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

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!

• Note that $\frac{\partial}{\partial y} \Big(\tfrac{x}{r}\Big)=x\frac{\partial}{\partial y} \Big(\tfrac{1}{r}\Big)$ and then you can use the chain rule once more. – Alex Sep 15 '16 at 1:23

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): $$\frac{\partial ^2r}{\partial y \partial x } = \frac{\partial}{\partial y} \left(\frac{\partial \sqrt{x^2+y^2}}{\partial x} \right)= \frac{\partial}{\partial y}\bigg(\frac{2x}{2r}\bigg)= \frac{\partial}{\partial y}\bigg(\frac{x}{r}\bigg) \\ =x\frac{\partial}{\partial y}\bigg(\frac{1}{r}\bigg)=-\frac{x}{r^2}\frac{\partial r}{\partial y}=-\frac{x}{r^2}\frac{\partial \sqrt{x^2+y^2}}{\partial y}= \\ =-\frac{x}{r^2}\frac{2y}{2r}=-\frac{xy}{r^3}$$
• That makes sense mostly. I'm not too good at this stuff. Why is it that you have to bring the $x$ out in the fourth step? – amazonprime Sep 15 '16 at 1:58
• because it is -as a variable- independent of $y$ and the differentiation is w.r.t. $y$. – KonKan Sep 15 '16 at 1:59
$$\frac{0r-xr_y}{r^2}$$
Note $r_y=\frac{y}r$