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Homework Help: Help on logarithmic differentiation problem

  1. Jan 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Use logarithmic differentiation to find the derivative of the function.

    2. Relevant equations

    y = (sin(x))^(ln(x))

    3. The attempt at a solution

    I guessing the first step is raise both sides to "e", but so far I have only completed problems by taking the natural log of both sides.

    (1) e^(y) = e^(sin(x))^(ln(x))

    (2) d/dx (e^y) = d/dx (e^(sin(x))^(ln(x)))

    (3) Can I start moving exponents in front of "e" to use the product rule, if so, do I bring "sin(x)^(ln(x)) down?, or am I on the completely wrong track?
  2. jcsd
  3. Jan 23, 2009 #2


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    Hi crm08! :smile:
    = e^(sin(x)*ln(x)) :wink:
  4. Jan 23, 2009 #3
    Oh I gotcha, so now use:

    d/dx (e^g(x)) = e^g(x) * g'(x), where g(x) = (sin(x))(ln(x)), and use product rule for g'(x)

    I think I'm on the right track
  5. Jan 23, 2009 #4
    use ln not e, since [tex]\ln{x^a} =a \ln{x}[/tex]
    So that [tex]\ln{y} = \ln( \sin{x}^{\ln{x}} ) = \ln(x) \ln(sin{x})[/tex]
  6. Jan 23, 2009 #5
    tiny-tim, I'm confused, why would [tex]e^y = e^{(\sin(x)^{\ln(x)})}[/tex] equals [tex]e^{(\sin(x)*\ln(x))}[/tex]
  7. Jan 23, 2009 #6
    ok i'm getting

    (1/y) * dy/dx = (ln(x))*(1/sin(x))*(cos(x)) + (ln(sin(x))*(1/x)

    = y*[(ln(x))*(tan(x)) + ((ln(sin(x))) / x)]

    am I getting any closer?
  8. Jan 23, 2009 #7

    Gib Z

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    It wouldn't. In general, (a^b)^c = a^(bc), but not a^(b^c). Take an example using a=3, b=2, c=3.

    (a^b)^c = 729= a^(bc). a^(b^c) = 3^8, different.

    I'm sure tiny-tim just got a bit confused.

    As for crm08 - the tan should be a cot, and you should not use an equals signs when you actually multiplied by y, but other than that its looking sweet.
  9. Jan 23, 2009 #8


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    mmm … all those ^s :redface:

    it'll be a lot easier if everyone uses the X2 tag (just above the reply box) :wink:

    let's see, it should have been …

    sinxlnx = (eln(sinx))lnx = eln(sinx)*lnx
  10. Jan 23, 2009 #9
    ahhh, so you were referring to y and not [tex]e^y[/tex] that equals [tex]e^{\ln(sinx)*\ln{x}}[/tex].
    That makes sense now!
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