Product rule for partial differentiation

Question

If given a function $f(x,y)$ that can be re-expressed as $g(\rho, \phi), $then by the chain rule $$\frac{\partial f}{\partial x}= \frac{\partial f}{\partial \phi}\frac{\partial \phi}{\partial x}+\frac{\partial f}{\partial \rho}\frac{\partial \rho}{\partial x}.$$ If we have to find $\partial^2 f \over \partial x^2$, is there a product rule for partial differentiation that says $$\frac{\partial^2 f}{\partial x^2}= (\frac{\partial}{\partial x}\frac{\partial f}{\partial \phi})\frac{\partial \phi}{\partial x}+\frac{\partial f}{\partial \phi}(\frac{\partial}{\partial x} \frac{\partial \phi}{\partial x})+(\frac{\partial }{\partial x}\frac{\partial f}{\partial \rho})\frac{\partial \rho}{\partial x} + \frac{\partial f}{\partial \rho}(\frac{\partial}{\partial x}\frac{\partial \rho}{\partial x})?$$ I understand the proof for product rule in normal differentiation, as in like ${d \over dx}u(x)h(x)$, but I'm not sure if we can say the same for partial differentiation. If we can, why?

Answer 1

If $e_j$ is the $j$-th unit vector, then the definition of the partial derivative gives \begin{align*} \partial_j \big(f(x)g(x)\big) &= \lim_{h\to 0} \frac{f(x+he_j)g(x+he_j) - f(x)g(x)}{h}\\ &= \lim_{h \to 0} \frac{f(x+he_j)g(x+he_j) - f(x+he_j) g(x) + f(x+he_j) g(x) + f(x)g(x)}{h}\\ &= \lim_{h\to 0} \left(f(x+he_j)\frac{g(x+he_j)-g(x)}{h} + g(x) \frac{f(x+he_j) - f(x)}{h} \right)\\ &= f(x) \partial_jg(x) + g(x)\partial_jf(x), \end{align*} assuming, of course, that our functions are differentiable in the first place.