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I am having trouble with the following problem. I feel that I do understand the multivariable chain rule in general, but applying it here is more difficult. I am lost on where to start. Any help would be appreciated.

Let $f(x,y)$ be a twice continously differentiable function, and let $x=r\cos\theta$, $y=r\sin\theta$.

Show that

$$\frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2} =\frac{\partial^2 f}{\partial r^2} +\frac{1}{r^2} \frac{\partial^2 f}{\partial \theta^2} +\frac{1}{r} \frac{\partial f}{\partial r}$$

P.S. I apologize if my formatting is poor. This is my first time on this site.

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  • $\begingroup$ Can you find any of the partial derivatives involved? $\endgroup$ – Git Gud Mar 24 '14 at 21:21
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    $\begingroup$ Look here: math.ucdavis.edu/~saito/courses/21C.w11/polar-lap.pdf $\endgroup$ – user76568 Mar 24 '14 at 21:30
  • $\begingroup$ @GitGud I am unsure of where to start. I can take ∂f/∂r and I get ∂f/∂x[cosθ]+∂f/dy[sinθ] but get sort of stuck there. $\endgroup$ – Zach Mar 24 '14 at 21:34
  • $\begingroup$ @Dror thanks, taking a look now! $\endgroup$ – Zach Mar 24 '14 at 21:34
  • $\begingroup$ @Zach Just differentiate again with respect to $r$ and to the same for the others. Note that you can't find what $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ are, so you just drag them along in the computations and hope that in the end you can prove the equality. $\endgroup$ – Git Gud Mar 24 '14 at 21:36
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Here we actually compute the Right hand side and show that it is equivalent to the left.

Firstly two variable chain rule: $\frac{\partial}{\partial t}(h(g(t,s),f(t,s)))= \frac{\partial h}{\partial g}\frac{\partial g}{\partial t}+\frac{\partial h}{\partial f}\frac{\partial f}{\partial t}$

Now in our context we have $f(x(r,\theta),y(r,\theta))$, so:

$\frac{\partial f}{\partial r}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}=\frac{\partial f}{\partial x}\cos(\theta)+\frac{\partial f}{\partial y}\sin(\theta)$

$\frac{\partial^2 f}{\partial r^2}=\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial r}\cos(\theta)+\frac{\partial^2 f}{\partial y^2}\frac{\partial y}{\partial r}\sin(\theta)=\frac{\partial^2 f}{\partial x^2}\cos^2(\theta)+\frac{\partial^2 f}{\partial y^2}\sin^2(\theta)$

$\frac{\partial f}{\partial \theta}=\frac{\partial f}{\partial x}(-r\sin\theta)+\frac{\partial f}{\partial y}(r\cos\theta)$

$\frac{\partial^2 f}{\partial \theta^2} = -\frac{\partial f}{\partial x}r\cos(\theta)-\frac{\partial f}{\partial y}r\sin(\theta)+r^2\cos^2\theta\frac{\partial^2 f}{\partial x^2}+r^2\sin^2\theta\frac{\partial^2 f}{\partial y^2}$

So:

RHS = $\frac{\partial^2 f}{\partial x^2}\cos^2(\theta)+\frac{\partial^2 f}{\partial y^2}\sin^2(\theta)+\frac{1}{r^2}(-\frac{\partial f}{\partial x}r\cos(\theta)-\frac{\partial f}{\partial y}r\sin(\theta)+r^2\sin^2\theta\frac{\partial^2 f}{\partial x^2}+r^2\cos^2\theta\frac{\partial^2 f}{\partial y^2})$

$+\frac{1}{r}(\frac{\partial f}{\partial x}\cos(\theta)+\frac{\partial f}{\partial y}\sin(\theta))=\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial x^2}$

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