[proofplan]
We prove each algebra operation by computing its first-order expansion at an arbitrary point $a\in U$. The sum rule follows by adding the differentiability expansions for $f$ and $g$, while the product rule follows by expanding the product and checking that the leftover terms are $o(|h|)$. For the quotient, we first prove that $1/g$ is $C^1$ by a direct reciprocal expansion, and then multiply by $f$. The continuity of the displayed gradients follows from the continuity of $f$, $g$, $\nabla f$, and $\nabla g$, together with the nonvanishing of $g$ in the quotient case.
[/proofplan]
[step:Compute the derivative of the sum from the two first-order expansions]
Throughout the proof, for $x\in\mathbb R^n$ and $r>0$, let $B(x,r):=\{y\in\mathbb R^n:|y-x|<r\}$ denote the open Euclidean ball. For a scalar function $E$ defined near $0\in\mathbb R^n$, the notation $E(h)=o(|h|)$ as $h\to 0$ means $|E(h)|/|h|\to 0$, and $E(h)=O(|h|)$ as $h\to 0$ means that there exist constants $C>0$ and $\delta>0$ such that $|E(h)|\le C|h|$ whenever $0<|h|<\delta$.
Let $a\in U$. Since $U$ is open, there exists $\rho>0$ such that $B(a,\rho)\subset U$. For $h\in \mathbb R^n$ with $0<|h|<\rho$, differentiability of $f$ and $g$ at $a$ gives remainders
\begin{align*}
r_f: B(0,\rho)\to \mathbb R
\end{align*}
and
\begin{align*}
r_g: B(0,\rho)\to \mathbb R
\end{align*}
such that
\begin{align*}
f(a+h)=f(a)+\nabla f(a)\cdot h+r_f(h)
\end{align*}
and
\begin{align*}
g(a+h)=g(a)+\nabla g(a)\cdot h+r_g(h),
\end{align*}
with
\begin{align*}
\frac{|r_f(h)|}{|h|}\to 0
\end{align*}
and
\begin{align*}
\frac{|r_g(h)|}{|h|}\to 0
\end{align*}
as $h\to 0$.
Define the sum function $S:U\to \mathbb R$ by $S(x)=f(x)+g(x)$. Adding the two expansions gives
\begin{align*}
S(a+h)=S(a)+(\nabla f(a)+\nabla g(a))\cdot h+r_f(h)+r_g(h).
\end{align*}
Since
\begin{align*}
\frac{|r_f(h)+r_g(h)|}{|h|}\le \frac{|r_f(h)|}{|h|}+\frac{|r_g(h)|}{|h|}\to 0,
\end{align*}
the function $S$ is differentiable at $a$ and
\begin{align*}
\nabla S(a)=\nabla f(a)+\nabla g(a).
\end{align*}
Because $a\in U$ was arbitrary, $f+g$ is differentiable on $U$ and the displayed gradient identity holds on $U$.
[/step]
[step:Expand the product and isolate the first-order part]
Define the product function $P:U\to \mathbb R$ by $P(x)=f(x)g(x)$. Fix $a\in U$ and choose $\rho>0$ with $B(a,\rho)\subset U$. Use the same differentiability expansions at $a$, and define
\begin{align*}
\alpha(h):=\nabla f(a)\cdot h+r_f(h)
\end{align*}
and
\begin{align*}
\beta(h):=\nabla g(a)\cdot h+r_g(h).
\end{align*}
Then $f(a+h)=f(a)+\alpha(h)$ and $g(a+h)=g(a)+\beta(h)$, so
\begin{align*}
P(a+h)-P(a)=f(a)\beta(h)+g(a)\alpha(h)+\alpha(h)\beta(h).
\end{align*}
Substituting the definitions of $\alpha$ and $\beta$ gives
\begin{align*}
P(a+h)=P(a)+(f(a)\nabla g(a)+g(a)\nabla f(a))\cdot h+R_P(h),
\end{align*}
where
\begin{align*}
R_P(h):=f(a)r_g(h)+g(a)r_f(h)+\alpha(h)\beta(h).
