Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 15/refcontrol

Higher derivatives

The derivative ${}f'$ of a differentiable function is also called the first derivative of ${}f$ . The zeroth derivative is the function itself. Higher derivatives are defined recursively.

Definition

Let ${}I\subseteq \mathbb {R}$ denote an interval,MDLD/interval (R) and let

f\colon I\longrightarrow \mathbb {R}

be a function.MDLD/function The function ${}f$ is called ${}n$ -times differentiable, if it is ${}(n-1)$ -times differentiable, and the ${}(n-1)$ -th derivative, that is ${}f^{(n-1)}$ , is also differentiable.MDLD/differentiable (R) The derivative

{}f^{(n)}(x):=(f^{(n-1)})'(x)\,

is called the

{}n

-th derivative of

{}f

.

The second derivative is written as ${}f^{\prime \prime }$ , the third derivative as ${}f^{\prime \prime \prime }$ . If a function is ${}n$ -times differentiable, then we say that the derivatives exist up to order ${}n$ . A function ${}f$ is called infinitely often differentiable, if it is ${}n$ -times differentiable for every ${}n$ .

A differentiable function is continuous due to Corollary 14.6 , but its derivative is not necessarily so. Therefore, the following concept is justified.

Definition

Let ${}I\subseteq \mathbb {R}$ be an interval,MDLD/interval (R) and let

f\colon I\longrightarrow \mathbb {R}

be a function.MDLD/function The function ${}f$ is called continuously differentiable, if ${}f$ is differentiable MDLD/differentiable (R) and its derivative MDLD/derivative (R) ${}f'$ is

continuous.MDLD/continuous (R)

A function is called ${}n$ -times continuously differentiable, if it is ${}n$ -times differentiable, and its ${}n$ -th derivative is continuous.

Extrema of functions

We investigate now, with the help of the methods from differentiability, when a differentiable function

f\colon I\longrightarrow \mathbb {R} ,

where ${}I\subseteq \mathbb {R}$ denotes an interval, has a (local) extremum, and how the growing behavior looks like.

TheoremTheorem 15.3 change

Let

f\colon {]a,b[}\longrightarrow \mathbb {R}

be a function MDLD/function which attains in ${}c\in {]a,b[}$ a local extremum,MDLD/local extremum (R) and is differentiable MDLD/differentiable (R) there. Then ${}f'(c)=0$ holds.

Proof

We may assume that ${}f$ attains a local maximum in ${}c$ . This means that there exists an ${}\epsilon >0$ , such that ${}f(x)\leq f(c)$ holds for all ${}x\in [c-\epsilon ,c+\epsilon ]$ . Let ${}{\left(s_{n}\right)}_{n\in \mathbb {N} }$ be a sequence with ${}c-\epsilon \leq s_{n}<c$ , tending to ${}c$ ("from below“). Then ${}s_{n}-c<0$ , and so ${}f(s_{n})-f(c)\leq 0$ , and therefore the difference quotient

{}{\frac {f(s_{n})-f(c)}{s_{n}-c}}\geq 0\,.

Due to Lemma 7.12 , this relation carries over to the limit, which is the derivative. Hence, ${}f'(c)\geq 0$ . For another sequence ${}{\left(t_{n}\right)}_{n\in \mathbb {N} }$ with ${}c+\epsilon \geq t_{n}>c$ , we get

{}{\frac {f(t_{n})-f(c)}{t_{n}-c}}\leq 0\,.

Therefore, also ${}f'(c)\leq 0$ and thus ${}f'(c)=0$ .

\Box

We remark that the vanishing of the derivative is only a necessary, but not a sufficient, criterion for the existence of an extremum. The easiest example for this phenomenon is the function ${}\mathbb {R} \rightarrow \mathbb {R} ,x\mapsto x^{3}$ , which is strictly increasing and its derivative is zero at the zero point. We will provide a sufficient criterion in Corollary 15.9 below, see also Theorem 17.4 .

The mean value theorem

TheoremTheorem 15.4 change

Let ${}a<b$ , and let

f\colon [a,b]\longrightarrow \mathbb {R}

be a continuous MDLD/continuous (R) function, which is differentiable MDLD/differentiable (R) on ${}]a,b[$ , and such that ${}f(a)=f(b)$ . Then there exists some ${}c\in {]a,b[}$ , such that

{}f'(c)=0\,.

Proof

The statement is true if ${}f$ is constant. So suppose that ${}f$ is not constant. Then there exists some ${}x\in {]a,b[}$ , such that ${}f(x)\neq f(a)=f(b)$ . Let's say that ${}f(x)$ has a larger value. Due to Theorem 11.13 , there exists some ${}c\in [a,b]$ , where the function attains its maximum.MDLD/maximum (R) This point is not on the border. For this ${}c$ , we have ${}f'(c)=0$ , due to Theorem 15.3 .

