Definition 2.1.1. Suppose $y$ is a quantity that depends on another quantity $x$. The difference quotient
\[\frac{\Delta y}{\Delta x}=\frac{f(x_2)-f(x_1)}{x_2-x_1}\]is called the average rate of change of $y$ with respect to $x$ over the interval $[x_1,x_2]$. Alternatively, we can also consider the average rate of change based on a single point $(a,f(a))$ as follows:
\[\frac{f(x)-f(a)}{x-a}.\]Taking limit to this, we get the instantaneous rate of change of $y$ with respect to $x$ at $(a,f(a))$ as follows:
\[\left.\frac{dy}{dx}\right\vert_{x=a}=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}.\]Definition 2.1.2. Let $f$ be a real-valued function. If
\[\lim_{x \to a} \frac{f(x)-f(a)}{x-a}=L\]holds, we define the derivative of $f$ at $x=a$ to be $L$, denoted by $f’(a)$. It means the instantaneous rate of change of the output with respect to the input at $(a,f(a))$. In this case, $f$ is said to be differentiable at $x=a$. More generally, if $I$ is an open interval and $f$ is differentiable at every point of $I$, then $f$ is said to be differentiable on $I$.
Remark 2.1.3. Let $f$ be a function. When we draw the graph of $f$ on the Cartesian coordinate plane, the slope of the tangent line to $f$ at $x=a$ is equal to $f’(a)$.
Example 2.1.4. Let $f:\mathbb R \to \mathbb R$ be the function such that $f(x)=\vert x \vert$. Where is $f$ differentiable?
Solution. 어떤 $a$에서 “$L$ is a real number such that
가 참”인지를 묻는 것이다. definition of absolute value로부터 조건별로 정의되어 있음을 알 수 있다. 따라서 경우를 나누는 수밖에 없겠다. 다만, 모든 경우를 볼 필요는 없다. $h\ne 0$가 충분히 $0$에 가깝다고 가정하고 $\vert a+h \vert$를 보자. $a>0$일 때에는 $a+h>0$임을 알 수 있다. 그러면 대칭적으로 $a<0\Rightarrow a+h< 0$일 것이다. 그리고 $a=0$이라면 $h$의 부호에 따라 결정된다. 따라서 전체 값은 이렇게 정리된다:
\[\frac{\vert a+h\vert -\vert a\vert}{h}= \begin{cases} 1 & \text{if }a > 0, \text{ or } a=0 \text{ and } h>0,\\ -1 & \text{if }a < 0, \text{ or } a=0 \text{ and } h<0. \end{cases}\]Definition 2.1.2에 의하여 $a\ne 0$일 때만이 the case이다.
Theorem 2.1.5. If $f$ is differentiable at $x=a$, then $f$ is continuous at $x=a$.
Proof. Since $f$ is differentiable at $x=a$,
Also, $\lim_{x\to a}{[x-a]}=0$. Therefore, by the product law for limits,
\[\lim_{x \to a}{\left[\frac{f(x)-f(a)}{x-a}\times(x-a)\right]}=L\times 0=0.\]By the way, we have
\[f(x)-f(a)=\frac{f(x)-f(a)}{x-a}\times (x-a),\]when $x\ne a$. Thus it follows that $\lim_{x\to a}{[f(x)-f(a)]}=0$, by Theorem 1.3.11. And, we have $\lim_{x\to a}{[f(a)+(f(x)-f(a))]}=f(a)$. Finally, since $f(x) = f(a) + (f(x)-f(a))$, $\lim_{x\to a}f(x)=f(a)$, by Theorem 1.3.11. Therefore $f$ is continuous at $x=a$. $\square$
Proof 2. Let
Let $W$ be a neighborhood of $f(a)$, and let $W_0$ be a neighborhood such that $W_0\subseteq W-f(a)$. Choose $\varepsilon>0$ such that $(-\varepsilon, \varepsilon)\subseteq W_0$. Since $f$ is differentiable at $x=a$ there exists a punctured neighborhood $V$ of $a$ such that $\phi(V)\subseteq (L-1, L+1)$, so $\vert \phi(x)\vert < \vert L \vert +1$. Choose a neighborhood $T$ such that
\[x\in T\Rightarrow |x-a|<\frac{\varepsilon}{\vert L\vert +1}.\]Then for $x\in V\cap T$, $x\ne a$, we have $f(x)-f(a)=\phi(x)(x-a)\in(-\varepsilon,\varepsilon)\subseteq W_0$, hence $f(x)\in W$. For $x=a$, $f(a)\in W$ automatically. Therefore $f$ is continuous at $a$. $\square$
Definition 2.1.6. A function $f’:D\to \mathbb R$ is the derivative of $f$ if there exists a function $f$ such that
\[f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\]for every input $x\in D$ where $D=\lbrace x\in \mathbb R : f’(x) \text{ exists}\rbrace$.
Example 2.1.7. Let $f:\mathbb R \to \mathbb R$ be a function such that $f(x)=x^3-x$. Find the derivative of $f$.
Solution. By the Definition 2.1.6,
Because the limit exists for all $x\in \mathbb R$, $\operatorname{dom}(f’)=\mathbb R$.