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

Higher derivatives

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

## Definition

Let ${\displaystyle {}I\subseteq \mathbb {R} }$ denote an interval, and let

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

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

${\displaystyle {}f^{(n)}(x):=(f^{(n-1)})'(x)\,}$
is called the ${\displaystyle {}n}$-th derivative of ${\displaystyle {}f}$.

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

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

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

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

continuous.

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

Extrema of functions

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

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

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

## Theorem

Let

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

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

### Proof

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

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

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

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

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

${\displaystyle \Box }$

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 ${\displaystyle {}\mathbb {R} \rightarrow \mathbb {R} ,x\mapsto x^{3}}$, 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 ${\displaystyle {}a, and let

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

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

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

### Proof

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

${\displaystyle \Box }$

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

Let ${\displaystyle {}a, and let

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

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

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

### Proof

We consider the auxiliary function

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

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

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

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

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

## Corollary

Let

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

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

### Proof

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

${\displaystyle \Box }$

## Theorem

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

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

be a

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

### Proof

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

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

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

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

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

${\displaystyle \Box }$

## Corollary

A real polynomial function

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

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

### Proof

${\displaystyle \Box }$

## Corollary

Let ${\displaystyle {}I}$ denote a real interval,

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

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

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

### Proof

${\displaystyle \Box }$

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 ${\displaystyle {}b>a}$, and suppose that

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

are continuous functions which are differentiable on ${\displaystyle {}]a,b[}$ and such that

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

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

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

### Proof

The statement

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

follows from Theorem 15.4 . We consider the auxiliary function

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

We have

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

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

Rearranging proves the claim.

${\displaystyle \Box }$

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

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

## Corollary

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

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

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

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

exists. Then also the limit

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

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

### Proof

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

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

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

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

${\displaystyle \Box }$

## Example

The polynomials

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

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

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

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

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

Footnotes
1. The interior of a real interval ${\displaystyle {}I\subseteq \mathbb {R} }$

is the interval without the boundaries.

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