Derivatives of component inverse functions

Question

I might have missed the point of the following questions. Anyone kindly give a suggestion?

Let $f:\mathbb{R}_\mathbf{x}^3\to\mathbb{R}_\mathbf{y}^3$ and $g:\mathbb{R}_\mathbf{y}^3\to\mathbb{R}_\mathbf{x}^3$ be $\mathscr{C}^1$ inverse functions. Show that

$$ \frac{\partial g_1}{\partial y_1}=\frac{1}{J} \frac{\partial (f_2 , f_3)}{\partial (x_2, x_3)},J=\frac{\partial (f_1 , f_2 , f_3)}{\partial (x_1 , x_2 , x_3)} $$

And obtain similar formulas for the other derivatives of the component functions of $g$.

A similar question in terms of implicit functions:

If the equations $f(x,y,z)=0$, $g(x,y,z)=0$ can be solved for $y$ and $z$ as differentiable functions of $x$, show that \begin{eqnarray*} \frac{dy}{dx}=\frac{1}{J}\frac{\partial(f,g)}{\partial(z,x)}, & & \frac{dz}{dx}=\frac{1}{J}\frac{\partial(f,g)}{\partial(x,y)}, \end{eqnarray*} where $J=\frac{\partial(f,g)}{\partial(y,z)}$.

which I have just solved as below.

Answer 1

In general, if $f\circ g=\cal I$, where $\cal I:\mathbb R^n\to\mathbb R^n$, is the identity, then, accoriding to the Chain Rule for vector valued functions $$ \cal I=D\cal I=D(f\circ g)(x) =Df\big(g(x)\big)\cdot Dg(x), $$ where $Dh$ is the differential of $h$, and hence $$ 1=\det \cal I=\det Df\big(g(x)\big)\cdot \det Dg(x). $$ Thus $$ \det Dg(x)=\frac{1}{\det Df\big(g(x)\big)}. $$

Answer 2

I myself tried the second question just now.

Define $f(x,y(x),z(x))=0$ and $g(x,y(x),z(x))=0$. Then by the chain rule, \begin{eqnarray*} \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial z}\frac{dz}{dx}=0 & \implies & \frac{\partial f}{\partial x}\frac{\partial g}{\partial z}+\frac{\partial f}{\partial y}\frac{\partial g}{\partial z}\frac{dy}{dx}+\frac{\partial f}{\partial z}\frac{\partial g}{\partial z}\frac{dz}{dx}=0\\ \frac{\partial g}{\partial x}+\frac{\partial g}{\partial y}\frac{dy}{dx}+\frac{\partial g}{\partial z}\frac{dz}{dx}=0 & \implies & \frac{\partial f}{\partial z}\frac{\partial g}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial g}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial z}\frac{\partial g}{\partial z}\frac{dz}{dx}=0 \end{eqnarray*} So \begin{eqnarray*} & & -\left(\frac{\partial f}{\partial z}\frac{\partial g}{\partial x}-\frac{\partial f}{\partial x}\frac{\partial g}{\partial z}\right)+\left(\frac{\partial f}{\partial y}\frac{\partial g}{\partial z}-\frac{\partial f}{\partial z}\frac{\partial g}{\partial y}\right)\frac{dy}{dx}=0\\ & \implies & -\frac{\partial(f,g)}{\partial(z,x)}+\frac{\partial(f,g)}{\partial(y,z)}\frac{dy}{dx}=0\\ & \implies & \frac{dy}{dx}=\frac{1}{\frac{\partial(f,g)}{\partial(y,z)}}\frac{\partial(f,g)}{\partial(z,x)}=\frac{1}{J}\frac{\partial(f,g)}{\partial(z,x)} \end{eqnarray*}

It can be solved similarly for $\frac{dz}{dx}$.