Mathematical Analysis
Leture 11
6.1 Definition and basic properties
Definition 6.1.1. Let $I$ be an interval, let $f \colon I \to \R$ be a function, and let $c \in I$. If the limit \begin{equation*} L := \lim_{x \to c} \frac{f(x)-f(c)}{x-c} \end{equation*} exists, then we say $f$ is differentiable at $c$, that $L$ is the derivative of $f$ at $c$, and write $f'(c) := L$.
6.1 Definition and basic properties
If $f$ is differentiable at all $c \in I$, then we simply say that $f$ is differentiable, and then we obtain a function $f' \colon I \to \R$.
The expression $\displaystyle\frac{f(x)-f(c)}{x-c}$ is called the difference quotient.
6.1 Definition and basic properties
Graphical interpretation of the derivative
6.1 Definition and basic properties
Equivalent definition of the derivative
Let $I$ be an interval, let $f \colon I \to \R$ be a function, and let $c \in I$. If the limit \begin{equation*} L := \lim_{h \to 0} \frac{f(c+h)-f(c)}{h} \end{equation*} exists, then we say $f$ is differentiable at $c$, that $L$ is the derivative of $f$ at $c$, and write $f'(c) := L$.
6.1 Definition and basic properties
Equivalent definition of the derivative
$f'(c)$ $\ds =\lim_{x \to c} \frac{f(x)-f(c)}{x-c} $ $\ds =\lim_{h \to 0} \frac{f(c+h)-f(c)}{h}$
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Example 6.1.1.
Let $f(x) :=x^2$ defined on the whole real line.
For $c \in \R$ arbitrary and $x\neq c$, we have
$\ds \frac{x^2-c^2}{x-c}$
$\ds =
\frac{(x+c)(x-c)}{x-c} $
$\ds =
x+c . $
Hence $\;\ds f'(c) $
$\ds =
\lim_{x\to c} \frac{x^2-c^2}{x-c}$
$\ds=
\lim_{x\to c}\, (x+c) $
$\ds=
2c. $
Example 6.1.2.
Let $\,f(x) := ax + b\,$ for numbers $\,a, b \in \R.$
Again for $\,c \in \R$ arbitrary and $x\neq c,$ we have
$\ds\frac{f(x)-f(c)}{x-c} $
$\ds =
\frac{ax + b - (ac + b)}{x-c} $
$\ds =
\frac{a(x-c)}{x-c} $
$\ds =
a. $
Hence $\;\ds f'(c) $
$\ds
=\lim_{x\to c}
\frac{f(x)-f(c)}{x-c}$
$\ds=
\lim_{x\to c}
a $
$\ds=
a. $
6.1 Definition and basic properties
Theorem 6.1.1. Let $f \colon I \to \R$ be differentiable at $c \in I,$ then it is continuous at $c.$
6.1 Definition and basic properties
Theorem 6.1.1. Let $f \colon I \to \R$ be differentiable at $c \in I,$ then it is continuous at $c.$
Proof. We know that $\ds\lim_{x\to c}\frac{f(x)-f(c)}{x-c} = f'(c)$ and $\ds \lim_{x\to c}(x-c) = 0$ exist. Moreover $\ds f(x)-f(c)$ $\ds = \left( \frac{f(x)-f(c)}{x-c} \right) (x-c) .$ Therefore, the limit of $f(x)-f(c)$ exists, and
$\ds\lim_{x\to c} \bigl( f(x)-f(c) \bigr)$
$\ds=
\left(\lim_{x\to c} \frac{f(x)-f(c)}{x-c} \right)
\left(\lim_{x\to c} (x-c) \right) $
$\ds=
f'(c) \cdot 0 $
$=0.$
Hence $\lim\limits_{x\to c} f(x) = f(c),$ and $f$ is continuous at $c. \;\bs$
6.1 Definition and basic properties
Theorem 6.1.1. Let $f \colon I \to \R$ be differentiable at $c \in I,$ then it is continuous at $c.$
Note: There exists a continuous function that is not differentiable at any point. 🤯 This is known as the Weierstrass function.
