Second Order Implicit Partial Derivatives

wceagles10 · #1 $Old$ June 1st, 2009, 12:55 AM

This problem is doing my head in...

If $z$ is defined implicitly as a function of $x$ and $y$ by the equation

$xze^y + yz^3 = 1$ ,

find the values of $dz/dx, dz/dy$ and $d^2z/dxdy$ when $x=1$ and $y=0$ .

I've got $dz/dy$ and $dz/dx$ although these were worked out fairly quickly without double checking, but irregardless I can't for the life of me figure out how to get $d^2z/dxdy$ , and no amount of googling it or reading the textbook seems to yield any more information.

$dz/dy = (-z^3-xze^y)/(3yz^2 + xe^y)$

$dz/dx = -(ze^y)/(xe^y+3yz^2)$

Any help would be hugely appreciated, as this question is part of my exam preparation, and knowing my luck, a similar question is likely to come up in the actual exam.

Moo · #2 $Old$ June 1st, 2009, 01:17 AM

Hello,

$\frac{d^2z}{dxdy}$ means that you differentiate $\frac{dz}{dy}$ with respect to x. (maybe for some people it's the other way round : diff dz/dx with respect to y...that will be depending on how you've been taught)

Don't forget that z is still a function of x. So if you differentiate $\frac{dz}{dy}$ , you'll have some terms containing $\frac{dz}{dx}$ , which you will substitute by what you found.

wceagles10 · #3 $Old$ June 1st, 2009, 01:46 AM

It shouldn't matter which way they're differentiated, [ie $dz/dy$ with respect to $x$ , or $dz/dx$ with respect to $y$ ], since the answers would be equal, would they not?

Either way, I have tried differentiating $dz/dx$ with respect to $y$ , and - as you mentioned - got an answer with a $dz/dy$ term in it. So you think the way to go would be to just sub in the original answer for $dz/dy$ ? It sounds like a sensible idea, and one I had considered, but for some reason didn't think it'd work [although I had become quite frustrated with the problem by the time I got around to trying it, so maybe didn't give it a proper go].

Cheers for the response - I'll try it and let you know how I go.

Moo · #4 $Old$ June 1st, 2009, 02:00 AM

Quote:

Originally Posted by wceagles10 $View Post$

It shouldn't matter which way they're differentiated, [ie $dz/dy$ with respect to $x$ , or $dz/dx$ with respect to $y$ ], since the answers would be equal, would they not?

Oops, yes... I guess so. I don't exactly remember, but you must be right.

Quote:

Either way, I have tried differentiating $dz/dx$ with respect to $y$ , and - as you mentioned - got an answer with a $dz/dy$ term in it. So you think the way to go would be to just sub in the original answer for $dz/dy$ ? It sounds like a sensible idea, and one I had considered, but for some reason didn't think it'd work [although I had become quite frustrated with the problem by the time I got around to trying it, so maybe didn't give it a proper go].

Yes, you just have to sub.

For the second derivative... It may be easier to work on $e^y \left(z+x\frac{dz}{dy}\right)+3yz^2\frac{dz}{dy}+z^3=0$ , taking its derivative wrt x.
In general, I prefer applying the product and addition rules of differentiation rather than the quotient rule...less mistakes. It's as you wish !