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Differerntiating Exponential Functions

  1. Jun 28, 2005 #1
    I really really need help with this one. This was a bonus question for one of my previous exams. I have no idea how to work through it.

    Differentiate the function:

    [tex]y= ( \ln x ) ^\cos x[/tex]
  2. jcsd
  3. Jun 28, 2005 #2


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    Do you know about logarithmic differentiation...?If you don't,are you willing to learn? :smile:

  4. Jun 28, 2005 #3
    Logarithmic differentiation:

    [tex]y = a^x[/tex]

    [tex]\ln y = \ln (a^x) = x\ln a[/tex]

    Then by the chain rule: [tex]\frac{1}{y} \frac{dy}{dx} = \ln a[/tex] (only true if a is a constant. If a = a(x), then you need to apply the chain and product rules to the RHS).

    So, [tex]\frac{dy}{dx} = y\ln a = a^x \ln a[/tex]
  5. Jun 28, 2005 #4
    Is this even close to the right answer? Or do I need to also multiply by the derivative of the exponent?

    [tex] ( \ln x )^\cos x [/tex] [tex] ( \ln ( \ln x) ) [/tex]
  6. Jun 28, 2005 #5


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    I misses a few terms.You have to use the Leibniz rule of differetiantion of products of functions.

  7. Jun 28, 2005 #6
    Nylex, question asked is the differential of



    [tex]y=\ln{\left (x^\cos{x} \right )}[/tex]
  8. Jun 28, 2005 #7


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    He didn't go for the function in the OP.He exemplified what is meant by logarithmic differentiation.

  9. Jun 28, 2005 #8
    True, but it was implied that you can use the [itex]\ln{\left (x^a\right )}=a\ln{x}[/itex] rule directly for the problem at hand.
  10. Jun 28, 2005 #9


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    In all The usual formulas elementary calculus students are expected to learn why is
    Rarely included. It is quite nice and it is easy to see using the chain rule.
    Or Individual function differential operators.
    where (Du)f(u,v)=Dx f(u(x),v(y))|y=x
    (Dv)f(u,v)=Dx f(u(y),v(x))|y=x
    Dx y=0
  11. Jul 4, 2005 #10
    If this is the bonus on the exams... your teacher is really nice..

    my bonus question on the test was
    Limit(x^x, x, 0)...
  12. Jul 4, 2005 #11


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    These should not be bonus questions.
    x^x:=exp(x log(x))
    x^x~1+x log(x)~1+|x|
    Also le Hopitals rule works nicely.
    What I want to know is did you mean the directed limit? You need complex numbers to consider x^x for x<0.
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