Results 1 to 6 of 6

Math Help - Dirivitive using the power "e"

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    46

    Derivitive using the power "e"

    Can anyone help me with a few problems please?

    I'm not sure how to find the derivative of a function like this.. 5x^e+15e^x

    also, what is the difference between (d/dt) and (d/dx). I found on (d/dt) that I can solve normally, however (d/dx) asks for a different answer?
    Last edited by skyslimit; April 28th 2009 at 12:07 PM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by skyslimit View Post
    Can anyone help me with a few problems please?

    I'm not sure how to find the derivative of a function like this.. 5x^e+15e^x
    e is a constant. so for x^e just use the power rule to differentiate it. as for e^x its derivative is itself.

    also, what is the difference between (d/dt) and (d/dx). I found on (d/dt) that I can solve normally, however (d/dx) asks for a different answer?
    i suppose you are talking in the context of implicit differentiation?

    anyway, the notation is interpreted thus: \frac d{dx} mean you are taking the derivative with respect to x. an analogous interpretation holds for \frac d{dt}

    example:

    differentiate implicitly with respect to x

    x^2 + y^2 = 2x

    \Rightarrow 2x~\frac {dx}{dx} + 2y~\frac {dy}{dx} = 2~\frac {dx}{dx}

    so we have

    2x + 2y~\frac {dy}{dx} = 2

    (we put dx/dx to say we took the "derivative of an x-term with respect to x" and dy/dx to say we took the "derivative of a y-term with respect to x")


    if we differentiated the same equation with respect to t we would get

    2x~\frac {dx}{dt} + 2y~\frac {dy}{dt} = 2~\frac {dx}{dt}

    (we put dx/dt to say we took the "derivative of an x-term with respect to t" and dy/dt to say we took the "derivative of a y-term with respect to t")

    see the difference?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    Posts
    46
    Thank you.

    If I'm not mistaken, are you saying the only difference of the two equations are the variable. So i should be able to find the derivitive of a similar eqaution and find the same answer?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by skyslimit View Post
    Thank you.

    If I'm not mistaken, are you saying the only difference of the two equations are the variable. So i should be able to find the derivitive of a similar eqaution and find the same answer?
    the difference is the variable, but that difference can have a lot of repercussions. in general, the answer would not be the same. and it depends on what you are solving for. there are more unknowns in the latter than the former
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2008
    Posts
    46
    would you mind showing how i can solve for d/dx using an equation like e^(3x^2+5x)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by skyslimit View Post
    would you mind showing how i can solve for d/dx using an equation like e^(3x^2+5x)
    recall, \frac d{dx} e^u = u' e^u, where u is a function of x

    thus, \frac d{dx} e^{3x^2 + 5x} = \bigg[ \frac d{dx}(3x^2 + 5x) \bigg] \cdot e^{3x^2 + 5x}

    can you continue?

    and next time, post a new question in a new thread
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 16th 2011, 01:08 AM
  2. Replies: 2
    Last Post: June 4th 2011, 12:11 PM
  3. Replies: 2
    Last Post: April 24th 2011, 07:01 AM
  4. Replies: 1
    Last Post: October 25th 2010, 04:45 AM
  5. Replies: 1
    Last Post: June 4th 2010, 10:26 PM

Search Tags


/mathhelpforum @mathhelpforum