


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    the function [itex]x^2e^x[/itex] is a product of two functions. So what rule do you need to use, to take the derivative?
  5. Feb 25, 2013 #4
    Can't believe I didn't notice the use of the product rule. As I was progressing through the question, I wasn't sure of one new thing: is two [itex]e^{x}[/itex] equal to [itex]e^{x}[/itex] or [itex]2e^{x}[/itex]? I reckon it's the latter, but the answers are in the simplest form and doesn't seem to fit; unless I've made another minute error.
  6. Feb 25, 2013 #5
    ##e^x + e^x = 2e^x## if that is what you're asking. Pretty standard stuff...
  7. Feb 25, 2013 #6
    Well try to look at x^2e^x as u=x^2 and v=e^x

    so (uv)'=u'v+v'u. Now just compute that and you will find your derivative.

    You should get 2xe^x+x^2e^x=x(2e^x+xe^x)=xe^x(x+2)
  8. Feb 25, 2013 #7
    Yeah that's what I thought. But I had to check as I was getting problems. It would be my working-out. We were doing some different than standard stuff with Euler's Number today, so I wondered if it applied to this situation as well.
  9. Feb 25, 2013 #8
    Did you differently simplify it? The answer in my book: [itex]xe^{x}(2+x)[/itex]
  10. Feb 25, 2013 #9
    No it's the same look at my post.
  11. Feb 25, 2013 #10
    Oh whoops, I somehow didn't see that segment ...
  12. Feb 25, 2013 #11
    Thankyou everyone for your help.

    P.S How do I mark this topic as [SOLVED] ?
  13. Feb 25, 2013 #12


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    There's no solved prefixes on the forums.
  14. Feb 26, 2013 #13
    Oh ok. Well, thank you!

    P.S Turns out that our teacher accidentally jumped ahead in the course. Explaining why I seemed to ask silly questions, that I'm now able to since we had the grounding today.
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