# Doubt chain rule for second order partial derivative

I am doing an exercise in which I am given a function $$z(x,y)$$ and I am introduced the "change of variables" $$x=u\cos v$$ and $$y=u\sin v$$. I am asked to calculate $$\frac{\partial z}{\partial u}$$ and applying chain rule I have done $$\frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u}=\frac{\partial z}{\partial x}\cos v+\frac{\partial z}{\partial y}\sin v$$ but with respect to second order partial derivative, it could be obtained like this?:

$$\frac{\partial^2 z}{\partial u^2}=\frac{\partial }{\partial u}(\frac{\partial z}{\partial u})=\frac{\partial }{\partial x}(\frac{\partial z}{\partial u})\frac{\partial x}{\partial u}+\frac{\partial }{\partial y}(\frac{\partial z}{\partial u})\frac{\partial y}{\partial u}=\frac{\partial }{\partial x}(\frac{\partial z}{\partial x}\cos v+\frac{\partial z}{\partial y}\sin v)\frac{\partial x}{\partial u}+\frac{\partial }{\partial y}(\frac{\partial z}{\partial x}\cos v+\frac{\partial z}{\partial y}\sin v)\frac{\partial y}{\partial u}$$

And if that is correct, when doing the derivatives, if for example $$\frac{\partial x}{\partial v}=-u\sin v$$, then we have that $$\frac{\partial v}{\partial x}=-\frac{1}{u\sin v}$$? Thanks for your help.

• The formula for $\frac{\partial^\color{red}{2}z}{\partial u^2}$ is correct. What has this to do with $\frac{\partial x}{\partial v}$ ? Nov 2, 2022 at 16:02

We can use the notation $$z_{x}=\frac{\partial z}{\partial x}$$.
We have, \begin{align*} z_{uu}&=(z_{x}x_{u}+z_{y}y_{u})_{u}\\ &=(z_{x})_{u}x_{u}+z_{x}x_{uu}+(z_{y})_{u}y_{u}+z_{y}y_{uu}\\ &=\underbrace{(z_{xx}x_{u}+z_{xy}y_{u})}_{(z_{x})_{u}}x_{u}+z_{x}x_{uu}+\underbrace{(z_{yx}x_{u}+z_{yy}y_{u})}_{(z_{y})_{u}}y_{u}+z_{y}y_{uu}. \end{align*}Since $$z$$ is suppose to be $$C^{2}$$, then $$z_{xy}=z_{yx}$$, so $$z_{uu}=z_{xx}x^{2}_{u}+2z_{xy}x_{u}y_{u}+z_{yy}y^{2}_{u}+z_{x}x_{uu}+z_{y}y_{uu}$$ Now, notice that $$x_{u}=(u+\cos v)_{u}=1,\quad y_{u}=(u+\sin v)_{u}=1, x_{uu}=0, \quad y_{uu}=0$$
Partial differentials cannot be flipped the same way that total differentials can, because the numerator and the denominator have different meanings. The $$\partial v$$ in $$\frac{\partial v}{\partial x}$$ is a DIFFERENT $$\partial v$$ than the one in $$\frac{\partial x}{\partial v}$$.