\end{align*}
The first two terms in $R_P(h)$ are $o(|h|)$. Also, $\alpha(h)=O(|h|)$ and $\beta(h)=O(|h|)$ as $h\to 0$, because $r_f(h)=o(|h|)$ and $r_g(h)=o(|h|)$. Hence $\alpha(h)\beta(h)=O(|h|^2)=o(|h|)$. Therefore
\begin{align*}
\frac{|R_P(h)|}{|h|}\to 0.
\end{align*}
Thus $P$ is differentiable at $a$ and
\begin{align*}
\nabla P(a)=f(a)\nabla g(a)+g(a)\nabla f(a).
\end{align*}
Since $a\in U$ was arbitrary,
\begin{align*}
\nabla(fg)=f\nabla g+g\nabla f
\end{align*}
on $U$.
[guided]
We want to prove the product rule without hiding any differentiability step. Recall that $B(x,r):=\{y\in\mathbb R^n:|y-x|<r\}$ denotes the open Euclidean ball. Also, $E(h)=o(|h|)$ means $|E(h)|/|h|\to 0$, while $E(h)=O(|h|)$ means that $|E(h)|\le C|h|$ for all sufficiently small nonzero $h$ and for some constant $C>0$. Fix $a\in U$. Since $U$ is open, we can choose $\rho>0$ such that $a+h\in U$ whenever $|h|<\rho$. Differentiability of $f$ and $g$ at $a$ means that there are remainders
\begin{align*}
r_f:B(0,\rho)\to \mathbb R
\end{align*}
and
\begin{align*}
r_g:B(0,\rho)\to \mathbb R
\end{align*}
satisfying
\begin{align*}
f(a+h)=f(a)+\nabla f(a)\cdot h+r_f(h)
\end{align*}
and
\begin{align*}
g(a+h)=g(a)+\nabla g(a)\cdot h+r_g(h),
\end{align*}
with
\begin{align*}
\frac{|r_f(h)|}{|h|}\to 0
\end{align*}
and
\begin{align*}
\frac{|r_g(h)|}{|h|}\to 0.
\end{align*}
Define
\begin{align*}
\alpha(h):=\nabla f(a)\cdot h+r_f(h)
\end{align*}
and
\begin{align*}
\beta(h):=\nabla g(a)\cdot h+r_g(h).
\end{align*}
These are the increments of $f$ and $g$ from the base point $a$, namely $f(a+h)=f(a)+\alpha(h)$ and $g(a+h)=g(a)+\beta(h)$. The useful point of this notation is that the product expansion becomes ordinary algebra:
\begin{align*}
f(a+h)g(a+h)-f(a)g(a)=f(a)\beta(h)+g(a)\alpha(h)+\alpha(h)\beta(h).
\end{align*}
The first-order terms are obtained by replacing $\alpha(h)$ and $\beta(h)$ with their linear parts:
\begin{align*}
f(a)\beta(h)+g(a)\alpha(h)=(f(a)\nabla g(a)+g(a)\nabla f(a))\cdot h+f(a)r_g(h)+g(a)r_f(h).
\end{align*}
Thus
\begin{align*}
f(a+h)g(a+h)=f(a)g(a)+(f(a)\nabla g(a)+g(a)\nabla f(a))\cdot h+R_P(h),
\end{align*}
where
\begin{align*}
R_P(h):=f(a)r_g(h)+g(a)r_f(h)+\alpha(h)\beta(h).
\end{align*}
It remains to check that this remainder is negligible compared with $|h|$. The terms $f(a)r_g(h)$ and $g(a)r_f(h)$ are $o(|h|)$ because $f(a)$ and $g(a)$ are fixed [real numbers](/page/Real%20Numbers). For the mixed term, differentiability gives $r_f(h)=o(|h|)$ and $r_g(h)=o(|h|)$, so $\alpha(h)=O(|h|)$ and $\beta(h)=O(|h|)$ as $h\to 0$. Therefore $\alpha(h)\beta(h)=O(|h|^2)$, and hence $\alpha(h)\beta(h)=o(|h|)$. Consequently
\begin{align*}
\frac{|R_P(h)|}{|h|}\to 0.
\end{align*}
This proves that $fg$ is differentiable at $a$ and that
\begin{align*}
\nabla(fg)(a)=f(a)\nabla g(a)+g(a)\nabla f(a).
