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Differentiation problem

  1. Oct 12, 2004 #1
    can anyone tell me how to differentiate this function f(x)=2sin(x)+2x^x?
     
  2. jcsd
  3. Oct 12, 2004 #2

    matt grime

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    x^x is the same as exp{xlogx}

    you can now differentiate using the basic results you've been taught.
     
  4. Oct 14, 2004 #3
    Hello, dbzgtjh!
    First of all, we must know how to differentiate this function:
    [f(x)]^g(x) = h(x), so, ln{[f(x)]^g(x)} = ln h(x), is equal to
    g(x)lnf(x) = ln h(x)
    differentiating implicitly, we have, g'(x)lnf(x) + g(x)f'(x)/f(x) = h'(x)/h(x),
    therefore,
    h'(x) = h(x) [ g'(x)lnf(x) + g(x)f'(x)/f(x) ]
    In this case g(x) = x = f(x), so
    h(x) = x^x, so h'(x) = (x^x)(lnx + 1)
    Therefore,
    the derivative of function f is f'(x) = 2cosx + 2(x^x)(lnx + 1).
     
    Last edited: Oct 15, 2004
  5. Oct 14, 2004 #4

    arildno

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    Dearly Missed

    This is incorrect; you must apply the chain rule on the last term.
     
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