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Find the derivative of a function

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  1. Sep 13, 2016 #1
    1. The problem statement, all variables and given/known data
    E(t) = 100te1+cos((2π*t)/365

    What is the derivative of this function?

    2. Relevant equations


    3. The attempt at a solution

    100*e1+cos((2π*t)/365) + 100t*e1+cos((2π*t)/365) * -(2π/365)sin(2πt/365)

    I have tried to use the rules for derivative of products, and also used the chain rule.
     
    Last edited: Sep 13, 2016
  2. jcsd
  3. Sep 13, 2016 #2

    Simon Bridge

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    The sine should not be inside the exponent - example:
    ##\frac{d}{dt}e^{\sin(t)} = e^{\sin(t)}\cos(t)##

    Note:
    It can make for better formatting to use ##\exp[f(t)]## instead of ##e^{f(t)}## when ##f(t)## is quite involved.
     
  4. Sep 13, 2016 #3
    But other than that, is it correct? So can I put:

    100e1+cos(2πt)/365) (1 - (2π/365)sin(2πt/365) t) ?
     
  5. Sep 13, 2016 #4

    Simon Bridge

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    ##\frac{d}{dt}\big[1+\cos(2\pi t/365) \big] \neq 1-(2\pi/365)\sin(2\pi t/365)##
    The main trick with using chain rule is to keep careful track of what is what.

    Try ##g = e^{u(v(t))}## so ##\frac{dg}{dt} = e^{u(v(t))}\frac{du}{dv}\frac{dv}{dt}## and work out du/dv and dv/dt separately.

    [no - got it right first time ...]
     
    Last edited: Sep 13, 2016
  6. Sep 13, 2016 #5

    vela

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    That's not what he's doing.

    This is correct, though it's more common to put the factor of ##t## in front of the sin. That way there's no ambiguity it's multiplying the function rather than the argument of the function.
     
  7. Sep 13, 2016 #6

    Simon Bridge

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    Oh I'm having fun reading the notation ... like reading "*-" as a "-" instead of "multiplied by the negative of". My bad:
    $$\frac{d}{dt}E(t) = 100\left[1-(2\pi /365)t\sin(2\pi t/365) \right]e^{1+\cos(2\pi t/365)}$$ ... something like that?.
     
  8. Sep 13, 2016 #7
    Yes, that is what I have now. Is it done? Or can I simplify it? :)
     
  9. Sep 13, 2016 #8

    Simon Bridge

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    I'd stop there.
     
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