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

Differentiability

In this section, we consider functions

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

where ${\displaystyle {}D\subseteq \mathbb {R} }$ is a subset of the real numbers. We want to explain what it means that such a function is differentiable in a point ${\displaystyle {}a\in D}$. The intuitive idea is to look at another point ${\displaystyle {}x\in D}$, and to consider the secant, given by the two points ${\displaystyle {}(a,f(a))}$ and ${\displaystyle {}(x,f(x))}$, and then to let "${\displaystyle {}x}$ move towards ${\displaystyle {}a}$“. 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 ${\displaystyle {}D\subseteq \mathbb {R} }$ be a subset, ${\displaystyle {}a\in D}$ a point, and

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

a function. For ${\displaystyle {}x\in D}$, ${\displaystyle {}x\neq a}$, the number

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

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

${\displaystyle {}a}$ and ${\displaystyle {}x}$.

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

## Definition

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

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

a function. We say that ${\displaystyle {}f}$ is differentiable in ${\displaystyle {}a}$ if the limit

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

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

${\displaystyle f'(a).}$

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

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

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

## Example

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

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

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

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

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

## Example

We consider the function

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

The difference quotient for ${\displaystyle {}a}$ and ${\displaystyle {}a+h}$ is

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

The limit of this, as ${\displaystyle {}h}$ tends to ${\displaystyle {}0}$, is ${\displaystyle {}2a}$. The derivative of ${\displaystyle {}f}$ in ${\displaystyle {}a}$ is therefore ${\displaystyle {}f'(a)=2a}$.

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)

## Theorem

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

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

a function. Then ${\displaystyle {}f}$ is differentiable in ${\displaystyle {}a}$ if and only if there exists some ${\displaystyle {}s\in \mathbb {R} }$ and a function

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

such that ${\displaystyle {}r}$ is continuous in ${\displaystyle {}a}$, ${\displaystyle {}r(a)=0}$, and such that

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

### Proof

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

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

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

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

Because of differentiability, the limit

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

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

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

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

${\displaystyle \Box }$

The affine-linear function

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

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

## Corollary

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

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

a function. Then ${\displaystyle {}f}$ is also continuous in ${\displaystyle {}a}$.

### Proof

This follows immediately from Theorem 14.5 .

${\displaystyle \Box }$

Rules for differentiable functions

## 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 Theorem 14.5 , 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 Lemma 10.11 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 Corollary 14.6 , 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 Theorem 14.5 , 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 Theorem 11.7 , 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 Lemma 8.1 .

${\displaystyle \Box }$

## 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 Theorem 14.9 , 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 Theorem 14.9 , 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.

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

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

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

${\displaystyle f'\colon I\longrightarrow \mathbb {R} ,x\longmapsto f'(x),}$
is called the derivative of ${\displaystyle {}f}$.