In this answer there's a comment which says
That's not wrong; that's a perfectly valid method. You get the derivative of any expression with respect to $x$ as the sums of all the derivatives with respect to the individual instances of $x$ while holding all other instances constant. – joriki
This blew my mind. How do you prove this?
$$dF(u,v,w)=\frac{\partial F}{\partial u}du+\frac{\partial F}{\partial v}dv+\frac{\partial F}{\partial w}dw.$$ Then $$\frac{dF(u,v,w)}{dx}=\frac{\partial F}{\partial u}\frac{du}{dx}+\frac{\partial F}{\partial v}\frac{dv}{dx}+\frac{\partial F}{\partial w}\frac{dw}{dx}.$$ Taking $u=v=w=x$,
$$\frac{dF(x,x,x)}{dx}=\frac{\partial F}{\partial u}\Big|_{(x,x,x)}+\frac{\partial F}{\partial v}\Big|_{(x,x,x)}+\frac{\partial F}{\partial w}\Big|_{(x,x,x)}.$$
Example:
$$\frac{\partial u^v}{\partial u}=v\ u^{v-1},$$ $$\frac{\partial u^v}{\partial v}=\ln u\ v^u,$$ $$(x^x)'=x\ x^{x-1}+\ln x\ x^x.$$
