[proofplan]
We prove each rule directly from the definition of differentiability at a point. Openness of $U$ lets us restrict to small increments $h$ with $a+h \in U$, and differentiability gives a linear part plus a remainder that is negligible compared with $|h|$. Addition and scalar multiplication follow by adding or scaling these expansions. For the product rule, we multiply the two scalar expansions, collect the linear terms, and estimate every remaining term as $o(|h|)$.
[/proofplan]
[step:Choose increments that stay inside the open domain]
For $x_0 \in \mathbb{R}^m$ and $r > 0$, let $B(x_0,r) := \{x \in \mathbb{R}^m : |x-x_0| < r\}$ denote the open ball. Since $U \subset \mathbb{R}^m$ is open and $a \in U$, there exists $\rho > 0$ such that $B(a,\rho) \subset U$. Define the increment ball $V := B(0,\rho) \subset \mathbb{R}^m$. Then $a+h \in U$ for every $h \in V$, so all difference quotients below are taken only for such increments $h$.
For a map $F: U \to \mathbb{R}^k$ differentiable at $a$, where $k \in \mathbb{N}$, the definition gives a [linear map](/page/Linear%20Map) $DF_a: \mathbb{R}^m \to \mathbb{R}^k$ and a remainder map
\begin{align*}
r_F: V \to \mathbb{R}^k, \qquad r_F(h) := F(a+h)-F(a)-DF_a(h)
\end{align*}
such that
\begin{align*}
\lim_{\substack{h \to 0, h \neq 0}} \frac{|r_F(h)|}{|h|} = 0.
\end{align*}
[/step]
[step:Add the linear approximations for $f$ and $g$]
Apply the preceding differentiability expansion to $f,g: U \to \mathbb{R}^n$. Thus there are remainder maps $r_f,r_g: V \to \mathbb{R}^n$ such that
\begin{align*}
f(a+h)=f(a)+Df_a(h)+r_f(h)
\end{align*}
and
\begin{align*}
g(a+h)=g(a)+Dg_a(h)+r_g(h)
\end{align*}
for every $h \in V$, with $|r_f(h)|/|h| \to 0$ and $|r_g(h)|/|h| \to 0$ as $h \to 0$, $h \neq 0$.
Define the pointwise sum of linear maps $A: \mathbb{R}^m \to \mathbb{R}^n$ by
\begin{align*}
A(h) := Df_a(h)+Dg_a(h).
\end{align*}
Since sums of linear maps are linear, $A$ is linear. Adding the two expansions gives
\begin{align*}
(f+g)(a+h)=(f+g)(a)+A(h)+r_f(h)+r_g(h).
\end{align*}
The triangle inequality gives
\begin{align*}
\frac{|r_f(h)+r_g(h)|}{|h|} \leq \frac{|r_f(h)|}{|h|}+\frac{|r_g(h)|}{|h|}.
\end{align*}
The right-hand side tends to $0$, so $r_f+r_g=o(|h|)$. Therefore $f+g$ is differentiable at $a$ and $D(f+g)_a=A=Df_a+Dg_a$.
[/step]
[step:Scale the linear approximation for $f$]
Let $\lambda \in \mathbb{R}$. Using the differentiability expansion for $f$, we have
\begin{align*}
(\lambda f)(a+h)=(\lambda f)(a)+(\lambda Df_a)(h)+\lambda r_f(h),
\end{align*}
where the linear map $\lambda Df_a: \mathbb{R}^m \to \mathbb{R}^n$ is defined by $(\lambda Df_a)(h):=\lambda Df_a(h)$. Since scalar multiples of linear maps are linear, $\lambda Df_a$ is linear. Moreover,
\begin{align*}
\frac{|\lambda r_f(h)|}{|h|}=|\lambda|\frac{|r_f(h)|}{|h|}\to 0
\end{align*}
as $h \to 0$, $h \neq 0$. Hence $\lambda f$ is differentiable at $a$ and $D(\lambda f)_a=\lambda Df_a$.
[/step]
[step:Multiply the scalar expansions and isolate the linear part]
Apply the differentiability expansion to $u,v: U \to \mathbb{R}$. Thus there are remainder maps $r_u,r_v: V \to \mathbb{R}$ such that
\begin{align*}
u(a+h)=u(a)+Du_a(h)+r_u(h)
\end{align*}
and
\begin{align*}
v(a+h)=v(a)+Dv_a(h)+r_v(h),
\end{align*}
with $|r_u(h)|/|h| \to 0$ and $|r_v(h)|/|h| \to 0$ as $h \to 0$, $h \neq 0$.
Define $L: \mathbb{R}^m \to \mathbb{R}$ by
\begin{align*}
L(h):=u(a)Dv_a(h)+v(a)Du_a(h).
\end{align*}
Since $Du_a$ and $Dv_a$ are linear maps and $u(a),v(a) \in \mathbb{R}$ are scalars, $L$ is linear. Expanding the product gives
\begin{align*}
(uv)(a+h)-u(a)v(a)-L(h)=u(a)r_v(h)+v(a)r_u(h)+Du_a(h)Dv_a(h)+Du_a(h)r_v(h)+Dv_a(h)r_u(h)+r_u(h)r_v(h).
