Find $\frac{\partial^2f}{\partial r\partial s}$ for $f(x(r,s),y(r,s))$.

Question

Given a function of $x$ and $y$, $f(x,y)$, where $x$ and $y$ are both implicitly functions of $r$ and $s$, how can I find the second mixed partial derivative $\frac{\partial^2f}{\partial r\partial s}$ of $f$?

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Differentiating first by $s$ via basic chain rule gives

$$\frac{\partial f}{\partial s} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial s}$$

But next differentiating by $r$ is where I get stuck. I know $\frac{\partial x}{\partial s}$ and $\frac{\partial r}{\partial s}$ are just constants with respect to $r$, which simplifies things, but I can't figure out how to differentiate $\frac{\partial f}{\partial x}$ or $\frac{\partial f}{\partial y}$ with respect to $r$. How do I continue from here?

Answer 1

You have to use the rule of product when you derivate again.

Notice that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ keep being functions of $x$ and $y$, so there you have to use chain rule again.

Denoting $\frac{\partial^2 f}{\partial r\partial s}$ as $f_{sr}$, \begin{equation} f_{sr} = (f_xx_s)_r + (f_yy_s)_r= (f_{xx}x_r+f_{xy}y_r)x_{s}+f_xx_{sr}+(f_{yx}x_r+f_{yy}y_r)y_{s}+f_yy_{sr} \end{equation}