6.1 Definition and basic properties
Weierstrass function
|
$$ \displaystyle f(x):=\sum_{n=0}^{\infty}a^n\cos(b^n\pi x)\, $$ where $\,0\lt a\lt 1,\,$ and $\,b\,$ is a positive odd integer. |
6.1 Definition and basic properties
Weierstrass function
|
$ \displaystyle f(x):=\sum_{n=0}^{\infty}a^n\cos(b^n\pi x) $ |
Plot made with MathCell |
6.1 Definition and basic properties
Weierstrass function
Plot made with PGFplts.net
3D terrain
Source: Elevated by Inigo Quilez
6.1 Definition and basic properties
Theorem 6.1.2. (Linearity) Let $I$ be an interval, let $f \colon I \to \R$ and $g \colon I \to \R$ be differentiable at $c \in I,$ and let $\alpha \in \R.$
- Define $h \colon I \to \R$ by $h(x) := \alpha f(x).$ Then $h$ is differentiable at $c$ and \[h'(c) = \alpha f'(c).\]
- Define $h \colon I \to \R$ by $h(x) := f(x) + g(x).$ Then $h$ is differentiable at $c$ and \[h'(c) = f'(c) + g'(c).\]
6.1 Definition and basic properties
Proof. Define $h(x) := f(x)+g(x).$ For $x \in I$, $x \not= c,$ we have
$\ds \frac{h(x)-h(c)}{x-c} $
$\ds =
\frac{\bigl(f(x) + g(x)\bigr) - \bigl(f(c) + g(c)\bigr)}{x-c}$
$\ds \qquad\quad=
\frac{f(x) - f(c)}{x-c}
+
\frac{g(x) - g(c)}{x-c} .$
6.1 Definition and basic properties
Proof. Define $h(x) := f(x)+g(x).$ For $x \in I$, $x \not= c,$ we have
$\ds \frac{h(x)-h(c)}{x-c} $ $\ds = \frac{f(x) - f(c)}{x-c} + \frac{g(x) - g(c)}{x-c} .$
Now we take the limit in both sides. That is
$\ds\lim_{x\to c} \frac{h(x)-h(c)}{x-c} $ $\ds= \lim_{x\to c} \frac{f(x) - f(c)}{x-c} +\lim_{x\to c} \frac{g(x) - g(c)}{x-c} .$
Therefore, $h$ is differentiable at $c$ and $h'(c) = f'(c) + g'(c).$
The proof of item 1 is left as an exercise. 📝
6.1 Definition and basic properties
Product rule
Theorem 6.1.3. Let $I$ be an interval, let $f \colon I \to \R$ and $g \colon I \to \R$ be functions differentiable at $c.$ If $h \colon I \to \R$ is defined by \begin{equation*} h(x) := f(x) g(x) , \end{equation*} then $h$ is differentiable at $c$ and \begin{equation*} h'(c) = f(c) g'(c) + f'(c) g(c) . \end{equation*}
📝 Hint: Note that $\, f(x)g(x)-f(c)g(c)= f(x)(g(x)-g(c)) + (f(x)-f(c))g(c) .$
6.1 Definition and basic properties
Quotient rule
Theorem 6.1.4. Let $I$ be an interval, let $f \colon I \to \R$ and $g \colon I \to \R$ be differentiable at $c$ and $g(x) \not= 0$ for all $x \in I.$ If $h \colon I \to \R$ is defined by \begin{equation*} h(x) := \frac{f(x)}{g(x)}, \end{equation*} then $h$ is differentiable at $c$ and \begin{equation*} h'(c) = \frac{f'(c) g(c) - f(c) g'(c)}{{\bigl(g(c)\bigr)}^2} . \end{equation*}
Proof of Theorem 6.1.4. - Quotient rule
Since $g$ is differentiable at $c,$ it is continuous at that point. Then since $g(c)\neq 0,$ there must be an interval $J\subseteq I$ such that $g(x)\neq 0 ,\,\forall x\in J.$ So for $x\in J, x\neq c$ we have
$\ds \frac{h(x)-h(c)}{x-c}$ $\ds = \frac{\dfrac{f(x)}{g(x)}- \dfrac{f(c)}{g(c)}}{x-c}$ $\ds = \frac{f(x)g(c)-f(c)g(x)}{g(x)g(c)(x-c)} $
$\ds =\frac{f(x)g(c)-f(c)g(c)+f(c)g(c)-f(c)g(x)}{g(x)g(c)(x-c)}\quad\;$
$\ds = \frac{1}{g(x)g(c)} \Bigg[\Bigg.$ $\ds \frac{f(x)-f(c)}{x-c}g(c)$ $\ds - \,f(c)\frac{g(x)-g(c)}{x-c}$ $\ds \Bigg. \Bigg]$
Proof of Theorem 6.1.4. - Quotient rule
$\ds \frac{h(x)-h(c)}{x-c} = \frac{1}{g(x)g(c)} \left[ \frac{f(x)-f(c)}{x-c}g(c)- f(c)\frac{g(x)-g(c)}{x-c} \right]$
Using the fact that $g$ is continuous at $c$ and the differentiability of $f$ and $g$ at $c,$ we get
$\ds h'(c)$ $\ds =\lim_{x \to c}\frac{h(x)-h(c)}{x-c}\qquad\qquad\qquad \qquad\qquad$
$\ds = \frac{1}{g(c)g(c)} \left[\, f'(c)g(c) - f(c) g'(c) \right] \qquad$
$\ds = \frac{f'(c)g(c)- f(c) g'(c)}{\big(g(c)\big)^2} .$ $\;\bs \qquad\qquad\,$
6.1 Definition and basic properties
Chain rule
Theorem 6.1.5. Let $I_1, I_2$ be intervals, let $g \colon I_1 \to I_2$ be differentiable at $c \in I_1$, and $f \colon I_2 \to \R$ be differentiable at $g(c)$. If $h \colon I_1 \to \R$ is defined by \begin{equation*} h(x) := \left(\,f \circ g\right) (x) = f\bigl(g(x)\bigr) , \end{equation*} then $h$ is differentiable at $c$ and \begin{equation*} h'(c) = f'\bigl(g(c)\bigr)g'(c) . \end{equation*}
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