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.

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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

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1 Answer

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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).\begin{align}

f(a, b, c) = (ab, ac).

\end{align}

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