\Box

This theorem is called Theorem of Rolle.

The following theorem is called Mean value theorem. It says that if a function describes a differentiable one-dimensional movement, then the average velocity is obtained at least once as the instantaneous velocity.

Theorem Create referencenumber

Let ${}a<b$ , and let

f\colon [a,b]\longrightarrow \mathbb {R}

be a continuous function MDLD/continuous function (R) which is differentiable MDLD/differentiable (R) on ${}]a,b[$ . Then there exists some ${}c\in {]a,b[}$ , such that

{}f'(c)={\frac {f(b)-f(a)}{b-a}}\,.

Proof

We consider the auxiliary function

g\colon [a,b]\longrightarrow \mathbb {R} ,x\longmapsto g(x):=f(x)-{\frac {f(b)-f(a)}{b-a}}(x-a).

This function is also continuous MDLD/continuous (R) and differentiable MDLD/differentiable (R) in ${}]a,b[$ . Moreover, we have ${}g(a)=f(a)$ and

{}g(b)=f(b)-(f(b)-f(a))=f(a)\,.

Hence, ${}g$ fulfills the conditions of Theorem 15.4 , and therefore there exists some ${}c\in {]a,b[}$ , such that ${}g'(c)=0$ . Because of the rules for derivatives, we obtain

{}f'(c)={\frac {f(b)-f(a)}{b-a}}\,.

\Box

CorollaryCorollary 15.6 change

Let

f\colon {]a,b[}\longrightarrow \mathbb {R}

be a differentiable function MDLD/differentiable function (R) such that ${}f'(x)=0$ for all ${}x\in {]a,b[}$ . Then ${}f$ is constant.

Proof

If ${}f$ is not constant, then there exists some ${}x<x'$ such that ${}f(x)\neq f(x')$ . Then there exists, due to the mean value theorem, some ${}c$ , ${}x<c<x'$ , such that ${}f'(c)={\frac {f(x')-f(x)}{x'-x}}\neq 0$ , which contradicts the assumption.

\Box

TheoremTheorem 15.7 change

Let ${}I\subseteq \mathbb {R}$ be an open interval,MDLD/open interval (R) and let

f\colon I\longrightarrow \mathbb {R}

be a

differentiable function.MDLD/differentiable function (R) Then the following statements hold.

The function ${}f$ is increasingMDLD/increasing (function) (decreasing) on ${}I$ , if and only if ${}f'(x)\geq 0$ ( ${}f'(x)\leq 0$ ) holds for all ${}x\in I$ .
If ${}f'(x)\geq 0$ holds for all ${}x\in I$ , and ${}f'$ has only finitely many zeroes,MDLD/zeroes (function) then ${}f$ is strictly increasing.MDLD/strictly increasing (function)
If ${}f'(x)\leq 0$ holds for all ${}x\in I$ , and ${}f'$ has only finitely many zeroes, then ${}f$ is strictly decreasing.MDLD/strictly decreasing (function)

Proof

(1). It is enough to prove the statements for increasing functions. If ${}f$ is increasing and ${}x\in I$ , then the difference quotient MDLD/difference quotient (R) fulfills

{}{\frac {f(x+h)-f(x)}{h}}\geq 0\,

for every ${}h$ with ${}x+h\in I$ . This estimate carries over to the limit as ${}h\rightarrow 0$ , and this limit is ${}f'(x)$ .
Suppose now that the derivative is ${}\geq 0$ . We assume, in order to obtain a contradiction, that there exist two points ${}x<x'$ in ${}I$ with ${}f(x)>f(x')$ . Due to the mean value theorem, there exists some ${}c$ with ${}x<c<x'$ and

{}f'(c)={\frac {f(x')-f(x)}{x'-x}}<0\,,

which contradicts the condition.
(2). Suppose now that ${}f'(x)>0$ holds with finitely many exceptions. We assume that ${}f(x)=f(x')$ holds for two points ${}x<x'$ . Since ${}f$ is increasing, due to the first part, it follows that ${}f$ is constant on the interval ${}[x,x']$ . But then ${}f'=0$ on this interval, which contradicts the condition that ${}f'$ has only finitely many zeroes.

\Box

Corollary Create referencenumber

A real polynomial function MDLD/polynomial function (R)

f\colon \mathbb {R} \longrightarrow \mathbb {R}

of degree MDLD/degree (polynomial) ${}d\geq 1$ has at most ${}d-1$ local extrema,MDLD/local extrema (R) and one can partition the real numbers into at most ${}d$ intervals, on which ${}f$ is alternatingly strictly increasing MDLD/strictly increasing (real function) or strictly decreasing.MDLD/strictly decreasing (real function)

Proof

See Exercise 15.14 .

