Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 14/refcontrol

Differentiability

In this section, we consider functions

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

where ${}D\subseteq \mathbb {R}$ is a subset of the real numbers. We want to explain what it means that such a function is differentiable in a point ${}a\in D$. The intuitive idea is to look at another point ${}x\in D$, and to consider the secant, given by the two points ${}(a,f(a))$ and ${}(x,f(x))$, and then to let "${}x$ move towards ${}a$“. If this limiting process makes sense, the secants tend to become a tangent. However, this process only has a precise basis, if we use the concept of the limit of a function as defined earlier.

Definition

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

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

a function.MDLD/function (R) For ${}x\in D$, ${}x\neq a$, the number

${\frac {f(x)-f(a)}{x-a}}$

is called the difference quotient of ${}f$ for

${}a$ and ${}x$.

The difference quotient is the slope of the secant at the graph, running through the two points ${}(a,f(a))$ and ${}(x,f(x))$. For ${}x=a$, this quotient is not defined. However, a useful limit might exist for ${}x\rightarrow a$. This limit represents, in the case of existence, the slope of the tangent for ${}f$ in the point ${}(a,f(a))$.

Definition

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

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

a function.MDLD/function (R) We say that ${}f$ is differentiable in ${}a$ if the limitMDLD/limit (function R)

$\operatorname {lim} _{x\in D\setminus \{a\},\,x\rightarrow a}\,{\frac {f(x)-f(a)}{x-a}}$

exists. In the case of existence, this limit is called the derivative of ${}f$ in ${}a$, written

$f'(a).$

The derivative in a point ${}a$ is, if it exists, an element in ${}\mathbb {R}$. Quite often one takes the difference ${}h:=x-a$ as the parameter for this limiting process, that is, one considers

$\operatorname {lim} _{h\rightarrow 0}\,{\frac {f(a+h)-f(a)}{h}}.$

The condition ${}x\in D\setminus \{a\}$ translates then to ${}a+h\in D$, ${}h\neq 0$. If the Function ${}f$ describes a one-dimensional movement, meaning a time-dependent process on the real line, then the difference quotient ${}{\frac {f(x)-f(a)}{x-a}}$ is the average velocity between the (time) points ${}a$ and ${}x$ and ${}f'(a)$ is the instantaneous velocity in ${}a$.

Example Create referencenumber

Let ${}s,c\in \mathbb {R}$, and let

$\alpha \colon \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto sx+c,$

be an affine-linear function.MDLD/affine-linear function To determine the derivativeMDLD/derivative (R) in a point ${}a\in \mathbb {R}$, we consider the difference quotient

${}{\frac {(sx+c)-(sa+c)}{x-a}}={\frac {sx-sa}{x-a}}=s\,.$

This is constant and equals ${}s$, so that the limit of the difference quotient as ${}x$ tends to ${}a$ exists and equals ${}s$ as well. Hence, the derivative exists in every point and is just ${}s$. The slope of the affine-linear function is also its derivative.

Example Create referencenumber

We consider the functionMDLD/function

$f\colon \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto x^{2}.$

The difference quotientMDLD/difference quotient (R) for ${}a$ and ${}a+h$ is

${}{\frac {f(a+h)-f(a)}{h}}={\frac {(a+h)^{2}-a^{2}}{h}}={\frac {a^{2}+2ah+h^{2}-a^{2}}{h}}={\frac {2ah+h^{2}}{h}}=2a+h\,.$

The limitMDLD/limit (function R) of this, as ${}h$ tends to ${}0$, is ${}2a$. The derivativeMDLD/derivative (R) of ${}f$ in ${}a$ is therefore ${}f'(a)=2a$.

Linear approximation

We discuss a property which is equivalent with differentiability, the existence of a linear approximation. This formulation is important in many respects: It allows giving quite simple proofs of the rules for differentiable functions, one can use it to reduce differentiability to the continuity of an error function, it yields a model for approximation with polynomials of higher degree (quadratic approximation, Taylor expansion), and it allows a direct generalization to the higher-dimensional situation(in the second term)

TheoremTheorem 14.5 change

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

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

a function.MDLD/function (R) Then ${}f$ is differentiableMDLD/differentiable (R) in ${}a$ if and only if there exists some ${}s\in \mathbb {R}$ and a function

$r\colon D\longrightarrow \mathbb {R} ,$

such that ${}r$ is continuousMDLD/continuous (R) in ${}a$, ${}r(a)=0$, and such that

