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Log differentiation

  1. Jul 14, 2005 #1
    Using log differentiation, find dy/dx, in terms of x for the following:
    y = (cosx)^sinx

    any help wud be appreciated, im unsure of how to start this question, thx in advance
     
  2. jcsd
  3. Jul 14, 2005 #2

    dextercioby

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    Logarithmate and differentiate using the chain rule.

    Daniel.
     
  4. Jul 14, 2005 #3
    all i need help is in how to logarithmate it, if i understood that then i should be able to differentiate it easily
     
  5. Jul 14, 2005 #4

    Curious3141

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    [tex]a = e^{\ln{a}}[/tex] where 'ln' is natural log.
     
  6. Jul 14, 2005 #5

    Curious3141

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    Then use chain rule plus product rule. The other way is to differentiate implicitly after taking logs of both sides.
     
  7. Jul 14, 2005 #6

    Pyrrhus

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    Apply logarithm on both sides, and use the property of logs [itex] \ln a^b = b \ln a [/itex]
     
  8. Jul 14, 2005 #7
    ok. i think i got it

    y = (cosx)^sinx
    logy = log(cosx^sinx)
    logy=sinx(logcosx)
    (dy/dx)(1/(yln10)) = sinx(1/cosx(ln10) + log(cosx^cosx)
    (dy/dx)=[(tanx/ln10) + (log(cosx^cosx))](yln10)

    jus let me know if im on the right track. thanks for your help guys.
     
  9. Jul 14, 2005 #8

    Pyrrhus

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    I think so, but why this ln 10??

    [tex] \ln y = \sin x \ln (\cos x) [/tex]

    [tex] \frac{dy}{dx} \frac{1}{y} = \sin x \frac{-\sin x}{\cos x} + \cos x \\ln (\cos x) [/tex]

    [tex] \frac{dy}{dx} = (\cos x)^{\sin x}(\sin x \frac{-\sin x}{\cos x} + \cos x \ln (\cos x)) [/tex]
     
  10. Jul 14, 2005 #9
    isnt the derivative of log(base)x = 1/xln(base) ????
     
  11. Jul 14, 2005 #10

    Pyrrhus

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    Yes, you should have used natural logarithm to avoid that, but it's ok.
     
  12. Jul 14, 2005 #11
    how could u have used natural logs to avoid that?
     
  13. Jul 14, 2005 #12
    o nevermind stupid question lol. i should have converted log to ln then found the deriative.
     
  14. Jul 15, 2005 #13
    hah - "logarithmate"? I would prefer "logarithmize"
     
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