Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

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 $\frac{\partial}{\partial s}(2r)=0$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.