${}f(x)=f(a)+s\cdot (x-a)+r(x)(x-a)\,.$

Proof

If ${}f$ is differentiable, then we set

${}s:=f'(a)\,.$

Then the only possibility to fulfill the conditions for ${}r$ is

${}r(x)={\begin{cases}{\frac {f(x)-f(a)}{x-a}}-s{\text{ for }}x\neq a\,,\\0{\text{ for }}x=a\,.\end{cases}}\,$

Because of differentiability, the limit

${}\operatorname {lim} _{x\rightarrow a,\,x\in D\setminus \{a\}}r(x)=\operatorname {lim} _{x\rightarrow a,\,x\in D\setminus \{a\}}{\left({\frac {f(x)-f(a)}{x-a}}-s\right)}\,$

exists, and its value is ${}0$. This means that ${}r$ is continuous in ${}a$.
If ${}s$ and ${}r$ exist with the described properties, then for ${}x\neq a$ the relation

${}{\frac {f(x)-f(a)}{x-a}}=s+r(x)\,$

holds. Since ${}r$ is continuous in ${}a$, the limit on the left-hand side, for ${}x\rightarrow a$, exists.

$\Box$

The affine-linear function

$D\longrightarrow \mathbb {R} ,x\longmapsto f(a)+f'(a)(x-a),$

is called the affine-linear approximation. The constant function given by the value ${}f(a)$ can be considered as the constant approximation.

CorollaryCorollary 14.6 change

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

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

a

function.MDLD/function (R) Then ${}f$ is also continuous in ${}a$.

Proof

This follows immediately from Theorem 14.5 .

$\Box$

Rules for differentiable functions

An illustration of the product rule: the increment of the area is about the seize of the sum of the two products of the side length and the increment of the other side length. For the infinitesimal increment, the product of the two increments is irrelevant.

LemmaLemma 14.7 change

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

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

functionsMDLD/functions which are differentiableMDLD/differentiable (R)

in ${}a$. Then the following rules for differentiability holds.

1. The sum ${}f+g$ is differentiable in ${}a$, with

   ${}(f+g)'(a)=f'(a)+g'(a)\,.$

2. The product ${}f\cdot g$ is differentiable in ${}a$, with

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

3. For ${}c\in \mathbb {R}$, also ${}cf$ is differentiable in ${}a$, with

   ${}(cf)'(a)=cf'(a)\,.$

4. If ${}g$ has no zero in ${}a$, then ${}1/g$ is differentiable in ${}a$, with

   ${}{\left({\frac {1}{g}}\right)}'(a)={\frac {-g'(a)}{(g(a))^{2}}}\,.$

5. 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 Theorem 14.5 , 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 Lemma 10.11 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 Corollary 14.6 , 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.

TheoremTheorem 14.8 change

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 differentiableMDLD/differentiable (R) in ${}a$ and that ${}g$ is differentiable in ${}b:=f(a)$. Then also the compositionMDLD/composition

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

is differentiable in ${}a$, and its derivativeMDLD/derivative (R) is

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

Proof

Due to Theorem 14.5 , 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 ${}0$.

$\Box$

An illustration for the derivative of the inverse function. The graph of the inverse function is the reflection of the graph at the diagonal, and the tangent behaves accordingly.

TheoremTheorem 14.9 change

Let ${}D,E\subseteq \mathbb {R}$ denote intervals,MDLD/intervals (R) and let

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

be a bijectiveMDLD/bijective continuous function,MDLD/continuous function (R) with the inverse functionMDLD/inverse function

$f^{-1}\colon E\longrightarrow D.$

Suppose that ${}f$ is

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

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

holds.

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 sequenceMDLD/sequence (R) in ${}E\setminus \{b\}$, convergingMDLD/converging (R) to ${}b$. Because of Theorem 11.7 , 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 Lemma 8.1   (5).

$\Box$

Example Create referencenumber

The function

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

is the inverse functionMDLD/inverse function of the function ${}f$, given by ${}f(x)=x^{2}$ (restricted to ${}\mathbb {R} _{+}$). The derivativeMDLD/derivative (R) of ${}f$ in a point ${}a$ is ${}f'(a)=2a$. Due to Theorem 14.9 , 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 Create referencenumber

The function

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

is the inverse functionMDLD/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 Theorem 14.9 , 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.

The derivative function

So far, we have considered differentiability of a function in just one point, now we consider the derivative in general.

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 We say that ${}f$ is differentiable, if for every point ${}a\in I$, the derivativeMDLD/derivative (point R) ${}f'(a)$ of ${}f$ in ${}a$ exists. In this case, the mappingMDLD/mapping

$f'\colon I\longrightarrow \mathbb {R} ,x\longmapsto f'(x),$

is called the derivative of ${}f$.

<< | Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I | >> PDF-version of this lecture Exercise sheet for this lecture (PDF)