A Question About Multivariate Differentiation I am stuck on a practice problem which involves the multivariable chain rule I think.
$$F \colon \mathbb{R}^2 \to \mathbb{R}^2$$ $$g \colon \mathbb{R}^2 \to \mathbb{R}$$
$$ F(x,y) := (g(x, g(x, y)), g(g(x, y), y))$$
Show that $F$ is differentiable and find the derivative.
It seems as though this might involve the chain rule, but I'm not sure. Perhaps it involves the definition of a derivative.
Any help would be appreciated, and like I mentioned this isn't homework, just a practice problem.
Thanks again.
 A: You need to suppose that $g$ is differentiable. Then you're right, this is a chain rule question. To clear things up, you might define $h: \Bbb{R}^2\to\Bbb{R}^2,(x,y)\mapsto (x,g(x,y))$ and similarly $k$ so that $F$ becomes $(g\circ h,g\circ k)$. Now you may be able to see that $F$ is differentiable since each of its coordinate functions is a composition of two differentiable functions ($h$ and $k$ are differentiable in turn because each of their coordinate functions is differentiable.) 
Then in the simplest but least detailed notation, which assumes that you're comfortable with the chain rule, we get $DF=(Dg\circ Dh,Dg\circ Dk)$.
Let's flesh this out more concretely. $h'_{(x,y)}=\left(\begin{matrix} 1&0\\g_1(x,y)&g_2(x,y)\end{matrix}\right)$ where $g_1=\partial g/\partial x$ and similarly for $g_2$. In the same way we get $k'.$ Meanwhile we have to evaluate $g'$ at $h(x,y)$ to satisfy the chain rule, so we use $g'(h(x,y))=(g_1(x,g(x,y)), g_2(x,g(x,y))$ to compute $$Dg\circ Dh=(g_1(x,g(x,y))+g_2(x,g(x,y))g_1(x,y),g_2(x,g(x,y))g_2(x,y))$$ 
by matrix multiplication. You can get $F_2=Dg\circ Dk$ in the same way.
One more point: someone else, maybe a classmate of yours, posted a similar question that supposes $g$ is continuous. That's not enough: say $g(x,y)=|x|$. Then we'd have $F(x,y)=(|x|,|x|)$, which isn't differentiable.
A: Hint:  you need $g$ to be differentiable, not just continuous.  Then it is an application of the chain rule.  Let $g_x(x,y)=\frac {\partial g(x,y)}{\partial x}$ and $g_y(x,y)=\frac {\partial g(x,y)}{\partial y}$ and similarly for $F$.  Then the first coordinate of $F_x$ is $\frac {\partial g(x,g(x,y))}{\partial x}=g_x(x,g(x,y))+g_y(x,g(x,y))g_x(x,y)$
