Sums, scalar multiples, products and quotients: $(f+g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)+g^{\mathrm{\prime }}(c)$, $(\alpha f{)}^{\mathrm{\prime }}(c)=\alpha f^{\mathrm{\prime }}(c)$, $(fg{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)g(c)+f(c)g^{\mathrm{\prime }}(c)$, and $(f\mathrm{/}g{)}^{\mathrm{\prime }}(c)=(f^{\mathrm{\prime }}(c)g(c)-f(c)g^{\mathrm{\prime }}(c))\mathrm{/}g(c{)}^2$ when $g(c)\mathrm{\ne }0$

Let $A\subseteq \mathbb{R}$, let $c\in A$ be a limit point of $A$ (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$), let $f,g:A\to \mathbb{R}$ be differentiable at $c$ (The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set) and let $\alpha \in \mathbb{R}$. Then:

1. $f+g$ is differentiable at $c$ and $(f+g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)+g^{\mathrm{\prime }}(c)$;
2. $\alpha f$ is differentiable at $c$ and $(\alpha f{)}^{\mathrm{\prime }}(c)=\alpha f^{\mathrm{\prime }}(c)$;
3. $fg$ is differentiable at $c$ and $(fg{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)g(c)+f(c)g^{\mathrm{\prime }}(c)$;
4. if $g(c)\mathrm{\ne }0$ then, writing $A_0:=\{x\in A:g(x)\mathrm{\ne }0\}$, the point $c$ lies in $A_0$ and is a limit point of $A_0$, the quotient $(f\mathrm{/}g){\mathrm{\mid }}_{A_0}:A_0\to \mathbb{R}$, $x\mapsto f(x)\mathrm{/}g(x)$, is differentiable at $c$ as a function on $A_0$, and $((f\mathrm{/}g){\mathrm{\mid }}_{A_0}{)}^{\mathrm{\prime }}(c)=\frac{f^{\mathrm{\prime }}(c)g(c)-f(c)g^{\mathrm{\prime }}(c)}{g(c{)}^2}\mathrm{.}$

Each claim asserts two things: that the derivative on the left exists, and that it has the stated value. Both are proved.

Why claim 4 is stated on $A_0$. The function $f\mathrm{/}g$ is not defined where $g$ vanishes, and $g$ may vanish at points of $A$ far from $c$; restricting to $A_0$ is forced. That the restriction still has $c$ as a limit point, so that a derivative there means anything at all, is not free either, and it is the last claim of If ${\lim }_{x\to c}f(x)=L\mathrm{\ne }0$ then $\mathrm{\mid }f\mathrm{\mid }>\mathrm{\mid }L\mathrm{\mid }\mathrm{/}2$ on a punctured neighbourhood of $c$; in particular if $L>0$ then $f>L\mathrm{/}2>0$ there applied to $g$. The hypothesis is $g(c)\mathrm{\ne }0$, not "$g$ vanishes nowhere".

Everything is proved through Carathéodory's characterisation: $f$ is differentiable at $c$ if and only if there is $\phi :A\to \mathbb{R}$, continuous at $c$, with $f(x)-f(c)=\phi (x)(x-c)$ for every $x\in A$, and then $\phi$ is unique and $\phi (c)=f^{\mathrm{\prime }}(c)$. No difference quotient is estimated and no limit theorem beyond continuity is used, so no choice principle is spent. The four identities are four algebraic rearrangements of an increment, each followed by a reading of Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function.

Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$

The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set

If ${\lim }_{x\to c}f(x)=L\mathrm{\ne }0$ then $\mathrm{\mid }f\mathrm{\mid }>\mathrm{\mid }L\mathrm{\mid }\mathrm{/}2$ on a punctured neighbourhood of $c$; in particular if $L>0$ then $f>L\mathrm{/}2>0$ there

Carathéodory's characterisation: $f$ is differentiable at $c$ if and only if there is $\phi :A\to \mathbb{R}$, continuous at $c$, with $f(x)-f(c)=\phi (x)(x-c)$ for every $x\in A$, and then $\phi$ is unique and $\phi (c)=f^{\mathrm{\prime }}(c)$

Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function

• The product rearrangement in one line. The identity $fg-f(c)g(c)=(f-f(c))g+f(c)(g-g(c))$ splits the increment of a product into two increments, one multiplied by $g$ and one by a constant. It is the same identity that carries the product case of Sums, scalar multiples, products and quotients of function limits, the quotient under the hypothesis that the denominator limit is nonzero, read at the level of increments rather than of $\epsilon$; here the factor $g$ has to be continuous at $c$ rather than merely bounded near it, and A function differentiable at $c$ is continuous at $c$ is what supplies that.

• The reciprocal is the case $f\equiv 1$. Claim 4 then reads $((1\mathrm{/}g){\mathrm{\mid }}_{A_0}{)}^{\mathrm{\prime }}(c)=-g^{\mathrm{\prime }}(c)\mathrm{/}g(c{)}^2$, since $f^{\mathrm{\prime }}(c)=0$ for a constant $f$; nothing separate has to be proved, and the derivative of a negative integer power on this page is obtained exactly this way.

• Two hypotheses that look removable and are not. In claim 4 the hypothesis $g(c)\mathrm{\ne }0$ cannot be weakened to "$g$ is nonzero somewhere near $c$", because $c$ itself must lie in the smaller domain for a derivative there to be a statement about $c$; and the conclusion is about $(f\mathrm{/}g){\mathrm{\mid }}_{A_0}$, not about any extension of it to $A$, since no such extension is canonical.