OpenStudy (anonymous):

w=(xy^2)/(z). Let D_x denote differentiation with respect to x, D_y with respect to y, and D_z with respect to z. Thus, when we compute D_x w we consider y and z as constants, and similarly for D_y and D_z. Find D_zw

OpenStudy (anonymous):

I keep using the quotient rule, but my homework isn't taking it.

zepdrix (zepdrix):

$\Large\rm w=\frac{xy^2}{z}=\left(\frac{y^2}{z}\right)x$So when you're taking your partial with respect to x, you can think of that whole part in front as a constant.

zepdrix (zepdrix):

$\Large\rm w_x=D_x\left(c\right)x$

OpenStudy (anonymous):

Wait, is this grabbing the derivative? Because I had to find D_xw and D_yw, and I was able to get those.

zepdrix (zepdrix):

Oh you're only stuck on the D_zw portion?

OpenStudy (anonymous):

Yessir, sorry, I should have specified.

zepdrix (zepdrix):

$\Large\rm w=\frac{xy^2}{z}$$\Large\rm w=(xy^2)\frac{1}{z}$$\Large\rm w=(xy^2)z^{-1}$Taking our partial derivative with respect to z,$\Large\rm D_z\left[w\right]=D_z\left[(c)z^{-1}\right]$Again, we hold the other variables constant.

zepdrix (zepdrix):

So just uhhh, power rule, yes? :)

OpenStudy (anonymous):

Oh, it would just be power rule? I was doing quotient?

zepdrix (zepdrix):

You CAN do quotient rule if you want. It ends up giving you a little extra work though.

zepdrix (zepdrix):

$\Large\rm \frac{d}{dz}\frac{c}{z}=\frac{c'z-c (z)'}{z^2}=\frac{0z-c(1)}{z^2}$

zepdrix (zepdrix):

The important thing is that when you take the derivative of your $$\Large\rm (xy^2)$$ with respect to z, you remember that it's giving you zero. That's why I'm calling it C instead, so I don't forget.

OpenStudy (anonymous):

Ok, was wondering what C was, haha.

zepdrix (zepdrix):

XD

OpenStudy (anonymous):

So is that the final answer? I'm not sure I completely understand. would it be xy^2/z^2?

zepdrix (zepdrix):

(xy^2) is CONSTANT when we're taking our z derivative. We don't usually apply product and quotient rule with a constant and variable. $\Large\rm w=\color{orangered}{(xy^2)}z^{-1}$So we treat this whole orange part as a constant when we take our z derivative,$\Large\rm w=\color{orangered}{c}z^{-1}$$\Large\rm w_z=\color{orangered}{c}(-z^{-2})$

zepdrix (zepdrix):

$\Large\rm w_z=-\color{orangered}{(xy^2)}z^{-2}$

zepdrix (zepdrix):

Looks like you're missing a negative.

zepdrix (zepdrix):

Too confusing? :U

OpenStudy (anonymous):

that totally makes sense! Got the answer right. Thanks!

OpenStudy (anonymous):

So, is that a rule we can use in general to avoid using the quotient rule, is to take the denominator and raise it to the -1 power to bring it to the top, then use product/whatever else rule?

zepdrix (zepdrix):

Oh yes, I guess you need to make sure you're comfortable with these types of things:$\Large\rm \frac{1}{x}=x^{-1}$Which allows you to avoid the quotient rule with the 1 and x. The most important thing to realize though, is that your (xy^2) are not involved in the differentiation process when taking partial with respect to z. With partials, when the letter doesn't match our variable we just sort offfffffff ignore it. Bah that's a sloppy way of saying it I suppose lol

zepdrix (zepdrix):

If you're doing partials derivatives you REALLY should be comfortable with negative exponents at this point >.< I would hope so at least. Try to get comfortable with this "treating variables as constant idea", it takes a little getting used to.

OpenStudy (anonymous):

Thanks so much, that makes a lot of sense!