Dimensions of a differential matrix

Suppose that h:R2→Rh:\mathbb{R^2} \to \mathbb{R} is differentiable and that g:R3→Rg:\mathbb{R^3} \to \mathbb{R} is defined by g(a,b,c)=h(ab,ac)g(a,b,c)=h(ab,ac). How to find the differential matrix dimensions for the differential dhh?

My feeling is that the matrix should be a (2×3)(2 \times 3) matrix, since the function gg has two coordinate functions and three variables. How can I make sure I am correct? Thanks in advance for the help.




Use the chain rule.
– Jacky Chong
Oct 20 at 20:18



Did you mean to look for the matrix dimensions for DgDg or DhDh?
– copper.hat
Oct 20 at 20:51



In general, if f:A→Bf:A \to B, then Df(x):A→BDf(x):A \to B.
– copper.hat
Oct 20 at 20:53



I’m looking for dgg. But shouldn’t dgg and dhh have the same dimensions?
– George Oscar Bluth
Oct 20 at 20:55


1 Answer


Hint: Think of g=h∘fg= h\circ f where f:R3→R2f:\mathbb{R}^3\rightarrow \mathbb{R}^2 given by
f(a, b, c) = (ab, ac).

Then apply the chain rule to g=h∘fg = h \circ f.



By the chain rule we would have dgg= ∂h∂f∂f∂a+∂h∂f∂f∂b+∂h∂f∂f∂a+∂h∂f∂f∂c\frac{\partial h}{\partial f} \frac{\partial f}{\partial a} +\frac{\partial h}{\partial f} \frac{\partial f}{\partial b} + \frac{\partial h}{\partial f} \frac{\partial f}{\partial a} + \frac{\partial h}{\partial f} \frac{\partial f}{\partial c}, right?
– George Oscar Bluth
Oct 20 at 20:29