# product rule for partial derivative conversion

Working problems in Colley's Vector Calculus and I'm refreshing on partial derivatives, in particular the product rule and chain rules. I've come across a problem where we're identifying cartesian partials with polar partials under the identities $$\frac{\partial}{\partial{x}}=\cos{\theta}\frac{\partial}{\partial{r}}-\frac{\sin{\theta}}{r}\frac{\partial}{\partial{\theta}}$$ $$\frac{\partial}{\partial{y}}=\sin{\theta}\frac{\partial}{\partial{r}}+\frac{\cos{\theta}}{r}\frac{\partial}{\partial{\theta}}$$ The example I'm working asks me to find both $\frac{\partial^2}{\partial{x^2}}$ and $\frac{\partial^2}{\partial{y^2}}$ using the second order partials $\frac{\partial^2}{\partial{r^2}},\frac{\partial^2}{\partial{\theta^2}},\frac{\partial^2}{\partial{r}\partial{\theta}},\frac{\partial}{\partial{r}},$ and $\frac{\partial}{\partial{\theta}}$ So I start the problem as such; $$\frac{\partial^2}{\partial{x^2}}=\frac{\partial}{\partial{x}}\left(\frac{\partial}{\partial{x}}\right)=\frac{\partial}{\partial{x}}\left(\cos{\theta}\frac{\partial}{\partial{r}}-\frac{\sin{\theta}}{r}\frac{\partial}{\partial{\theta}}\right)$$And it's from here that I become confused. I believe it's time to use the product rule, but how do I go about using the product rule with regards to the variables involved?

EDIT: Is the next move to again substitute $\cos{\theta}\frac{\partial}{\partial{r}}-\frac{\sin{\theta}}{r}\frac{\partial}{\partial{\theta}}$ again and expand and condense?

I'll do one of the terms, and leave the other to you:

$$\frac{\partial}{\partial x}\left(\cos\theta\frac{\partial}{\partial r}\right)=\frac{\partial(\cos\theta)}{\partial x}\cdot\frac{\partial}{\partial r}+\cos\theta\cdot\frac{\partial}{\partial x}\frac{\partial}{\partial r}$$

Remembering that $\cos\theta=x/\sqrt{x^2+y^2}$ we find that

$$\frac{\partial(\cos\theta)}{\partial x}=\frac{y^2}{(x^2+y^2)^{3/2}}=\frac{\sin^2\theta}{r}$$

And using the identity you've provided, we have:

$$\frac{\partial}{\partial x}\frac{\partial}{\partial r}=\left(\cos{\theta}\frac{\partial}{\partial{r}}-\frac{\sin{\theta}}{r}\frac{\partial}{\partial{\theta}}\right)\frac{\partial}{\partial r}=\cos{\theta}\frac{\partial^2}{\partial{r}^2}-\frac{\sin{\theta}}{r}\frac{\partial^2}{\partial{\theta}\partial r}$$

Putting everything together we have:

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

Note: The method you've indicated in your edit should work as well. You can always do both methods to check your work.

• I really understand your method. When I worked my method, either through error or whatever, I can't get the single derivative terms for $\frac{\partial}{\partial{r}}$ and $\frac{\partial}{\partial{\theta}}$, but seeing your method was enlightening. I'm not sure I would have connected turning cosine into it's cartesian counterpart, perhaps after trial and error, but I see how this works now. Thank you! Commented Aug 8, 2013 at 17:17