Let $z$ be a function of $u$ and $v$ where $u=x+y$ and $v=3x-3y$.

I have previously shown, by the Chain Rule, that $$\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}= \left(\frac{\partial z}{\partial u}\right)^2 - 9 \left(\frac{\partial z}{\partial v}\right)^2.$$

Now, "assuming equality of mixed second-order partial derivatives", I must show that $$\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 2\frac{\partial^2 z}{\partial u^2} + 18\frac{\partial^2 z}{\partial v^2}.$$

Firstly, I don't understand what is meant by "equality of mixed second-order partial derivatives". Please could somebody explain this?

I think I must use the Chain Rule again to get this equation but I don't really understand how to do this. I just need a pointer to get me started. Thanks.


1 Answer 1


Equality of mixed second-order partial derivatives means $$\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial^2 z}{\partial y \partial x}.$$ You will probably need that because $$\frac{\partial^2 z}{\partial x \partial y} - \frac{\partial^2 z}{\partial y \partial x} = 0.$$ Continue using the chain rule.


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