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Partial Derivatives of Power Functions

  1. Apr 17, 2012 #1
    For a function such as

    w=5xy/z

    How would you find the partial derivative of w with respect to y or z? I've tried using basic logarithmic differentiation, but can't arrive at the correct answer. For reference, the correct answer is

    wy=5*(xy/z/z)*ln(x)
     
  2. jcsd
  3. Apr 17, 2012 #2

    micromass

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    Do you know how to differentiate things like

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

    (with a constant) or

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

    to x?? This problem is exactly the same as the one you've given
     
  4. Apr 17, 2012 #3
    To differentiate y=ax I would use logarithmic differentiation.

    1) Take the ln of both sides: ln(y)=ln(ax)

    2) Using properties of logs, this equals: ln(y)=x*ln(a)

    3) Differentiating each side: y'/y=ln(a)

    4) Solving for y': y'=y*ln(a)

    5) Substituting for y: y'=ax*ln(a)



    The derivative of y=ex is simply y'=ex*x', where x'=1.



    Not sure how knowing these helps solve this problem though...here's my attempt:

    w=5xy/z

    1) Taking ln of both sides: ln(w)=ln(5xy/z)

    2) Using properties of logs: ln(w)=(y/z)*ln(5x)

    However, ln(5xy/z) ≠ (y/z)*ln(5x)
     
  5. Apr 17, 2012 #4
    There are several ways you can get around this. One way is to take the logarithm of w/5 instead of w. Another is to use the fact that ln(ab) = ln(a) + ln(b). Yet a third way is to note that you already know how to differentiate h(y) = x^(y/z) and use the linearity of differentiation:in particular, the derivative of c*h(y) for any constant c is just c times the derivative of h(y).
     
  6. Apr 17, 2012 #5
    Thanks slider.

    I originally thought that ln(abx)=xln(ab), but apparently that's not true. Yet ln(ax)=xln(a). Why is this so?
     
  7. Apr 18, 2012 #6
    ln((ab)x)=xln(ab), but ln(a(bx))≠xln(ab). Does that make sense?
     
  8. Apr 18, 2012 #7
    Got it. Thanks!
     
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