Let $f = 0$ s.t. $x_n$ is implicitly function $\phi$ of $x_1, ..., x_{n-1}$. Prove a partial derivative of $\phi$ is a quotient of partials of $f$. I'm trying to prove this using Taylor polynomials in particular, although there is a quicker way. I was hoping someone could help me out complete the proof.
Let $f : \mathbb{R}^n \rightarrow \mathbb{R}$. Suppose $f(\vec a) = 0$ and $ \frac{\partial f (\vec a)}{\partial x_n} \neq 0$ and using the implicit function theorem let $x_n$ be implicitly defined from $f= 0$ by $x_1,...,x_{n-1} = \underline x$ such that $\phi(\underline x) = x_n$. Notice therefore that $\vec x = \langle \underline x, \phi(\underline x) \rangle$ .
Prove that $\frac{\partial \phi( {\underline a)}}{\partial x_j}= -\frac{\frac{\partial f(\vec a)}{\partial x_j}}{{\frac{\partial f(\vec a)}{\partial x_n}}}$.
Notice that, using the first-order Taylor expansion for $f$ at $\vec a$, we get:
$$f(\vec x) = f_{x_1}(\vec a)\Delta x_1 + \cdots + f_{x_n}(\vec a)\Delta x_n + \epsilon(\vec x)$$
Since $\frac{\partial \phi( {\underline a)}}{\partial x_j}$ amounts only to a change in $\Delta x_n$ and $\Delta x_j$, and setting the expression equal to $0$, we get: $$f(\vec x) = f_{x_j}(\vec a)\Delta x_j + f_{x_n}(\vec a)\Delta x_n + \epsilon(\vec x) = 0$$
Ignoring the error function, we get approximately: $$f_{x_j}(\vec a)\Delta x_j + f_{x_n}(\vec a)\Delta x_n + \approx 0$$
$$\frac{\Delta x_n}{\Delta x_j} \approx -\frac{f_{x_j}(\vec a)}{f_{x_n}(\vec a)}$$
Notice that $\Delta x_n$ with respect to $\Delta x_j$ is nearly the definition of $\frac{\partial \phi (\underline a)}{\partial x_j}$. What exactly would I need to get  $\frac{\partial \phi( {\underline a)}}{\partial x_j}= -\frac{\frac{\partial f(\vec a)}{\partial x_j}}{{\frac{\partial f(\vec a)}{\partial x_n}}}$ from this? Especially, how could I get rid of the approximation and the error function?
 A: This is an interesting question. To use Taylor polynomials, we assume $f$ is local $C^2$ around $\vec{a}$. Then, to calculate $\frac{\partial \phi\left(\underline{a}\right)}{\partial x_j}$, define
$$
\underline{a}^{j,\Delta}=\underline{a}+\Delta e^j\,,
$$
where $e^j\in\mathbb{R}^n$ and $e^j_i=\delta_{i,j}$. Thus, we need to calculate
$$
\frac{\partial \phi\left(\underline{a}\right)}{\partial x_j}=\lim_{\Delta\rightarrow0}\frac{\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)}{\Delta}\,.
$$
Now, we use Taylor polynomials. Because $f$ is local $C^2$ around $\vec{a}$ and $\phi$ is continuous, when $\Delta$ is small enough,
$$
\begin{aligned}
&\left|f((\underline{a}^{j,\Delta},\phi(\underline{a}^{j,\Delta}))-f((\underline{a},\phi(\underline{a}))-\partial_jf((\underline{a},\phi(\underline{a}))\Delta-\partial_nf((\underline{a},\phi(\underline{a}))\left(\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)\right)\right|\\
\leq& C\left(|\Delta|^2+|\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)|^2\right)\,.
