Implicit Function Theorem for Component Function Prove that id the functions $f_1,..,f_n$ in the statement of the implicit function theorem are assumed to be k times continuously differentiable (i.e., all partial derivatives of order k exist and are continuous), then the same is true of the component function $\gamma_1,.., \gamma_n$ of $\gamma.$
I know this theorem is true, but I don't get how you can relate to the componet function, $\gamma.$
 A: It follows from the formula for the (Fréchet) derivative of $\gamma$. Suppose $x^*,y^*,z^*$ are given points such that $g(x^*,y^*) = z^*$, where $g$ is $C^k$. If $\gamma$ is $C^1$ and solves (as the IFT gives you)
$$g(x,\gamma(x)) = z^*$$
Then by the Chain Rule,
$$d\gamma(x) = -[\partial_yg(x,\gamma(x))]^{-1}\partial_xg(x,\gamma(x))$$
Looking at this equation, we see that everything on the right hand side is indeed $C^1$; hence $d\gamma$ is $C^1$, so $\gamma$ is $C^2$. But by the same argument, we get $d\gamma$ is actually $C^2$, so in fact $\gamma$ is $C^3$. And so it continues, until you run out of continuous derivatives of $g$.
In coordinate form, we get by Chain rule on $z^* = g(x,\gamma(x))$ 
$$ \mathbf{0} = J_xg(x,\gamma(x)) + J_yg(x,\gamma(x))∇ \gamma(x)$$
where $\mathbf{0}$ is a zero matrix of appropriate size, $J_yg(a,b)$ is the Jacobian matrix of $g$ at $(a,b)$, considered as a function of $y$ only, and similarly with $J_xg(a,b)$. This gives us
$$∇ \gamma(x) = \begin{pmatrix}\frac{\partial\gamma}{\partial x_1}(x) \\ \vdots \\ \frac{\partial\gamma}{\partial x_n}(x) \end{pmatrix}  = -[J_yg(x,\gamma(x))]^{-1}J_xg(x,\gamma(x))$$
Now apply the same argument as before: everything on the right hand side (partials of g, $\gamma$) and all matrix operations (inversion) are $C^1$; hence the left hand side is $C^1$. But then this means that $\gamma$ is $C^2$. Repeating the argument inductively, we see that $\gamma$ is $C^k$.
