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Homework Help: Distributing the e term first

  1. Aug 3, 2010 #1
    1. The problem statement, all variables and given/known data
    e3x(3sin x-cos x)


    3. The attempt at a solution

    e3x3(3sin x-cos x)+(3cos x+sin x)e3x=10e3xsin x.

    Is that right?
     
  2. jcsd
  3. Aug 3, 2010 #2

    berkeman

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    Staff: Mentor

    Re: derivative

    I don't think so, at least not the intermediate steps.

    Could you try distributing the e term first, and then take the derivative of the two resulting terms, using the chain rule on each?
     
  4. Aug 3, 2010 #3
    Re: derivative

    OK, so let me then get rid of parenthesis. I will show you my intermediate steps.

    e3x3sinx-e3xcosx=
    3e3xsinx+e3x3cosx-(3e3xcosx+e3x(-sinx))=
    10e3xsin x
     
  5. Aug 3, 2010 #4

    berkeman

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    Staff: Mentor

    Re: derivative

    I'm not seeing the chain rule being used. Remember, it's the first term multiplied by the derivative of the second term, plus the second term multiplied by the derivative of the first term....

    http://en.wikipedia.org/wiki/Chain_rule

    .
     
  6. Aug 3, 2010 #5

    berkeman

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    Staff: Mentor

    Re: derivative

    Also, you should be careful to distinguish between the initial terms, and the derivative that you are solving for. Write it like this:

    (e3x3sinx-e3xcosx)' =

    Or

    d/dx(e3x3sinx-e3xcosx) =
     
  7. Aug 3, 2010 #6
    Re: derivative

    I am sorry, I don't quite understand what you are saying. I performed chain rule to the best of my knowledge. I don't know what else to do.
     
  8. Aug 3, 2010 #7

    lurflurf

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    Homework Helper

    Re: derivative

    Don't worry about berkeman's confusing, unclear, and not helpful post. Your work is quite correct, but it may help to use a few more steps until you have more practice.

    product rule
    [e3x(3sin x-cos x)]'=(e3x)'(3sin x-cos x)+e3x(3sin x-cos x)'
    chain rule and difference rule
    =3e3x(3sin x-cos x)+e3x(3sin' x-cos' x)
    distribute
    =e3x(9sin x-3cos x)+e3x(3cos x+sin x)
    gather like terms (exponential)
    =e3x(9sin x-3cos x+3cos x+sin x)
    gather like terms (sine and cosine)
    =e3x(10sin x)
    final answer
    =10e3xsin x
     
  9. Aug 3, 2010 #8

    berkeman

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    Re: derivative

    Sorry if I was confusing. Should I have been using the term product rule instead of the more general chain rule? It was the intermediate steps that I was trying to get laid out.
     
  10. Aug 3, 2010 #9

    Mark44

    Staff: Mentor

    Re: derivative

    Both rules need to be used in this problem. The product rule is used first, because the function is a product - e3x(3sin x-cos x). Then, to differentiate e3x, the chain rule is called for.
     
  11. Aug 3, 2010 #10
    Re: derivative

    Does it mean that my calculation is correct?
     
  12. Aug 3, 2010 #11

    Mark44

    Staff: Mentor

    Re: derivative

    Yes.
     
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