$G(x,y)=(x^2+y,xy+x^2)$, $P=(a,b)$ is G differentiable at P? calculate dG(P)
Attempt:
I think I understand how to find dG(P) it is just, $dG(P) = (2a,b)$ correct?
I am needing help how I show that G is differentiable at P.
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Since $G: \mathbb{R}^2 \to \mathbb{R}^2$, you must have $DG(P): \mathbb{R}^2 \to \mathbb{R}^2$, so the above cannot be correct as it must be a $2 \times 2$ matrix. – copper.hat Mar 10 '15 at 05:12
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@copper.hat I see I think I calculated dG(P) incorrectly it is supposed to be 2a+b, meaning it goes from $\mathbb R^2 \rightarrow \mathbb R$. Is there a way to show it is not differentiable by any other means? – hobbit Mar 10 '15 at 05:23
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It is differentiable, but the derivative is a $2 \times 2$ matrix. – copper.hat Mar 10 '15 at 05:24
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which derivative are you talking about? Gradient, or divergence? or exterior derivative ? – mastrok Mar 10 '15 at 06:43
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@mastrok I am positive the question is talking about the gradient. Agh, I am getting gradient and divergence mixed up. – hobbit Mar 10 '15 at 22:30