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

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, and let

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

be a 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. 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, and let

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

be a function. The function ${}f$ is called continuously differentiable, if ${}f$ is differentiable and its derivative ${}f'$ is

continuous.

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.

Theorem

Let

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

be a function which attains in ${}c\in {]a,b[}$ a local extremum, and is differentiable 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

Theorem

Let ${}a<b$, and let

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

be a continuous function, which is differentiable 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. 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 mean value theorem means that, for every secant, there exists a parallel tangent.

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

Let ${}a<b$, and let

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

be a continuous function which is differentiable 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 and differentiable 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

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

$\Box$

Corollary

Let

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

be a differentiable function 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.

$\Box$

Theorem

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

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

be a

differentiable function. Then the following statements hold.

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

Proof

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

A real polynomial function

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

of degree ${}d\geq 1$ has at most ${}d-1$ local extrema, and one can partition the real numbers into at most ${}d$ intervals, on which ${}f$ is alternatingly strictly increasing or

strictly decreasing.

Proof

See Exercise 15.14 .

$\Box$

Corollary

Let ${}I$ denote a real interval,

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

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

holds. Then the following statements hold.

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

Theorem

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

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

are continuous functions which are differentiable 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.

L’Hospital (1661-1704)

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

Corollary

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

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

are continuous functions, which are differentiable 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

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

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 in ${}I\setminus \{a\}$, converging 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 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 ${}w$.

$\Box$

Example

The polynomials

$3x^{2}-5x-2{\text{ and }}x^{3}-4x^{2}+x+6$

have both a zero for ${}x=2$. It is therefore not immediately clear whether the limit

$\operatorname {lim} _{x\rightarrow 2}\,{\frac {3x^{2}-5x-2}{x^{3}-4x^{2}+x+6}}$

exists. Applying twice L'Hôpital's rule, we get the existence and

${}\operatorname {lim} _{x\rightarrow 2}\,{\frac {3x^{2}-5x-2}{x^{3}-4x^{2}+x+6}}=\operatorname {lim} _{x\rightarrow 2}\,{\frac {6x-5}{3x^{2}-8x+1}}={\frac {7}{-3}}=-{\frac {7}{3}}\,.$

Footnotes

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