# Real functions/Differentiability/Rules/Section
## Lemma

Let $D\subseteq \mathbb {R}$ be a subset, $a\in D$ a point, and
    $f,g\colon D\longrightarrow \mathbb {R}$
functions which are differentiable
$a$
* The sum $f+g$ is differentiable in $a$, with
    $(f+g)'(a)=f'(a)+g'(a)\,.$
* The product $f\cdot g$ is differentiable in $a$, with
    $(f\cdot g)'(a)=f'(a)g(a)+f(a)g'(a)\,.$
* For $c\in \mathbb {R}$, also $cf$ is differentiable in $a$, with
    $(cf)'(a)=cf'(a)\,.$
* If $g$ has no zero in $a$, then $1/g$ is differentiable in $a$, with
    ${\left({\frac {1}{g}}\right)}'(a)={\frac {-g'(a)}{(g(a))^{2}}}\,.$
* If $g$ has no zero in $a$, then $f/g$ is differentiable in $a$, with
    ${\left({\frac {f}{g}}\right)}'(a)={\frac {f'(a)g(a)-f(a)g'(a)}{(g(a))^{2}}}\,.$

### 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).
$\Box$
These rules are called sum rule, product rule, quotient rule. The following statement is called chain rule.

## Theorem

Let $D,E\subseteq \mathbb {R}$ denote subsets, and let
    $f\colon D\longrightarrow \mathbb {R}$
and
    $g\colon E\longrightarrow \mathbb {R}$
be functions with $f(D)\subseteq E$. Suppose that $f$ is differentiable in $a$ and that $g$ is differentiable in $b:=f(a)$. Then also the composition
    $g\circ f\colon D\longrightarrow \mathbb {R}$
is differentiable in $a$, and its derivative is
    $(g\circ f)'(a)=g'(f(a))\cdot f'(a)\,.$

### Proof

Due to fact, one can write
    $f(x)=f(a)+f'(a)(x-a)+r(x)(x-a)\,$
and
    $g(y)=g(b)+g'(b)(y-b)+s(y)(y-b)\,.$
Therefore,
    ${\begin{aligned}g(f(x))&=g(f(a))+g'(f(a))(f(x)-f(a))+s(f(x))(f(x)-f(a))\\&=g(f(a))+g'(f(a)){\left(f'(a)(x-a)+r(x)(x-a)\right)}+s(f(x)){\left(f'(a)(x-a)+r(x)(x-a)\right)}\\&=g(f(a))+g'(f(a))f'(a)(x-a)+{\left(g'(f(a))r(x)+s(f(x))(f'(a)+r(x))\right)}(x-a).\end{aligned}}$
The remainder function
    $t(x):=g'(f(a))r(x)+s(f(x))(f'(a)+r(x))\,$
is continuous in $a$ with value $0$.
$\Box$

## Theorem

Let $D,E\subseteq \mathbb {R}$ denote intervals, and let
    $f\colon D\longrightarrow E\subseteq \mathbb {R}$
be a bijective continuous function, with the inverse function
    $f^{-1}\colon E\longrightarrow D.$
$f$
differentiable in $a\in D$ with $f'(a)\neq 0$. Then also the inverse function $f^{-1}$ is differentiable in $b:=f(a)$, and
    $(f^{-1})'(b)={\frac {1}{f'(f^{-1}(b))}}={\frac {1}{f'(a)}}\,$

### Proof

We consider the difference quotient
    ${\frac {f^{-1}(y)-f^{-1}(b)}{y-b}}={\frac {f^{-1}(y)-a}{y-b}}\,,$
and have to show that the limit for $y\rightarrow b$ exists, and obtains the value claimed. For this, let ${\left(y_{n}\right)}_{n\in \mathbb {N} }$ denote a sequence in $E\setminus \{b\}$, converging to $b$. Because of fact, the function $f^{-1}$ is continuous. Therefore, also the sequence with the members $x_{n}:=f^{-1}(y_{n})$ converges to $a$. Because of bijectivity, $x_{n}\neq a$ for all $n$. Thus
    $\lim _{n\rightarrow \infty }{\frac {f^{-1}(y_{n})-a}{y_{n}-b}}=\lim _{n\rightarrow \infty }{\frac {x_{n}-a}{f(x_{n})-f(a)}}={\left(\lim _{n\rightarrow \infty }{\frac {f(x_{n})-f(a)}{x_{n}-a}}\right)}^{-1}\,,$
where the right-hand side exists, due to the condition, and the second equation follows from fact.
$\Box$

## Example

The function
    $f^{-1}\colon \mathbb {R} _{+}\longrightarrow \mathbb {R} _{+},x\longmapsto {\sqrt {x}},$
is the inverse function of the function $f$, given by $f(x)=x^{2}$ (restricted to $\mathbb {R} _{+}$). The derivative of $f$ in a point $a$ is $f'(a)=2a$. Due to fact, for $b\in \mathbb {R} _{+}$, the relation
    ${\left(f^{-1}\right)}'(b)={\frac {1}{f'(f^{-1}(b))}}={\frac {1}{2{\sqrt {b}}}}={\frac {1}{2}}b^{-{\frac {1}{2}}}\,$
holds. In the zero point, however, $f^{-1}$ is not differentiable.

## Example

The function
    $f^{-1}\colon \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto x^{\frac {1}{3}},$
is the inverse function of the function $f$, given by $f(x)=x^{3}$. The derivative of $f$ in $a$ is $f'(a)=3a^{2}$, which is different from $0$ for $a\neq 0$. Due to fact, we have for $b\neq 0$ the relation
    ${\left(f^{-1}\right)}'(b)={\frac {1}{f'(f^{-1}(b))}}={\frac {1}{3{\left(b^{\frac {1}{3}}\right)}^{2}}}={\frac {1}{3}}b^{-{\frac {2}{3}}}\,.$
In the zero point, however, $f^{-1}$ is not differentiable.