Partial Derivatives of Power Functions

  • Thread starter SwaGGeReR
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  • #1
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Main Question or Discussion Point

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)
 

Answers and Replies

  • #2
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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
 
  • #3
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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)
 
  • #4
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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).
 
  • #5
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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?
 
  • #6
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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?
ln((ab)x)=xln(ab), but ln(a(bx))≠xln(ab). Does that make sense?
 
  • #7
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Got it. Thanks!
 

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