Results 1 to 3 of 3

Math Help - [SOLVED] HELP !! defferentiation exponents

  1. #1
    demi
    Guest

    [SOLVED] HELP !! defferentiation exponents

    can someone please help me with this problem....

    .e^(x/2)
    ----------
    ..x^(5)
    can someone please give me an idea of how to do this problem... thank you!!
    Last edited by MathGuru; January 21st 2006 at 02:12 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by demi
    can someone please help me with this problem....

    .e^(x/2)
    ----------
    ..x^(5)
    can someone please give me an idea of how to do this problem... thank you!!
    Lets assume that you are asking about the derivative of:

    \frac{e^{x/2}}{x^5}.
    Well this can be done in a number of ways, you could use the quotient rule,
    or the product rule to find this derivative. Now I don't have that good a
    memory, and so never bother with the quotient rule, always using the product rule:

    \frac{d}{dx} f(x)g(x)=\frac{df}{dx}g(x)+f(x)\frac{dg}{dx}.

    Here we let:

    f(x)=e^{x/2},

    and:

    g(x)=\frac{1}{x^5}.

    Now:

    \frac{d}{dx}f=\frac{d}{dx}e^{x/2}=\frac{1}{2}e^{x/2},

    and:

    \frac{d}{dx}g=\frac{d}{dx}\left\{ \frac{1}{x^5} \right\}=-5\frac{1}{x^6}.

    So now you have enough to assemble the derivative

    RonL
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by demi
    can someone please help me with this problem....

    .e^(x/2)
    ----------
    ..x^(5)
    can someone please give me an idea of how to do this problem... thank you!!
    Differentiation with respect to x.

    d/dx [u/v] = [v*du/dx -u*dv/dx] / v^2 ------------***

    d/dx [e^(x/2) / x^5]
    = {[(x^5)*e^(x/2) *(1/2)] -[e^(x/2) *(5x^4)]} / [(x^5)^2]
    = {(1/2)(x^5)e^(x/2) -5(x^4)e^(x/2)} / x^10
    = [e^(x/2)][(x^5 /2) -(5x^4)]/ x^10
    = [e^(x/2)][1 /(2x^5) -5/(x^6)] ---------------------(i)
    = [e^(x/2)][{[1 /(2x^5)]*(x/x)} -{[5/(x^6)]*(2/2)]}
    = [e^(x/2)][x /(2x^6) -10/(2x^6)]
    = [e^(x/2)][(x -10) /(2x^6)]
    = (1/2)[e^(x/2)][(x -10) /(x^6)] ------------answer.


    --------------------------------------
    d/dx [u*v] = v*du/dx +u*dv/dx ----------***

    d/dx [e^(x/2) / x^5]
    = d/dx [(x^(-5))(e^(x/2))]
    = e^(x/2) *[-5 x^(-6)] +[x^(-5) *[e^(x/2) *(1/2)]
    = [e^(x/2)][-5/(x^6) +(1 /(2x^5)]
    = [e^(x/2)][1/(2x^5) -5/(x^6)] ------------same as (i) above.
    Etc...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Defferentiation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 3rd 2011, 11:36 AM
  2. Implicit Defferentiation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 7th 2010, 03:59 AM
  3. defferentiation
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: February 11th 2010, 03:51 AM
  4. Replies: 5
    Last Post: February 3rd 2009, 07:37 PM
  5. [SOLVED] Simplyify using the law of exponents
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 26th 2009, 09:12 AM

/mathhelpforum @mathhelpforum