Real functions/Differentiability/Rules/Section

Lemma

Let ${\displaystyle {}D\subseteq \mathbb {R} }$ be a subset, ${\displaystyle {}a\in D}$ a point, and

${\displaystyle f,g\colon D\longrightarrow \mathbb {R} }$

functions which are differentiable

in ${\displaystyle {}a}$. Then the following rules for differentiability holds.
1. The sum ${\displaystyle {}f+g}$ is differentiable in ${\displaystyle {}a}$, with
${\displaystyle {}(f+g)'(a)=f'(a)+g'(a)\,.}$
2. The product ${\displaystyle {}f\cdot g}$ is differentiable in ${\displaystyle {}a}$, with
${\displaystyle {}(f\cdot g)'(a)=f'(a)g(a)+f(a)g'(a)\,.}$
3. For ${\displaystyle {}c\in \mathbb {R} }$, also ${\displaystyle {}cf}$ is differentiable in ${\displaystyle {}a}$, with
${\displaystyle {}(cf)'(a)=cf'(a)\,.}$
4. If ${\displaystyle {}g}$ has no zero in ${\displaystyle {}a}$, then ${\displaystyle {}1/g}$ is differentiable in ${\displaystyle {}a}$, with
${\displaystyle {}{\left({\frac {1}{g}}\right)}'(a)={\frac {-g'(a)}{(g(a))^{2}}}\,.}$
5. If ${\displaystyle {}g}$ has no zero in ${\displaystyle {}a}$, then ${\displaystyle {}f/g}$ is differentiable in ${\displaystyle {}a}$, with
${\displaystyle {}{\left({\frac {f}{g}}\right)}'(a)={\frac {f'(a)g(a)-f(a)g'(a)}{(g(a))^{2}}}\,.}$

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 differentiability of ${\displaystyle {}g}$ in ${\displaystyle {}a}$, the right-hand factor converges to ${\displaystyle {}g'(a)}$.
(5) follows from (2) and (4).

${\displaystyle \Box }$

These rules are called sum rule, product rule, quotient rule. The following statement is called chain rule.

Theorem

Let ${\displaystyle {}D,E\subseteq \mathbb {R} }$ denote subsets, and let

${\displaystyle f\colon D\longrightarrow \mathbb {R} }$

and

${\displaystyle g\colon E\longrightarrow \mathbb {R} }$

be functions with ${\displaystyle {}f(D)\subseteq E}$. Suppose that ${\displaystyle {}f}$ is differentiable in ${\displaystyle {}a}$ and that ${\displaystyle {}g}$ is differentiable in ${\displaystyle {}b:=f(a)}$. Then also the composition

${\displaystyle g\circ f\colon D\longrightarrow \mathbb {R} }$

is differentiable in ${\displaystyle {}a}$, and its derivative is

${\displaystyle {}(g\circ f)'(a)=g'(f(a))\cdot f'(a)\,.}$

Proof

Due to fact, one can write

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

and

${\displaystyle {}g(y)=g(b)+g'(b)(y-b)+s(y)(y-b)\,.}$

Therefore,

{\displaystyle {}{\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

${\displaystyle {}t(x):=g'(f(a))r(x)+s(f(x))(f'(a)+r(x))\,}$

is continuous in ${\displaystyle {}a}$ with value ${\displaystyle {}0}$.

${\displaystyle \Box }$

Theorem

Let ${\displaystyle {}D,E\subseteq \mathbb {R} }$ denote intervals, and let

${\displaystyle f\colon D\longrightarrow E\subseteq \mathbb {R} }$

be a bijective continuous function, with the inverse function

${\displaystyle f^{-1}\colon E\longrightarrow D.}$
Suppose that ${\displaystyle {}f}$ is

differentiable in ${\displaystyle {}a\in D}$ with ${\displaystyle {}f'(a)\neq 0}$. Then also the inverse function ${\displaystyle {}f^{-1}}$ is differentiable in ${\displaystyle {}b:=f(a)}$, and

${\displaystyle {}(f^{-1})'(b)={\frac {1}{f'(f^{-1}(b))}}={\frac {1}{f'(a)}}\,}$

holds.

Proof

We consider the difference quotient

${\displaystyle {}{\frac {f^{-1}(y)-f^{-1}(b)}{y-b}}={\frac {f^{-1}(y)-a}{y-b}}\,,}$

and have to show that the limit for ${\displaystyle {}y\rightarrow b}$ exists, and obtains the value claimed. For this, let ${\displaystyle {}{\left(y_{n}\right)}_{n\in \mathbb {N} }}$ denote a sequence in ${\displaystyle {}E\setminus \{b\}}$, converging to ${\displaystyle {}b}$. Because of fact, the function ${\displaystyle {}f^{-1}}$ is continuous. Therefore, also the sequence with the members ${\displaystyle {}x_{n}:=f^{-1}(y_{n})}$ converges to ${\displaystyle {}a}$. Because of bijectivity, ${\displaystyle {}x_{n}\neq a}$ for all ${\displaystyle {}n}$. Thus

${\displaystyle {}\lim _{n\rightarrow \infty }{\frac {f^{-1}(y_{n})-a}{y_{n}-b}}=\lim _{n\rightarrow \infty }{\frac {x_{n}-a}{f(x_{n})-f(a)}}={\left(\lim _{n\rightarrow \infty }{\frac {f(x_{n})-f(a)}{x_{n}-a}}\right)}^{-1}\,,}$

where the right-hand side exists, due to the condition, and the second equation follows from fact.

${\displaystyle \Box }$

Example

The function

${\displaystyle f^{-1}\colon \mathbb {R} _{+}\longrightarrow \mathbb {R} _{+},x\longmapsto {\sqrt {x}},}$

is the inverse function of the function ${\displaystyle {}f}$, given by ${\displaystyle {}f(x)=x^{2}}$ (restricted to ${\displaystyle {}\mathbb {R} _{+}}$). The derivative of ${\displaystyle {}f}$ in a point ${\displaystyle {}a}$ is ${\displaystyle {}f'(a)=2a}$. Due to fact, for ${\displaystyle {}b\in \mathbb {R} _{+}}$, the relation

${\displaystyle {}{\left(f^{-1}\right)}'(b)={\frac {1}{f'(f^{-1}(b))}}={\frac {1}{2{\sqrt {b}}}}={\frac {1}{2}}b^{-{\frac {1}{2}}}\,}$

holds. In the zero point, however, ${\displaystyle {}f^{-1}}$ is not differentiable.

Example

The function

${\displaystyle f^{-1}\colon \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto x^{\frac {1}{3}},}$

is the inverse function of the function ${\displaystyle {}f}$, given by ${\displaystyle {}f(x)=x^{3}}$. The derivative of ${\displaystyle {}f}$ in ${\displaystyle {}a}$ is ${\displaystyle {}f'(a)=3a^{2}}$, which is different from ${\displaystyle {}0}$ for ${\displaystyle {}a\neq 0}$. Due to fact, we have for ${\displaystyle {}b\neq 0}$ the relation

${\displaystyle {}{\left(f^{-1}\right)}'(b)={\frac {1}{f'(f^{-1}(b))}}={\frac {1}{3{\left(b^{\frac {1}{3}}\right)}^{2}}}={\frac {1}{3}}b^{-{\frac {2}{3}}}\,.}$

In the zero point, however, ${\displaystyle {}f^{-1}}$ is not differentiable.