\end{aligned}
$$
Using $f((\underline{a}^{j,\Delta},\phi(\underline{a}^{j,\Delta}))=0$, $f((\underline{a},\phi(\underline{a}))=0$, when $\Delta$ is small enough
$$
\left|\partial_jf((\underline{a},\phi(\underline{a}))\Delta-\partial_nf((\underline{a},\phi(\underline{a}))\left(\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)\right)\right|
\leq C\left(|\Delta|^2+|\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)|^2\right)\,,
$$
which implies
$$\tag{1}
\left|\frac{\partial_jf((\underline{a},\phi(\underline{a}))}{\partial_nf((\underline{a},\phi(\underline{a}))}-\frac{\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)}{\Delta}\right|
\leq \frac{C}{\partial_nf((\underline{a},\phi(\underline{a}))}\left(|\Delta|+\frac{|\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)|^2}{\Delta}\right)\,.
$$
Thus
$$
\limsup_{\Delta\rightarrow0}\left|\frac{\partial_jf((\underline{a},\phi(\underline{a}))}{\partial_nf((\underline{a},\phi(\underline{a}))}-\frac{\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)}{\Delta}\right|
\leq \frac{C}{\partial_nf((\underline{a},\phi(\underline{a}))}\left(\limsup_{\Delta\rightarrow0}\frac{|\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)|^2}{\Delta}\right)\,.
$$
Now, we consider two cases. First, if
$$
\limsup_{\Delta\rightarrow0}\frac{|\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)|}{\Delta}<\infty\,,
$$
we have
$$
\limsup_{\Delta\rightarrow0}\frac{|\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)|^2}{\Delta}=0\,,
$$
which implies
$$
\limsup_{\Delta\rightarrow0}\left|\frac{\partial_jf((\underline{a},\phi(\underline{a}))}{\partial_nf((\underline{a},\phi(\underline{a}))}-\frac{\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)}{\Delta}\right|=0\,.
$$
Second, if
$$
\limsup_{\Delta\rightarrow0}\frac{|\phi\left(\underline{a}^{j,\Delta}\right)-\phi\left(\underline{a}\right)|}{\Delta}=\infty\,,
$$
we have a sequence ${\Delta_k}^\infty_{k=1}$ and $\lim_{k\rightarrow\infty}\Delta_k=0$ such that
$$
k<\frac{|\phi\left(\underline{a}^{j,\Delta_k}\right)-\phi\left(\underline{a}\right)|}{\Delta_k}<k+1\,.
$$
Go back (1)
$$
\begin{aligned}
&k-\left|\frac{\partial_jf((\underline{a},\phi(\underline{a}))}{\partial_nf((\underline{a},\phi(\underline{a}))}\right|\\
\leq &\left|\frac{\partial_jf((\underline{a},\phi(\underline{a}))}{\partial_nf((\underline{a},\phi(\underline{a}))}-\frac{\phi\left(\underline{a}^{j,\Delta_k}\right)-\phi\left(\underline{a}\right)}{\Delta_k}\right|\\
\leq &\frac{C}{\partial_nf((\underline{a},\phi(\underline{a}))}\left(|\Delta_k|+\frac{|\phi\left(\underline{a}^{j,\Delta_k}\right)-\phi\left(\underline{a}\right)|^2}{\Delta_k}\right)\\
\leq &\frac{C}{\partial_nf((\underline{a},\phi(\underline{a}))}\left(|\Delta_k|+|\phi\left(\underline{a}^{j,\Delta_k}\right)-\phi\left(\underline{a}\right)|(k+1)\right)\,,
\end{aligned}
$$
which implies for any $k>0$,
$$
1-\frac{\left|\frac{\partial_jf((\underline{a},\phi(\underline{a}))}{\partial_nf((\underline{a},\phi(\underline{a}))}\right|}{k}
\leq  \frac{C}{\partial_nf((\underline{a},\phi(\underline{a}))}\left(\frac{|\Delta_k|}{k}+|\phi\left(\underline{a}^{j,\Delta_k}\right)-\phi\left(\underline{a}\right)|\frac{k+1}{k}\right)\,.
$$
Taking $k\rightarrow\infty$ on both sides, since $\phi$ is continuous, we obtain
$$
1\leq 0\,.
$$
This is impossible, which implies the second case is impossible.
