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If $z = f(x,y)$ has continuous second order partial derivatives and $x = r^2 + s^2$ and $y = 2rs$, find $$ \frac{\partial^2 z}{\partial r^2} $$ So, $$ \frac{\partial^2 z}{\partial r^2} = \frac{\partial}{\partial r} \left(\frac{\partial z}{\partial r}\right) = \frac{\partial}{\partial r}\left(2r \frac{\partial z}{\partial x} + 2s \frac{\partial z}{\partial y}\right) = 2\frac{\partial z}{\partial x} + 2r\frac{\partial}{\partial r}\left(\frac{\partial z}{\partial x}\right) + 2s\frac{\partial }{\partial r}\left(\frac{\partial z}{\partial y}\right) $$

I understand that the product rule has been used to attain the last equality, however, this would imply that $$\frac{\partial }{\partial r} (2s) = 0$$ Why? Can I not write $s$ as a function of $r$ by rearranging the relations I have at the top of the page? Many thanks

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1 Answer 1

CAF, no, the way the functions have been described in the question imply that:

  • $z$ is a function of $x$ & $y$,

  • $x$ is a function of $r$ & $s$, and

  • $y$ is a function of $r$ & $s$.

The question does not say that $s$ is a function of $r$ (or vice versa). So we can view them to be independent variables. So $\frac {\partial}{\partial r}(2s)=0$.

If you wanted to find $\frac {\partial ^2z}{\partial s^2}$, you would also use that $\frac{\partial}{\partial s}(2r)=0$.

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Why can I not write say $s = (+/-)\sqrt{x - r^2}$? –  CAF Dec 17 '12 at 16:20
Anyone? Thanks! –  CAF Dec 17 '12 at 17:09

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