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Chain rule physics problem

  1. Oct 15, 2009 #1
    D e^(b*t*ln(t)) + ln(x) respect to t

    my answer:

    t^(b*t)*(ln(t)*b)+b + 1/x
     
    Last edited: Oct 15, 2009
  2. jcsd
  3. Oct 15, 2009 #2

    Mark44

    Staff: Mentor

    Re: Derivative

    No. For the exponential function, the chain rule looks like this: d/dt(eu) = eu*du/dt.

    Also, did you mean the ln(x) term to be ln(t)?
     
  4. Oct 15, 2009 #3
    Re: Derivative

    nope. It is ln(x)
     
  5. Oct 15, 2009 #4

    Mark44

    Staff: Mentor

    Re: Derivative

    And are you supposed to find the derivative with respect to t or is it the partial derivative with respect to t?

    Your D' notation doesn't mean anything, so I can't tell what you are trying to do.
     
  6. Oct 15, 2009 #5
    Re: Derivative

    derivative respect to t <:

    so:
    exp(b*t*ln(t))*(b*ln(t)+(b*t*1/t)+1/x ?

    = e^(b*t*ln(t) * (b*ln(t)+b)+1/x
     
    Last edited: Oct 15, 2009
  7. Oct 15, 2009 #6

    Mark44

    Staff: Mentor

    Re: Derivative

    Two things:
    1) d/dt(ln(x)) = 0. As far as t is concerned, x is a constant, so ln(x) is a constant, so d/dt(ln(x)) is 0.
    2) b*t*1/t can be simplified.

    Also, it would be good for you to be more explanatory in what you are doing. For this problem, that would mean indicating that you are taking the derivative of something, with respect to some variable, and then showing what you arrived at.

    For this problem, what I'm talking about is this:
    d/dt[e^(b*t*ln(t)) + ln(x)] = exp(b*t*ln(t)*(b*ln(t)+(b*t*1/t)+1/x

    Note that I copied your result verbatim, and the two points above still apply.
     
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