Proof of Derivative of a Sum and of a Difference

By An Open Interval is an Interval All of Whose Points Are Interior the set $(p,q)$ is an interval and $c$ is an interior point of it, so differentiability at $c$ is meaningful for each of $g$, $h$, $g+h$ and $g-h$. Write $L=g'(c)$ and $M=h'(c)$, and let $|\cdot|$ be the absolute value. Claim numbers below refer to Elementary Order Arithmetic in an Ordered Field and to Properties of the Absolute Value in an Ordered Field as indicated.

Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. By claim 8 of Elementary Order Arithmetic in an Ordered Field the element $\eta=\varepsilon\cdot2^{-1}$ satisfies $0<\eta$ and $\eta+\eta=\varepsilon$.

Differentiability of $g$ at $c$, applied with $\eta$, provides $\delta_{1}$ with $0<\delta_{1}$; differentiability of $h$ at $c$, applied with $\eta$, provides $\delta_{2}$ with $0<\delta_{2}$. By claim 9 there is $\delta$ with $\delta\le\delta_{1}$, $\delta\le\delta_{2}$, and $\delta$ equal to one of them; in either case $0<\delta$.

Let $k\in\mathbb{R}$ satisfy $0<|k|<\delta$ and $c+k\in(p,q)$. By claim 2 we have $|k|<\delta_{1}$ and $|k|<\delta_{2}$, so with

we have $|A|<\eta$ and $|B|<\eta$.

The sum. By the definition of $g+h$, the numerator $(g+h)(c+k)-(g+h)(c)$ equals $\bigl(g(c+k)-g(c)\bigr)+\bigl(h(c+k)-h(c)\bigr)$, and division by $k$ distributes over this sum in the field $\mathbb{R}$, so

Claim 5 of Properties of the Absolute Value in an Ordered Field gives $|A+B|\le|A|+|B|$, and claim 3 of Elementary Order Arithmetic in an Ordered Field gives $|A|+|B|<\eta+\eta=\varepsilon$; claim 2 gives $|A+B|<\varepsilon$.

Since $\varepsilon$ was arbitrary and $\delta$ was produced from it, the real number $L+M$ satisfies the defining condition of the derivative of $g+h$ at $c$. Hence $g+h$ is differentiable at $c$ with $(g+h)'(c)=L+M=g'(c)+h'(c)$.

The difference. Likewise the numerator $(g-h)(c+k)-(g-h)(c)$ equals $\bigl(g(c+k)-g(c)\bigr)+\bigl(-(h(c+k)-h(c))\bigr)$, so

By claim 5 of Properties of the Absolute Value in an Ordered Field and claim 2 of that lemma, $|A+(-B)|\le|A|+|-B|=|A|+|B|$, and the same estimate gives $|A+(-B)|<\varepsilon$. Hence $g-h$ is differentiable at $c$ with $(g-h)'(c)=L-M=g'(c)-h'(c)$.