# Using chain rule to derive 2nd derivative

1. May 1, 2013

### mrcleanhands

1. The problem statement, all variables and given/known data

Use $\frac{\partial z}{\partial r}=\cos\theta\frac{\partial z}{\partial x}+\sin\theta\frac{\partial z}{\partial y}$
and $\frac{\partial z}{\partial\theta}=-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y}$ to show that

$\frac{1}{r^{2}}\frac{\partial^{2}z}{\partial\theta^{2}}+\frac{1}{r} \frac{\partial z}{\partial r}=\sin^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}-2\sin\theta\cos\theta\frac{\partial ^ {2}z}{\partial x \partial y}+\cos^{2}\theta\frac{\partial ^ {2}z}{\partial y^{2}}$
2. Relevant equations

3. The attempt at a solution
$\frac{\partial z}{\partial\theta}\frac{\partial z}{\partial\theta}=\frac{\partial z}{\partial\theta}(-r\sin\theta\frac{\partial z}{\partial x}+r\cos\theta\frac{\partial z}{\partial y})$

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

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

$\frac{1}{r^{2}}\frac{\partial^{2}z}{\partial\theta^{2}}=\sin^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}-2\sin\theta\cos\theta\frac{\partial^{2}z}{\partial x\partial y}+\cos^{2}\theta\frac{\partial^{2}z}{\partial y^{2}}$

but $\frac{1}{r}\frac{\partial z}{\partial r}=\frac{1}{r}\cos\theta\frac{\partial z}{\partial x}+\frac{1}{r}\sin\theta\frac{\partial z}{\partial y}$

and if I add that I get:
$\frac{1}{r^{2}}\frac{\partial^{2}z}{\partial\theta^{2}}+\frac{1}{r} \frac{ \partial z}{\partial r}=\sin^{2}\theta\frac{\partial^{2}z}{\partial x^{2}}-2\sin\theta\cos\theta\frac{\partial^{2}z}{\partial x\partial y}+\cos^{2}\theta\frac{\partial^{2}z}{\partial y^{2}}+\frac{1}{r}\cos\theta\frac{\partial z}{\partial x}+\frac{1}{r}\sin\theta\frac{\partial z}{\partial y}$ so somehow $\frac{1}{r}\frac{\partial z}{\partial r}=\frac{1}{r}\cos\theta\frac{\partial z}{\partial x}+\frac{1}{r}\sin\theta\frac{\partial z}{\partial y}$ is supposed to be 0???

Last edited: May 1, 2013
2. May 1, 2013

### HallsofIvy

Staff Emeritus
What? When I saw "$\frac{dz}{d\theta}\frac{dz}{d\theta}$" above I started to write "No, that's the wrong notation- that means the product" but now it appears that you really are just multiplying the first derivative with itself. You do understand that this is NOT what "[itex]\frac{d^2z}{d\theta^2}" means, don't you?