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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

    dextercioby

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

    Daniel.
     
  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

    dextercioby

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

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

    [tex]y=(\ln{x})^\cos{x}[/tex]

    not

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

    dextercioby

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

    Daniel.
     
  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

    lurflurf

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    In all The usual formulas elementary calculus students are expected to learn why is
    (u^v)'=v*(u^(v-1))*v'+(u^v)*log(u)*v'
    Rarely included. It is quite nice and it is easy to see using the chain rule.
    Or Individual function differential operators.
    D(u^v)=(Du+Dv)(u^v)=(Du)(u^v)+(Dv)(u^v)
    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

    lurflurf

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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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