\end{align*}
Since $a$ was arbitrary, the product derivative formula holds throughout $U$.
[/guided]
[/step]
[step:Show that the reciprocal is $C^1$ wherever $g$ is nonzero]
Assume now that $g(x)\neq 0$ for every $x\in U$. Define the reciprocal function $H:U\to \mathbb R$ by $H(x)=1/g(x)$. Fix $a\in U$. Since $g$ is continuous at $a$ and $g(a)\neq 0$, there exists $\rho>0$ such that $B(a,\rho)\subset U$ and $g(a+h)\neq 0$ for all $h\in B(0,\rho)$.
For $0<|h|<\rho$,
\begin{align*}
H(a+h)-H(a)=\frac{g(a)-g(a+h)}{g(a+h)g(a)}.
\end{align*}
Using
\begin{align*}
g(a+h)-g(a)=\nabla g(a)\cdot h+r_g(h),
\end{align*}
we obtain
\begin{align*}
H(a+h)-H(a)=-\frac{\nabla g(a)\cdot h}{g(a)^2}+R_H(h),
\end{align*}
where
\begin{align*}
R_H(h):=-\frac{r_g(h)}{g(a+h)g(a)}+\left(\nabla g(a)\cdot h\right)\left(\frac{1}{g(a)^2}-\frac{1}{g(a+h)g(a)}\right).
\end{align*}
Since $g(a+h)\to g(a)$ and $g(a)\neq 0$, the factor $1/(g(a+h)g(a))$ remains bounded near $h=0$, so the first term in $R_H(h)$ is $o(|h|)$. For the second term, the factor $\nabla g(a)\cdot h$ is $O(|h|)$ and
\begin{align*}
\frac{1}{g(a)^2}-\frac{1}{g(a+h)g(a)}=\frac{g(a+h)-g(a)}{g(a)^2g(a+h)}\to 0.
\end{align*}
Thus the second term is also $o(|h|)$. Hence
\begin{align*}
\frac{|R_H(h)|}{|h|}\to 0,
\end{align*}
so $H$ is differentiable at $a$ and
\begin{align*}
\nabla H(a)=-\frac{\nabla g(a)}{g(a)^2}.
\end{align*}
Because $a\in U$ was arbitrary,
\begin{align*}
\nabla(1/g)=-\frac{\nabla g}{g^2}
\end{align*}
on $U$.
[/step]
[step:Apply the product computation to the quotient]
Define the quotient function $Q:U\to \mathbb R$ by $Q(x)=f(x)/g(x)$. Since $Q=fH$ and both $f$ and $H$ are differentiable on $U$, the product computation gives
\begin{align*}
\nabla Q=f\nabla H+H\nabla f.
\end{align*}
Substituting $H=1/g$ and $\nabla H=-\nabla g/g^2$ yields
\begin{align*}
\nabla Q=f\left(-\frac{\nabla g}{g^2}\right)+\frac{1}{g}\nabla f.
\end{align*}
Combining the two terms over the common denominator $g^2$ gives
\begin{align*}
\nabla(f/g)=\frac{g\nabla f-f\nabla g}{g^2}.
\end{align*}
[/step]
[step:Verify continuity of the computed gradients]
Since $f,g\in C^1(U)$, the maps
\begin{align*}
f,g:U\to \mathbb R
\end{align*}
and
\begin{align*}
\nabla f,\nabla g:U\to \mathbb R^n
\end{align*}
are continuous. Therefore $\nabla f+\nabla g$ is continuous, so $f+g\in C^1(U)$.
The maps $f\nabla g$ and $g\nabla f$ are continuous as products of continuous scalar and vector-valued maps, and hence $f\nabla g+g\nabla f$ is continuous. Therefore $fg\in C^1(U)$.
If $g$ is nonvanishing on $U$, then $1/g:U\to \mathbb R$ and $1/g^2:U\to \mathbb R$ are continuous, because the reciprocal map is continuous on $\mathbb R\setminus\{0\}$ and $g(U)\subset \mathbb R\setminus\{0\}$. Hence
\begin{align*}
\frac{g\nabla f-f\nabla g}{g^2}
\end{align*}
is continuous on $U$. Therefore $f/g\in C^1(U)$ with the stated derivative formula. This proves all assertions.
[/step]
