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Differentiation using chain/product rule

  1. Sep 23, 2012 #1
    Hi,

    Just a question on an example in a maths textbook. See attached image for question below.

    So, I understand that if you set u=sin(x) and v=e^-cos(x)
    f'(x)=u'.v + u.v'

    But I'm stuck looking at e^-cos(x), could it also be classified e^(w)?

    Also, the second step in differentiating the above equation seems to only differentiate the -cos(x) of the e^-cos(x)...is this because the derivative of e is simply e?
     

    Attached Files:

  2. jcsd
  3. Sep 23, 2012 #2

    chiro

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    Science Advisor

    Hey Baartzy89 and welcome to the forums.

    You have the right idea about the exponential: we have two properties.

    The first is that d/dx e^(x) = e^(x) and the chain rule which says d/dx f(g(x)) = g'(x)f'(g(x)), so you need to consider g(x) and the derivative of f(g(x)).
     
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