这是一份durham杜伦大学MATH2031-WE01作业代写的成功案例
Proof Fix $y \in V$ and choose indices $j$ and $k$ from ${1, \ldots, n}$. We need to show
$$
\lim {h \rightarrow 0} \frac{g{j}\left(\boldsymbol{y}+h \boldsymbol{E}{\boldsymbol{k}}\right)-g{j}(\boldsymbol{y})}{h}
$$
exists and that $g_{y_{j}}$ is continuous. For small enough h, the line segment $\left[y, y+h \boldsymbol{E}{k}\right]$ is in $V$. Let $$ \begin{aligned} &x=g(\boldsymbol{y})=f^{-1}(\boldsymbol{y}) \in U \ &\boldsymbol{z}=g\left(\boldsymbol{y}+h \boldsymbol{E}{k}\right)=f^{-1}\left(\boldsymbol{y}+h \boldsymbol{E}{k}\right) \in U \end{aligned} $$ Then, $\boldsymbol{x}$ and $z$ are distinct as $f^{-1}$ is $1-1$. Note the line segment $\left[x{,} z\right]$ is in $B\left(r_{,} x_{0}\right)$. Since $g$ is continuous on $V$, $\lim {h \rightarrow 0} g\left(\boldsymbol{y}+h \boldsymbol{E}{\boldsymbol{k}}\right)=g(\boldsymbol{y})$. But this says $\lim _{h \rightarrow 0} \boldsymbol{z}=g(\boldsymbol{y})=\boldsymbol{x}$
MATH2031-WE01 COURSE NOTES ：
$$
0=g_{x}^{0}\left(x-x_{0}\right)+g_{y}^{0}\left(\phi(x)-\phi\left(x_{0}\right)\right)+E_{g}\left(x, \phi(x), x_{0}, \phi\left(x_{0}\right)\right)
$$
Now divide through by $x-x_{0}$ to get
$$
0=g_{x}^{0}+g_{y}^{0}\left(\frac{\phi(x)-\phi\left(x_{0}\right)}{x-x_{0}}\right)+\frac{E_{g}\left(x, \phi(x), x_{0}, \phi\left(x_{0}\right)\right)}{x-x_{0}}
$$
Thus,
$$
g_{y}^{0}\left(\frac{\phi(x)-\phi\left(x_{0}\right)}{x-x_{0}}\right)=-g_{x}^{0}-\frac{E_{g}\left(x, \phi(x), x_{0}, \phi\left(x_{0}\right)\right)}{x-x_{0}}
$$
and assuming $g_{x}^{0} \neq 0$ and $g_{y}^{0} \neq 0$, we can solve to find
$$
\frac{\phi(x)-\phi\left(x_{0}\right)}{x-x_{0}}=-\frac{g_{x}^{0}}{g_{y}^{0}}-\frac{1}{g_{y}^{0}} \frac{E_{g}\left(x, \phi(x), x_{0}, \phi\left(x_{0}\right)\right)}{x-x_{0}}
$$