Partial Derivatives of Power Functions

[SwaGGeReR]

For a function such as

w=5x^y/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

w_y=5*(x^y/z/z)*ln(x)

 

[micromass]

Do you know how to differentiate things like

$$y=a^x$$

(with a constant) or

$$y=e^x$$

to x?? This problem is exactly the same as the one you've given

 

[SwaGGeReR]

To differentiate y=a^x I would use logarithmic differentiation.

1) Take the ln of both sides: ln(y)=ln(a^x)

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'=a^x*ln(a)



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



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

w=5x^y/z

1) Taking ln of both sides: ln(w)=ln(5x^y/z)

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

However, ln(5x^y/z) ≠ (y/z)*ln(5x)

 

[slider142]

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).

 

[SwaGGeReR]

Thanks slider.

I originally thought that ln(ab^x)=xln(ab), but apparently that's not true. Yet ln(a^x)=xln(a). Why is this so?

 

[lugita15]

Thanks slider.

I originally thought that ln(ab^x)=xln(ab), but apparently that's not true. Yet ln(a^x)=xln(a). Why is this so?

ln((ab)^x)=xln(ab), but ln(a(b^x))≠xln(ab). Does that make sense?

 

[SwaGGeReR]

Got it. Thanks!