I am having problems with the following exercise:
Exercise:
Let $\mathbf{h}: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ given by
$$ \mathbf{h}(x,y,z) = \begin{pmatrix} x^2 + (z-1)^2 -5 + e^{y-2} \\ ln\left(\frac{xz}{2}\right) \end{pmatrix} $$
(i) Show that $\mathbf{h}(2,2,1)=0$ and that $\mathbf{h} \in C^1(\mathbb{R}^2)$.
I have done this. I showed that the partial derivatives exist and that they are continous thus $\mathbf{h} \in C^1(\mathbb{R}^2)$.
I think I did the next exercise correct, but if I did something wrong please tell me to check it again or give a hint.
(ii) Show that one can apply the implicit function theorem in order to obtain some small enough $\epsilon >0 $ and a $C^1$ function $\mathbf{f}: (1-\epsilon, 1+\epsilon) \rightarrow \mathbb{R}^2$ such that
$$\mathbf{h}(\mathbf{f}(z), z) = (0,0), ~~~~\forall z \in (1-\epsilon, 1+\epsilon).$$
I have that $\mathbf{h}(2,2,1)=0$ so I need to have $\mathbf{f}(z)=(2,2)$. From the previous exercise I found out that the partialderivatives are
$$[D\mathbf{h}] = \begin{pmatrix} 2x & e^{y-2} & 2(z-1) \\ 1/x & 0 & 1/z \end{pmatrix}$$
and at $(2,2,1)$ I have that
$$\frac{\delta \mathbf{h}}{\delta x} (2,2,1) = \begin{pmatrix} 4 \\ 1/2 \end{pmatrix} \neq 0 $$
$$\frac{\delta \mathbf{h}}{\delta y} (2,2,1) = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \neq 0 $$
thus I can use the Implicit function theorem to find $\mathbf{f}(z)$.
The next exercise is the one causing me problems.
(iii) Find the Jacobi Matrix D\mathbf{f}.
How do I do this? By looking at the Implicit function theorem I am thinking that I need to use the following:
$$\frac{\delta \mathbf{f}}{\delta x} (1) = -\frac{\frac{\delta \mathbf{h}}{\delta z} (2,2,1)}{\frac{\delta \mathbf{h}}{\delta x}(2,2,1)}$$ and $$\frac{\delta \mathbf{f}}{\delta y} (1) = -\frac{\frac{\delta \mathbf{h}}{\delta z} (2,2,1)}{\frac{\delta \mathbf{h}}{\delta y}(2,2,1)}$$
so that $$[D\mathbf{f}](1) = \begin{pmatrix} \frac{\delta \mathbf{f}}{\delta x} (1) \\ \frac{\delta \mathbf{f}}{\delta y} (1) \end{pmatrix}$$
But how do I calculate $\frac{\delta \mathbf{f}}{\delta x} (1)$ and $\frac{\delta \mathbf{f}}{\delta y} (1)$? Because if I use the formula above I get a vector divided with a vector and this does not make sense to me.
Best regards Husky
If we define $\phi:\mathbb{R} \to \mathbb{R}^3$ by $\phi(z) = (f(z),z)^T$, we have, with slight abuse of notation, $D\phi(z) = \begin{bmatrix} Df(z) \\ I\end{bmatrix} $.
We know that $h \circ \phi(z) = 0$ around $z=1$, so $D (h \circ \phi)(z) = Dh(\phi(z)) D \phi(z) = 0$. Rewriting gives $\begin{bmatrix}{\partial h(\phi(z)) \over \partial x} & {\partial h(\phi(z)) \over \partial y} & {\partial h(\phi(z)) \over \partial z} \end{bmatrix} \begin{bmatrix} Df(z) \\ I\end{bmatrix} = 0$, or $\begin{bmatrix}{\partial h(\phi(z)) \over \partial x} & {\partial h(\phi(z)) \over \partial y} \end{bmatrix} Df(z) = - {\partial h(\phi(z)) \over \partial z}$. Then $Df(z) = - \begin{bmatrix}{\partial h(\phi(z)) \over \partial x} & {\partial h(\phi(z)) \over \partial y} \end{bmatrix}^{-1} {\partial h(\phi(z)) \over \partial z}$.