\Box

CorollaryCorollary 15.9 change

Let ${}I$ denote a real interval,MDLD/real interval

f\colon I\longrightarrow \mathbb {R}

a twice continuously differentiable MDLD/continuously differentiable (R) function,MDLD/function and ${}a\in I$ an inner point of the interval. Suppose that ${}f'(a)=0$

holds. Then the following statements hold.

If ${}f^{\prime \prime }(a)>0$ holds, then ${}f$ has an isolated local minimumMDLD/isolated local minimum (R) in ${}a$ .
If ${}f^{\prime \prime }(a)<0$ holds, then ${}f$ has an isolated local maximumMDLD/isolated local maximum (R) in ${}a$ .

Proof

See Exercise 15.15 .

\Box

We will encounter a more general statement in Theorem 17.4 .

General mean value theorem and L'Hôpital's rule

The following statement is called also the general mean value theorem.

TheoremTheorem 15.10 change

Let ${}b>a$ , and suppose that

f,g\colon [a,b]\longrightarrow \mathbb {R}

are continuous MDLD/continuous (R) functions which are differentiable MDLD/differentiable (R) on ${}]a,b[$ and such that

{}g'(x)\neq 0\,

for all ${}x\in {]a,b[}$ . Then ${}g(b)\neq g(a)$ , and there exists some ${}c\in {]a,b[}$ such that

{}{\frac {f(b)-f(a)}{g(b)-g(a)}}={\frac {f'(c)}{g'(c)}}\,.

Proof

The statement

{}g(a)\neq g(b)\,

follows from Theorem 15.4 . We consider the auxiliary function

{}h(x):=f(x)-{\frac {f(b)-f(a)}{g(b)-g(a)}}g(x)\,.

We have

{}{\begin{aligned}h(a)&=f(a)-{\frac {f(b)-f(a)}{g(b)-g(a)}}g(a)\\&={\frac {f(a)g(b)-f(a)g(a)-f(b)g(a)+f(a)g(a)}{g(b)-g(a)}}\\&={\frac {f(a)g(b)-f(b)g(a)}{g(b)-g(a)}}\\&={\frac {f(b)g(b)-f(b)g(a)-f(b)g(b)+f(a)g(b)}{g(b)-g(a)}}\\&=f(b)-{\frac {f(b)-f(a)}{g(b)-g(a)}}g(b)\\&=h(b).\end{aligned}}

Therefore, ${}h(a)=h(b)$ , and Theorem 15.4 yields the existence of some ${}c\in {]a,b[}$ with

{}h'(c)=0\,.

Rearranging proves the claim.

\Box

From this version, one can recover the mean value theorem, by taking for ${}g$ the identity.

For the computation of the limit of a function, the following method called L'Hôpital's rule helps.

Corollary Create referencenumber

Let ${}I\subseteq \mathbb {R}$ denote an open interval,MDLD/open interval (R) and let ${}a\in I$ denote a point. Suppose that

f,g\colon I\longrightarrow \mathbb {R}

are continuous functions,MDLD/continuous functions (R) which are differentiable MDLD/differentiable (R) on ${}I\setminus \{a\}$ , fulfilling ${}f(a)=g(a)=0$ , and with ${}g'(x)\neq 0$ for ${}x\neq a$ . Moreover, suppose that the limit MDLD/limit (real function)

{}w:=\operatorname {lim} _{x\rightarrow a}\,{\frac {f'(x)}{g'(x)}}\,

exists. Then also the limit

\operatorname {lim} _{x\rightarrow a}\,{\frac {f(x)}{g(x)}}

exists, and it also equals ${}w$ .

Proof

Because ${}g'$ has no zero in the interval and ${}g(a)=0$ holds, it follows, because of Theorem 15.4 , that ${}a$ is the only zero of ${}g$ . Let ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ denote a sequence MDLD/sequence (R) in ${}I\setminus \{a\}$ , converging MDLD/converging (R) to ${}a$ .

For every ${}x_{n}$ there exists, due to Theorem 15.10 , applied to the interval ${}I_{n}:=[x_{n},a]$ or ${}[a,x_{n}]$ , a ${}c_{n}$ (in the interior^[1] of ${}I_{n}$ ,) fulfilling

{}{\frac {f(x_{n})-f(a)}{g(x_{n})-g(a)}}={\frac {f'(c_{n})}{g'(c_{n})}}\,.

The sequence ${}{\left(c_{n}\right)}_{n\in \mathbb {N} }$ converges also to ${}a$ , so that, because of the condition, the right-hand side converges to ${}{\frac {f'(a)}{g'(a)}}=w$ . Therefore, also the left-hand side converges to ${}w$ , and, because of ${}f(a)=g(a)=0$ , this means that ${}{\frac {f(x_{n})}{g(x_{n})}}$ converges to ${}w$ .