\end{align*}
[guided]
We now prove the product rule from the differentiability expansions, keeping track of exactly which part is linear in $h$ and which part is a genuine error. Differentiability of $u$ at $a$ means that there is a linear map $Du_a: \mathbb{R}^m \to \mathbb{R}$ and a remainder map $r_u: V \to \mathbb{R}$ defined by
\begin{align*}
r_u(h):=u(a+h)-u(a)-Du_a(h)
\end{align*}
such that
\begin{align*}
\lim_{\substack{h \to 0, h \neq 0}} \frac{|r_u(h)|}{|h|}=0.
\end{align*}
Likewise, differentiability of $v$ at $a$ gives a linear map $Dv_a: \mathbb{R}^m \to \mathbb{R}$ and a remainder map $r_v: V \to \mathbb{R}$ defined by
\begin{align*}
r_v(h):=v(a+h)-v(a)-Dv_a(h)
\end{align*}
such that
\begin{align*}
\lim_{\substack{h \to 0, h \neq 0}} \frac{|r_v(h)|}{|h|}=0.
\end{align*}
Thus, for every $h \in V$,
\begin{align*}
u(a+h)=u(a)+Du_a(h)+r_u(h)
\end{align*}
and
\begin{align*}
v(a+h)=v(a)+Dv_a(h)+r_v(h).
\end{align*}
The only possible derivative of the product is the part of the product expansion that is linear in $h$. Multiplying the two expansions, the constant term is $u(a)v(a)$, and the linear terms are $u(a)Dv_a(h)$ and $v(a)Du_a(h)$. This motivates defining $L: \mathbb{R}^m \to \mathbb{R}$ by
\begin{align*}
L(h):=u(a)Dv_a(h)+v(a)Du_a(h).
\end{align*}
Because $Du_a$ and $Dv_a$ are linear maps, and multiplication by the fixed [real numbers](/page/Real%20Numbers) $u(a)$ and $v(a)$ preserves linearity, $L$ is linear.
Now subtract the constant term and this candidate linear term from the product:
\begin{align*}
(uv)(a+h)-u(a)v(a)-L(h)=u(a)r_v(h)+v(a)r_u(h)+Du_a(h)Dv_a(h)+Du_a(h)r_v(h)+Dv_a(h)r_u(h)+r_u(h)r_v(h).
\end{align*}
This identity is the key bookkeeping step. The first two terms contain the remainders directly, and the remaining four terms are products of quantities that are at least first order in $h$. The next step proves that the whole right-hand side is $o(|h|)$.
[/guided]
[/step]
[step:Estimate the product remainder as negligible compared with $|h|$]
Choose the standard basis vectors $e_1,\dots,e_m \in \mathbb{R}^m$. Define finite constants
\begin{align*}
M_u:=\sum_{i=1}^m |Du_a(e_i)|
\end{align*}
and
\begin{align*}
M_v:=\sum_{i=1}^m |Dv_a(e_i)|.
\end{align*}
For every $h=(h_1,\dots,h_m) \in \mathbb{R}^m$,
\begin{align*}
|Du_a(h)|\leq M_u |h|
\end{align*}
and
\begin{align*}
|Dv_a(h)|\leq M_v |h|.
\end{align*}
Set
\begin{align*}
R(h):=(uv)(a+h)-u(a)v(a)-L(h).
\end{align*}
Using the expansion from the previous step and the triangle inequality, for $h \in V$ with $h \neq 0$,
\begin{align*}
\frac{|R(h)|}{|h|}\leq |u(a)|\frac{|r_v(h)|}{|h|}+|v(a)|\frac{|r_u(h)|}{|h|}+M_uM_v|h|+M_u|r_v(h)|+M_v|r_u(h)|+\frac{|r_u(h)|}{|h|}|r_v(h)|.
\end{align*}
Since $r_u(h)=o(|h|)$ and $r_v(h)=o(|h|)$, the quotients $|r_u(h)|/|h|$ and $|r_v(h)|/|h|$ tend to $0$, and in particular the remainders $r_u(h)$ and $r_v(h)$ tend to $0$. Each term on the right-hand side therefore tends to $0$. Hence
\begin{align*}
\lim_{\substack{h \to 0, h \neq 0}} \frac{|R(h)|}{|h|}=0.
\end{align*}
Thus
\begin{align*}
(uv)(a+h)=u(a)v(a)+L(h)+o(|h|)
\end{align*}
as $h \to 0$. By the definition of differentiability, $uv$ is differentiable at $a$ and $D(uv)_a=L$. Therefore, for every $h \in \mathbb{R}^m$,
\begin{align*}
D(uv)_a(h)=u(a)Dv_a(h)+v(a)Du_a(h).
\end{align*}
This proves all asserted algebra rules.
[/step]
