Partial Derivatives: Computing fxx and fyy in terms of fu, fv, fuu, fuv, fvv

[Orphen89]

Homework Statement

Let f = f (u,v) and u = x + y , v = x - y .

Assume f to be twice differentiable and compute fxx and fyy f in terms of fu, fv, fuu, fuv fvv.

The Attempt at a Solution

First off, this is an assignment question. I really do hate cheating, but I need help with this because I couldn't attend any of my lectures this week. Unfortunately, these lectures covered second order partial derivatives, and because there are no lecture notes for this subject I really am screwed. While the actual calculus work isn't a problem for me, I am having trouble finding an equation to actually work with - in my textbook, on the internet etc. there is usually an equation for f(u,v) with which you can directly differentiate the equation with. However, this question does not provide it in that particular format, instead giving the variables u and v and their own equations.

If anyone can help me get an equation down for f=f(u,v) then I should be able to work out the rest from there. Thanks in advance.

 

[Pengwuino]

You do not need to know what f(u,v) or f(x,y) is! Remember the chain rule:

<br /> \frac{{\partial f(u(x,y),v(x,y))}}{{\partial x}} = \frac{{\partial f}}{{\partial u}} \cdot \frac{{du}}{{dx}} + \frac{{\partial f}}{{\partial v}} \cdot \frac{{dv}}{{dx}} = f_u \frac{{du}}{{dx}} + f_v \frac{{dv}}{{dx}}<br />

Further applications of the chain rule allow you to determine the second derivatives without knowing explicitly what f(u,v) is.

 

[Orphen89]

Thanks for the help! For some stupid reason I thought that I should ONLY get numbers for the equation - I completely forgot/ignored the "in terms of" part of the question. Only after you put down the chain rule did I realize what I had to do >_<