Given that $ r^2 = x^2 + y^2$, Compute $\frac{\partial ^2r}{\partial y \partial x } $

Question

I tried: $ \frac{\partial ^2r}{\partial y \partial x } = \frac{\partial}{\partial y} \left(\frac{\partial r}{\partial x} \right) $ which I know becomes $ \frac{\partial}{\partial y} \left(\frac{x}{r} \right) $. I then try to implicitly differentiate this with the quotient rule and get: $$ \frac{-y(r^2 +x^2)}{xr^3} $$

But after checking online, I know it should actually be much more simple:

$$ \frac{-xy}{r^3} $$

I'm not sure what I'm doing wrong. Any thoughts? Thank you!

Answer 1

Differentiating $r^2 = x^2 + y^2\Rightarrow r=\sqrt{x^2 + y^2} \ $, w.r.t. $x,y$ we get (by repeated use of the chain rule): $$ \frac{\partial ^2r}{\partial y \partial x } = \frac{\partial}{\partial y} \left(\frac{\partial \sqrt{x^2+y^2}}{\partial x} \right)= \frac{\partial}{\partial y}\bigg(\frac{2x}{2r}\bigg)= \frac{\partial}{\partial y}\bigg(\frac{x}{r}\bigg) \\ =x\frac{\partial}{\partial y}\bigg(\frac{1}{r}\bigg)=-\frac{x}{r^2}\frac{\partial r}{\partial y}=-\frac{x}{r^2}\frac{\partial \sqrt{x^2+y^2}}{\partial y}= \\ =-\frac{x}{r^2}\frac{2y}{2r}=-\frac{xy}{r^3} $$

Answer 2

Use quotient rule:

$$\frac{0r-xr_y}{r^2}$$

Note $r_y=\frac{y}r$