Differentiable function/D open in K/Rules/Fact/Proof

Proof

(1). We write ${}f$ and ${}g$ respectively with the objects which were formulated in fact, that is

${}f(x)=f(a)+s(x-a)+r(x)(x-a)\,$

and

${}g(x)=g(a)+{\tilde {s}}(x-a)+{\tilde {r}}(x)(x-a)\,.$

Summing up yields

${}f(x)+g(x)=f(a)+g(a)+(s+{\tilde {s}})(x-a)+(r+{\tilde {r}})(x)(x-a)\,.$

Here, the sum ${}r+{\tilde {r}}$ is again continuous in ${}a$, with value ${}0$.
(2). We start again with

${}f(x)=f(a)+s(x-a)+r(x)(x-a)\,$

and

${}g(x)=g(a)+{\tilde {s}}(x-a)+{\tilde {r}}(x)(x-a)\,,$

and multiply both equations. This yields

${}{\begin{aligned}f(x)g(x)&=(f(a)+s(x-a)+r(x)(x-a))(g(a)+{\tilde {s}}(x-a)+{\tilde {r}}(x)(x-a))\\&=f(a)g(a)+(sg(a)+{\tilde {s}}f(a))(x-a)\\&\,\,\,\,\,+(f(a){\tilde {r}}(x)+g(a)r(x)+s{\tilde {s}}(x-a)+s{\tilde {r}}(x)(x-a)+{\tilde {s}}r(x)(x-a)+r(x){\tilde {r}}(x)(x-a))(x-a).\end{aligned}}$

Due to fact for limits, the expression consisting of the last six summands is a continuous function, with value ${}0$ for ${}x=a$.
(3) follows from (2), since a constant function is differentiable with derivative ${}0$.
(4). We have

${}{\frac {{\frac {1}{g(x)}}-{\frac {1}{g(a)}}}{x-a}}={\frac {-1}{g(a)g(x)}}\cdot {\frac {g(x)-g(a)}{x-a}}\,.$

Since ${}g$ is continuous in ${}a$, due to fact, the left-hand factor converges for ${}x\rightarrow a$ to ${}-{\frac {1}{g(a)^{2}}}$, and because of the differentiability of ${}g$ in ${}a$, the right-hand factor converges to ${}g'(a)$.
(5) follows from (2) and (4).

To fact