> [!NOTE] > Theorem 3.3.4 (Local continuity of partial derivatives implies differentiability). Let $U \subset \mathbb{R}^n$ be open, $f: U \rightarrow \mathbb{R}^m$ and consider $a \in U$. Assume that all partial derivatives $\partial_j f_i(i \in\{1, \ldots, m\}$ and $j \in\{1, \ldots, n\}$ ) exist and are continuous at $a$. Then $f$ is differentiable at $a$. ###### Proof Due to Proposition 3.1.8, as in the proof of Theorem 3.3.1, it is sufficient to consider the case $m=1$. Since $U$ is open, there is $\varepsilon>0$ such that $a+h \in U$ for all $h$ such that $\max _i\left|h_i\right|<\varepsilon$. We can then write $ \begin{aligned} f(a+h)-f(a)= & f\left(a_1+h_1, a_2+h_2, \ldots, a_n+h_n\right)-f\left(a_1, a_2, \ldots, a_n\right) \\ = & f\left(a_1+h_1, a_2+h_2, \ldots, a_n+h_n\right)-f\left(a_1, a_2+h_2, \ldots, a_n+h_n\right) \\ & +f\left(a_1, a_2+h_2, \ldots, a_n+h_n\right)-f\left(a_1, a_2, \ldots, a_n\right) \\ = & f\left(a_1+h_1, a_2+h_2, \ldots, a_n+h_n\right)-f\left(a_1, a_2+h_2, \ldots, a_n+h_n\right) \\ & +f\left(a_1, a_2+h_2, \ldots, a_n+h_n\right)-f\left(a_1, a_2, a_3+h_3, \ldots, a_n+h_n\right) \\ & \vdots \\ & +f\left(a_1, \ldots, a_{n-1}, a_n+h_n\right)-f\left(a_1, \ldots, a_{n-1}, a_n\right) . \end{aligned} $ Recall that $\partial_1 f\left(x, a_2+h_2, \ldots, a_n+h_n\right)$ is the derivative of the function $x \mapsto g(x)=f\left(x, a_2+\right.$ $h_2, \ldots, a_n+h_n$ ). By assumption $g$ is differentiable on $\left(a_1-\varepsilon, a_1+\varepsilon\right)$. Thus by the mean value theorem applied to $g$ we obtain $ f\left(a_1+h_1, a_2+h_2, \ldots, a_n+h_n\right)-f\left(a_1, a_2+h_2, \ldots, a_n+h_n\right)=h_1 \partial_1 f\left(\theta_1, a_2+h_2, \ldots, a_n+h_n\right) $ for some $\theta_1$ between $a_1$ and $a_1+h_1$. Similarly the $i$-th term in the sum on the RHS of (3.13) equals $ h_i \partial_i f\left(a_1, \ldots, a_{i-1}, \theta_i, a_{i+1}+h_{i+1}, \ldots, a_n+h_n\right)=h_i \partial_i f\left(c_i\right) $ for some $\theta_i$ between $a_i$ and $a_i+h_i$ and we set $c_i:=\left(a_1, \ldots, a_{i-1}, \theta_i, a_{i+1}+h_{i+1}, \ldots, a_n+h_n\right)$. Note that $c_i \rightarrow a$ as $h \rightarrow 0$. We can then use the sum (3.13) with the above identification to estimate $ \begin{aligned} \frac{\left|f(a+h)-f(a)-\sum_{i=1}^n \partial_i f(a) h_i\right|}{\|h\|} & =\frac{\left|\sum_{i=1}^n\left(\partial_i f\left(c_i\right)-\partial_i f(a)\right) h_i\right|}{\|h\|} \\ & \leq \frac{1}{\|h\|} \sum_{i=1}^n\left|\partial_i f\left(c_i\right)-\partial_i f(a) \| h_i\right| \\ & \leq \sum_{i=1}^n\left|\partial_i f\left(c_i\right)-\partial_i f(a)\right| \end{aligned} $ Note that the final term tends to zero as $h \rightarrow 0$ since $c_i \rightarrow a$ and the partial derivatives of $f$ are continuous at $a$. We can thus define the linear map (as we would expect) $ D f(a)(h):=\sum_{i=1}^n \partial_i f(a) h_i $ and (3.14) can be rewritten as $ f(a+h)=f(a)+D f(a)(h)+R(a, h) $ with $ \lim _{\substack{h \rightarrow 0 \\ h \neq 0}} \frac{|R(a, h)|}{\|h\|}=0 $ This shows that $f$ is differentiable at $a$.