Today in my Calc 3 class, my math teacher did some really sketchy math to 'prove' the partial derivative chain rule. First, he had the functions $f(x,y)$, $x(u,v)$, and $y(u,v)$. He then took the differential $$df=f_xdx+f_ydy$$ $$df=f_x(x_udu+x_vdv)+f_y(y_udu+y_vdv)$$ To quote him, next he "converted everything to partials" (This is when I went "huh?") and then he divided by $\partial u$. $$\frac{\partial f}{\partial u}=f_x x_u \frac{\partial u}{\partial u} + f_x x_v \frac{\partial v}{\partial u}+f_y y_u\frac{\partial u}{\partial u}+ f_y y_v\frac{\partial v}{\partial u}$$ And then he argued that because we are taking derivatives with respect to $u$, all the partial $v$ are zero, and he also said that $\partial u/\partial u=1$, getting $$\frac{\partial f}{\partial u}=f_xx_u+f_yy_u$$ Is any of this reasoning correct? If not, what is a way to fix it without being overly complicated? If so, what is the justification for 1) the switch to partials and then 2) the dividing by partial u and then 3) the v's going to zero?
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He is correct that $\partial v/\partial u=0$ and $\partial u/\partial u=1$. But it should not be something he just says. It should be proven (which is easily done from the definitions, but still...) – Arthur Nov 18 '17 at 06:30
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@Arthur How can we switch from $df$ to $\partial f$? That doesn't seem very rigorous at all. – D.R. Nov 18 '17 at 16:44