# Unusual function format and its partial derivatives.

I came across a function of this format:
$z = f(u,v)$ where $u = x^2y^2$ and $v = 5x + 1$

Because this function is not in the same format of the ones I've seen before (explicit or implicit), I don't know how to find (or even to do a single solving step!) its partial derivatives.

Basically, I need to show that:
$\frac{\partial^2z}{\partial x\partial y} = 4xy\frac{\partial f}{\partial u} +4x^3y^3\frac{\partial^2f}{\partial u^2} +10x^2y\frac{\partial^2f}{\partial v \partial u}$

Anyone could tell me how I should approach this type of problem?
Thanks!

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@julien I believe that's part of his problem. –  Git Gud Feb 19 '13 at 22:18
@julien No. But I rarely understand analysts gibberish so what I see is irrelevant. –  Git Gud Feb 19 '13 at 22:20
I think you may need to quote more of the context ... –  Hagen von Eitzen Feb 19 '13 at 22:21
@wwwe Do you mean $u=x^2y^2$ and $v=5x+1$? –  Git Gud Feb 19 '13 at 22:37
yes, i will edit right away! –  wwwe Feb 19 '13 at 22:37

Yes this is a function. You can think of this as a composition of two functions

$z=f(u,v)=f(x^2y^2,5x+1)=g(x,y)$

and to answer your question, you need the chain rule for multivariable functions...applied twice. The chain rule in this case is

$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}$

and then similarly

$\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}.$

So first take the partial w.r.t $y$

$\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}=\frac{\partial f}{\partial u}\cdot2x^2y+\frac{\partial f}{\partial v}\cdot0=2x^2y\cdot\frac{\partial f}{\partial u}$

and then apply $\frac{\partial}{\partial x}$ to this quantity using the product and the chain rule and then simplify.

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I don't understand why should I apply the chain rule twice... And I'm not sure what df/du represents! –  wwwe Feb 23 '13 at 2:23
You need the chain rule twice because you are taking partials twice. And df/du means exactly what it means in your question, it is the instantaneous change in f with respect to u. –  Fixed Point Feb 23 '13 at 3:02