I am getting very confused when trying to find the partial derivative operators in polar co-ordinates. For example, I need to show that $\partial_{x}=\partial_{r}cos(\theta)-\frac{sin(\theta)}{r}\partial_{\theta}$, given that $x=rcos\theta, y=rsin\theta$

I started by using the chain rule to give that $\partial_x=\partial_r\frac{\partial r}{\partial x}+\partial_\theta\frac{\partial \theta}{\partial x}$ and then I went to fine the two partials I could and I am pretty sure this is where I went wrong but I don't understand why it doesn't work. $x=rcos\theta\\\frac{\partial x}{\partial r}=cos\theta\\\frac{\partial r}{\partial x}=\frac{1}{cos\theta}$

and similarly for $\theta$ I got that $\frac{\partial \theta}{\partial x}=\frac{-1}{rsin\theta}$ But I know what I should get and this doesn't get me there. I looked online and saw things with people using the fact that $tan\theta=\frac{y}{x}$ but I don't understand why my way doesn't seem to work and I am just getting confused.


1 Answer 1


As to why $\frac{\partial r}{\partial x}\ne\frac{1}{\frac{\partial x}{\partial r}}$ this question may be helpful (Partial derivatives inverse question).

Now to find $\partial_{x}$ and $\partial_{y}$, you need to use the inverse transformations $r=\sqrt{x^2+y^2}$ and $ \theta=arctan(\frac{y}{x})$

if we differentiate the expression for $r$, with respect to $x$ and $y$ we get $$\frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}=\frac{rcos(\theta)}{r}=cos\theta$$ and $$ \frac{\partial r}{\partial y}=\frac{y}{\sqrt{x^2+y^2}}=\frac{rsin(\theta)}{r}=sin\theta $$

and if we differentiate the expression for $\theta$, with respect to $x$ and $y$ we get $$\frac{\partial \theta}{\partial x}=\frac{1}{1+\frac{y^2}{x^2}}(\frac{-y}{x^2})=-\frac{y}{x^2+y^2}=-\frac{rsin(\theta)}{r^2}=-\frac{sin\theta}{r} $$ $$\frac{\partial \theta}{\partial x}=\frac{1}{1+\frac{y^2}{x^2}}(\frac{1}{x})=\frac{x}{x^2+y^2}=\frac{rcos(\theta)}{r^2}=\frac{cos\theta}{r} $$ you then substitute these expressions in to the chain rule expressions you have. Hopefully that helps


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