1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    trollcast

    User Avatar
    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

    BruceW

    User Avatar
    Homework Helper

    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

    trollcast

    User Avatar
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Derivates of e^x
  1. Derivatives of e^x (Replies: 3)

  2. Derivative of e^x (Replies: 2)

Loading...