# Second Order Implicit Partial Derivatives

• May 31st 2009, 10:55 PM
wceagles10
Second Order Implicit Partial Derivatives
This problem is doing my head in...

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

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

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

I've got $\displaystyle dz/dy$ and $\displaystyle dz/dx$ 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 $\displaystyle d^2z/dxdy$, and no amount of googling it or reading the textbook seems to yield any more information.

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

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

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.
• May 31st 2009, 11:17 PM
Moo
Hello,

$\displaystyle \frac{d^2z}{dxdy}$ means that you differentiate $\displaystyle \frac{dz}{dy}$ 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 $\displaystyle \frac{dz}{dy}$, you'll have some terms containing $\displaystyle \frac{dz}{dx}$, which you will substitute by what you found.
• May 31st 2009, 11:46 PM
wceagles10
It shouldn't matter which way they're differentiated, [ie $\displaystyle dz/dy$ with respect to $\displaystyle x$, or $\displaystyle dz/dx$ with respect to $\displaystyle y$], since the answers would be equal, would they not?

Either way, I have tried differentiating $\displaystyle dz/dx$ with respect to $\displaystyle y$, and - as you mentioned - got an answer with a $\displaystyle dz/dy$ term in it. So you think the way to go would be to just sub in the original answer for $\displaystyle dz/dy$? 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.
• Jun 1st 2009, 12:00 AM
Moo
Quote:

Originally Posted by wceagles10
It shouldn't matter which way they're differentiated, [ie $\displaystyle dz/dy$ with respect to $\displaystyle x$, or $\displaystyle dz/dx$ with respect to $\displaystyle y$], 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 $\displaystyle dz/dx$ with respect to $\displaystyle y$, and - as you mentioned - got an answer with a $\displaystyle dz/dy$ term in it. So you think the way to go would be to just sub in the original answer for $\displaystyle dz/dy$? 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 $\displaystyle e^y \left(z+x\frac{dz}{dy}\right)+3yz^2\frac{dz}{dy}+z ^3=0$, 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 !