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Finding the derivative of m(x)=-e^xcosx

  1. May 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the derivative of m(x)=-e^x*cosx at x=1 graphically and algebraically.

    2. Relevant equations



    3. The attempt at a solution

    So first i just attempted to find the derivative:

    m'(x)=-xe^(x-1)*(-sinx)----> is this correct?

    so to find at x=1 i just sub in 1.

    m'(1)=-(1)e^(1-1)*(-sin1)
    m'(1)=-1e^0*(-sin1)
    m'(1)=0.01745

    Is this correct? When i tried graphing the derivative i got a different answer so it made me second guess my original answer.

    Any help is very much appreciated, thank you.
     
  2. jcsd
  3. May 25, 2012 #2

    HallsofIvy

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    No, it's not. You are mis-using some basic properties of the derivative. In the first place the "product rule": the derivative of fg is (fg)'= f'g+ fg', not just " f'g' " as you have here.

    Second, the rule " (xn)'= n xn-1" applies only when the variable is the base and the exponent is a constant. With ex or, more generally, ax, for any positive a, that rule does not apply. Look up the derivative of ex sdecifically.

     
  4. May 25, 2012 #3
    Okay, thank you. I had not learned about the derivative of exponential functions but looked it up. So is -e^x still just -e^x?
     
  5. May 25, 2012 #4
    And as for using the product rule, would this be correct?

    m'(x)=-e^x*(cosx) + (-sinx)(-e^x)

    assuming my previous post is correct?
     
  6. May 25, 2012 #5

    HallsofIvy

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    Yes, f(x)= ex has the very nice property that its derivative is just f'(x)= ex again. That's why the number "e" has a special symbol.
     
  7. May 25, 2012 #6
    Okay, thanks very much for your help.
     
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