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How do you get the derivative of this?

  1. Sep 6, 2006 #1
    This is for my Circuits class, I am supposed to find the current i.

    If q = 5te^(-10^(3) t) Coulombs

    Find i (current) the formula for i is: i = dq/dt

    So what I did was try to find the derivative of q, which is dq

    So I brought out the constant 5, and I am left with te^(-10^(3) t)

    and I recognize this as a product of two functions so I tried the product rule.

    I let my first function f = t, and my second function g = e^(-10^(3) t)

    so f ' = 1 and g ' (using the chain rule) i got (e^(-10^(3) t) x '10^3)

    so i plugged it into the product role (fg)' = fg' + gf' and i get a weird answer.

    The given answer in the back of the book is :

    i = 5(1-10^3 t)e^(-10^3 t) Amps

    can someone help me please
  2. jcsd
  3. Sep 6, 2006 #2


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    Perhaps you just need to simplify? What does your current answer look like?

    PS: we have a homework section :wink:
  4. Sep 6, 2006 #3


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    Your general approach is correct. g' = e^(-10^(3) t) * -10^3

    When you plug this into the product rule, you get a "weird answer?" Are you sure that your weird answer is not, in fact, the correct answer in unsimplified form?

    - Warren
  5. Sep 6, 2006 #4

    George Jones

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    It looks to me that if you make a small correction in your calculation of g', then you will get the same answer as the book.
  6. Sep 6, 2006 #5


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    I presume that is a typo and it should be (e^(-10^(3)t) x (-10^3)) not

    How weird? You just said [itex]f= t, g'= -10^3 e^{-10^3t}, g= e^{-10^3t}, f'= 1[/itex] so [itex]fg'+ f'g= -10^3 te^{-10^3t}+ e^{-10^t}[/itex]
    What do you get if you factor [itex]e^{-10^3 t}[/itex] out of that?

    Last edited: Sep 6, 2006
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