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

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

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 $x\to a$ to $-\frac{1}{g(a{)}^2}$, and because of the differentiability of $g$ in $a$, the right-hand factor converges to $g^{\prime }(a)$.
(5) follows from (2) and (4).

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