# Partial Derivatives: Changing to Polar Coordinates

A function say $f$ of $x$, $y$ is away from the origin. This function can be written in polar coordinates as a function of $r$ and $\theta$. Now, if we know what $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$, how can we find $\frac{\partial f}{\partial r}$ and $\frac{\partial f}{\partial \theta}$ and vice versa. Additionally, if we know what $\frac{\partial^2 f}{\partial x^2}$, $\frac{\partial^2 f}{\partial x \partial y}$, and $\frac{\partial^2 f}{\partial y^2}$ is, how can we find $\frac{\partial^2 f}{\partial r^2}$, $\frac{\partial^2 f}{\partial r \partial \theta}$, and $\frac{\partial^2 f}{\partial \theta^2}$ and vice versa.

You can consider it to be a derivative of a composite function. Given $$f(x,y)$$ And any differentiable transformation $(x,y) \to (u,v)$ $$x = g(u,v),$$ $$y = h(u,v)$$ you can rewrite $$f(x,y) = f(g(u,v), h(u,v))$$ and so $$\frac{\partial f(x,y)}{\partial u} = \frac{\partial f(g(u,v),h(u,v))}{\partial u} = \frac{\partial f(x,y)}{\partial x} \frac{\partial g(u,v)}{\partial u} + \frac{\partial f(x,y)}{\partial y} \frac{\partial h(u,v)}{\partial u}$$ In particular, if you change from cartesian to polar coordinates $$x = r\cos\theta,$$ $$y = r\sin\theta$$ you get $$\frac{\partial f(x,y)}{\partial r} = \frac{\partial f(x,y)}{\partial x} \frac{\partial (r\cos\theta)}{\partial r} + \frac{\partial f(x,y)}{\partial y} \frac{\partial (r\sin\theta)}{\partial r} = \frac{\partial f(x,y)}{\partial x} \cos\theta + \frac{\partial f(x,y)}{\partial y} \sin\theta$$ Example: $$f(x,y) = x^2y = r^3\cos^2\theta\sin\theta$$ direct derivation will give you $$\frac{\partial f}{\partial r} = 3r^2\cos^2\theta\sin\theta$$ By the formula above you get: $$\frac{\partial f}{\partial r} = \frac{\partial f(x,y)}{\partial x} \cos\theta + \frac{\partial f(x,y)}{\partial y} \sin\theta = 2xy \cos\theta + x^2 \cos\theta= 2r^2\cos^2\theta\sin\theta + r^2\cos^2\theta\sin\theta=3r^2\cos^2\theta\sin\theta$$

This can be generalised to any number of variables and any differentiable transformation of coordinates.

You can use the fact that $x = rCos\theta$ and $y = rSin\theta$ and the Jacobian to transform it. Let me know if anything isn't clear.

For example if f is a function of x and y, you can express f in terms of $r$ and $\theta$ and then find those partial derivatives. Perhaps a more specific example might help.

e.g.

If $f(x, y) = xy$, then $f(r, \theta) = rCos\theta\times rSin\theta = r^2 Sin\theta Cos\theta = \frac{r^2}{2}Sin2\theta$

Once written in terms of $r$ and $\theta$, it's very straight forward to find those partial derivatives.

• I havent learned Jacobian yet.
– Anne
Jul 8, 2015 at 10:20
• I'll try to explain that further. Jul 8, 2015 at 10:21
• Can you post the specific example you're trying to solve? It'll be easier for me to deal with that. Jul 8, 2015 at 10:26
• Its not a problem. It is something I am having trouble understanding.
– Anne
Jul 8, 2015 at 10:27
• I've edited my post to explain more. Jul 8, 2015 at 10:29