Results 1 to 10 of 10

Math Help - Derivate f(x)=x^2*e*(-3x)

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    4

    Derivate f(x)=x^2*e*(-3x)

    Hi guys

    I've been trying to solve this for a little while now but I feel kinda lost and I thought you could maybe help me out.

    so

    f(x)=x^2*e^(-3x)

    f '(x)=?

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2009
    Posts
    263
    Quote Originally Posted by Blacklaw View Post
    Hi guys

    I've been trying to solve this for a little while now but I feel kinda lost and I thought you could maybe help me out.

    so

    f(x)=x^2*e^(-3x)

    f '(x)=?

    Thank you.
    <br />
f(x) = g(x)h(x) , then f'(x) = g'(x)h(x) + g(x)h'(x)

    take g(x) = x^2 and h(x) = e^{-3x}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    4
    Quote Originally Posted by dedust View Post
    <br />
f(x) = g(x)h(x) , then f'(x) = g'(x)h(x) + g(x)h'(x)

    take g(x) = x^2 and h(x) = e^{-3x}
    I get 2x*e^{-3x}+x^2*e^{-3x}

    But I know for a fact that the correct answer is 2x*e^{-3x}+x^2*e^{-3x}*(-3)

    Please explain how I get the (-3)?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Blacklaw View Post
    I get 2x*e^{-3x}+x^2*e^{-3x}

    But I know for a fact that the correct answer is 2x*e^{-3x}+x^2*e^{-3x}*(-3)

    Please explain how I get the (-3)?
    \frac{d}{dx}e^{-3x}=-3e^{-3x}

    You did not know the answer for a fact if it was what was in the back of the book. You know it for a fact if you know why it is so and now you should do.

    CB
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2010
    Posts
    4
    Quote Originally Posted by CaptainBlack View Post
    \frac{d}{dx}e^{-3x}=-3e^{-3x}

    You did not know the answer for a fact if it was what was in the back of the book. You know it for a fact if you know why it is so and now you should do.

    CB
    Does it always go like that? I thought the derivative of e^x was e^x?

    Thank you for you help, I truly appreciate it.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,035
    Thanks
    49
    Quote Originally Posted by Blacklaw View Post
    Does it always go like that?
    That's what the chain rule says.

    Just in case a picture helps...



    is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    In this case...






    PS:

    Just in case another picture doesn't hurt (and in view of comments below)...

    Wrapping the chain rule inside the (legs-uncrossed version of) ...



    ... the product rule, we get...




    __________________________________________
    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; January 6th 2010 at 10:05 AM. Reason: PS
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jan 2010
    Posts
    4
    I think I get it now.. Thank you very much for helping me out.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jan 2009
    Posts
    5
    f(x) = x^2 e^{-3x}
    f'(x) = x^2 e^{-3x} (-3) + e^{-3x} 2x By chain rule
    f'(x) = x e^{-3x} [ -3x + 2 ]

    Hope this helps.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by rshekhar.in View Post
    f(x) = x^2 e^{-3x}
    f'(x) = x^2 e^{-3x} (-3) + e^{-3x} 2x By chain rule
    f'(x) = x e^{-3x} [ -3x + 2 ]

    Hope this helps.
    What you label as "by chain riue" is mainly product rule and to a lesser extent chain rule

    CB
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Blacklaw View Post
    Does it always go like that? I thought the derivative of e^x was e^x?

    Thank you for you help, I truly appreciate it.
    The derivative of e^x is e^x but that is not what you have, what you should know is that the derivative of e^{ax} is ae^{ax} by the chain rule if you must, but you should know it because it will occur so often.

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivate
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 16th 2011, 06:30 AM
  2. Derivate
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 29th 2010, 04:55 PM
  3. [SOLVED] How to derivate these?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: November 16th 2010, 01:56 PM
  4. How to derivate this one?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 9th 2010, 06:00 PM
  5. Derivate.
    Posted in the Calculus Forum
    Replies: 10
    Last Post: July 6th 2006, 12:08 AM

Search Tags


/mathhelpforum @mathhelpforum