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

Proof

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

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

and

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

Summing up yields

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

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

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

and

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

and multiply both equations. This yields

{\displaystyle {}{\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 ${\displaystyle {}0}$ for ${\displaystyle {}x=a}$.
(3) follows from (2), since a constant function is differentiable with derivative ${\displaystyle {}0}$.
(4). We have

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

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