Skip to main content

Mathematical Analysis

Sam Macdonald

Section 5.1 The Derivative

Definition 5.1. Interior Point.

For a metric \(X\) and a set \(S\subseteq X\text{,}\) a point \(a\in S\) is called an interior point if there is some \(\varepsilon>0\) such that \(B_\varepsilon(a)\subseteq S\text{.}\)
In particular, if \(S\subseteq \mathbb{R},a\in S\) is an interior point iff exists \(\varepsilon>0\) such that \((a-\varepsilon,a+\varepsilon)\subseteq S\text{.}\)

Definition 5.2. Derivative.

Suppose \(S\subseteq \mathbb{R}\) and \(a\) is an interior point of \(S_0\text{.}\) We'll say a function \(f:S\rightarrow\mathbb{R}\) is differentiable at \(a\) if
\begin{equation*} f'(a):=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a} \end{equation*}
exists.
We need to show that \(\lim_{x\rightarrow a}f(x)=f(a)\text{.}\) If \(x\in S\setminus\{{a}\}\)
\begin{equation*} \begin{aligned} f(x) &=\frac{f(x)-f(a)}{x-a}(x-a)+f(a)\\ \implies \lim_{x\rightarrow a}f(x) &=f'(a)\cdot 0+f(a) =f(a). \end{aligned} \end{equation*}
(1), (2) are exercises
  1. For \(x\in S\setminus\{{a}\}\text{,}\)
    \begin{equation*} \begin{aligned} \frac{(fg)(x)-(fg)(a)}{x-a} =\frac{f(x)g(x)-f(a)g(a)}{x-a} =\left({\frac{f(x)-f(a)}{x-a}}\right)g(x)+f(a)\left({\frac{g(x)-g(a)}{x-a}}\right) \end{aligned} \end{equation*}
    Know
    \begin{equation*} \begin{aligned} \lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a} &=f'(a);\\ \lim_{x\rightarrow a}\frac{g(x)-g(a)}{x-a} &=g'(a) \end{aligned} \end{equation*}
    and since \(g\) is continuous at \(a\text{,}\)
    \begin{equation*} \lim_{x\rightarrow a}g(x)=g(a). \end{equation*}
    Therefore
    \begin{equation*} \begin{aligned} \lim_{x\rightarrow a}\frac{(fg)(x)-(fg)(a)}{x-a} =f'(a)g(a)+f(a)g'(a). \end{aligned} \end{equation*}
  2. To make sense of \(\left({\frac{f}{g}}\right)(a)\text{,}\) let \(T=\{{x\in S}\,:\,{g(x)\neq 0}\}\text{,}\) and we'll view \(\frac{f}{g}\) as a function \(\frac{f}{g}:T\rightarrow \mathbb{R}\text{.}\) Since \(g(a)\neq 0\) and \(g\) is continuous at \(a\text{,}\) there exists \(\delta>0\) such that \(g(x)\neq 0\) whenever \(x\in(a-\delta,a+\delta)\text{.}\) So \((a-\delta,a+\delta)\subseteq T\) and \(a\) is an interior point of \(T\text{.}\) The rest of the computation is similar to the one in (3).
\(n=0,1\) are exercises Assume the result holds for some \(n\geq 1\text{.}\) Set \(f(x)=x^{n+1},g(x)=x^n,h(x)=x\) so \(f=gh\text{.}\) Then
\begin{equation*} \begin{aligned} f'(x) =g'(x)h(x)+g(x)h'(x) =nx^{n-1}\cdot x+x^n\cdot 1 =(n+1)x^n. \end{aligned} \end{equation*}

Remark 5.6.

We could've defined derivatives of functions \(f:\mathbb{R}\rightarrow \mathbb{C}\) (or \(f:S\rightarrow\mathbb{C},S\subseteq \mathbb{R}\)) in the same way. Then \(f\) is differentiable at at point \(a\) iff \(\text{Re}(f)\) and \(\text{Im}(f)\) are, in which case
\begin{equation*} \begin{aligned} (\text{Re}(f))'(a)&=\text{Re}(f'(a));\\ (\text{Im}(f))'(a)&=\text{Im}(f'(a)). \end{aligned} \end{equation*}
One can also define derivatives on functions
\begin{equation*} f:\mathbb{C}\rightarrow\mathbb{C} \end{equation*}
in the same way
\begin{equation*} \begin{aligned} f'(a) =\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}, \end{aligned} \end{equation*}
but the theory is much different. In some sense, not many functions on \(\mathbb{C}\) are differentiable.
<Prev^TopNext>