Edouard Bisson (1856-1939) was a French painter.
We have sections
\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}
Derivatives of Real-Valued Functions.
Let \(f\) be a real-valued function defined on an open interval. Then, for any two distinct points \(x\) and \(c\) in \((a,b)\), we can form the difference quotient
\[\frac{f(x)-f(c)}{x-c}.\]
We keep \(c\) fixed and study the behavior of this quotient as \(x\rightarrow c\). Now, \(f\) is said to be differentiable at \(c\) when the limit
\[\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}\]
exists. In general, without considering the open intervals, we shall consider the accumulation point as follows.
Definition. Let \(f:D\rightarrow\mathbb{R}\) be a real-valued function defined on \(D\subseteq\mathbb{R}\), and let \(c\in D\) be a accumulation point of \(D\). The derivative of \(f\) at \(c\) is denoted and defined by
\begin{equation}{\label{maeq1}}\tag{1}
f'(c)=\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c},
\end{equation}
provided that this limit is finite. In this case, we also say that \(f\) is differentiable at \(c\). If \(f\) is differentiable at each point of \(D\), we say that \(f\) is differentiable on \(D\). \(\sharp\)
Let \(h=x-c\). Then (\ref{maeq1}) can be rewritten as
\[f'(c)=\lim_{h\rightarrow 0}\frac{f(c+h)-f(c)}{h}.\]
The function \(f’\) is called the first derivative of \(f\). Similarly, the \(n\)th derivative, denoted by \(f^{(n)}\), is defined to be the first derivative of \(f^{(n-1)}\) for \(n=2,3,\cdots\). Some other notations that are given below will also be used in the context.
\[f'(c)=Df(c)=\frac{df}{dx}(c)=\left .\frac{dy}{dx}\right |_{x=c}\]
with \(y=f(x)\). The concept of differentiability is stronger than that of continuity, which is shown below.
\begin{equation}{\label{map5}}\tag{2}\mbox{}\end{equation}
Proposition \ref{map5}. Let \(f:D\rightarrow\mathbb{R}\) be a real-valued function defined on \(D\subseteq\mathbb{R}\). Suppose that \(f\) is differentiable at \(c\). Then \(f\) must be continuous at \(c\).
Proof. Showing that \(f\) is continuous at \(c\) is equivalent to show
\[\lim_{x\rightarrow c}f(x)=f(c).\]
Now, we have
\begin{align*}\lim_{x\rightarrow c}\left [f(x)-f(c)\right ] & = \lim_{x\rightarrow c}\left [\left (f(x)-f(c)\right )\cdot\left (\frac{x-c}{x-c}\right )\right ]\\& = \lim_{x\rightarrow c}\left [\frac{f(x)-f(c)}{x-c}\cdot (x-c)\right ]\\& = \left [\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}\right ]\cdot\left [\lim_{x\rightarrow c}(x-c)\right ]\mbox{ (since \(f'(c)\) exists)}\\& = f'(c)\cdot 0 =0.\end{align*}
This completes the proof. \(\blacksquare\)
Example. Let \(f:\mathbb{R}\rightarrow\mathbb{R}\). Suppose that
\[|f(x)-f(p)|\leq |x-p|^{1+\alpha}\]
for any \(x,p\in\mathbb{R}\), where \(\alpha>0\). We are going to claim that \(f\) is a constant function. For any \(p\neq x\), we have
\[\frac{|f(x)-f(p)|}{|x-p|}<|x-p|^{\alpha}.\]
Therefore, we obtain
\begin{align*} \lim_{x\rightarrow p}\left |\frac{f(x)-f(p)}{x-p}\right | & =\lim_{x\rightarrow p}\frac{|f(x)-f(p)|}{|x-p|}\\ & =\lim_{x\rightarrow p}|x-p|^{\alpha}=0,\end{align*}
which implies
\[f'(p)=\lim_{x\rightarrow p}\frac{f(x)-f(p)}{x-p}=0\]
for any \(p\in\mathbb{R}\). This says that \(f\) is a constant function. \(\sharp\)
Definition. Let \(f\) be a real-valued function defined on a closed interval \(I\). We say that \(f\) has a righthand derivative at \(c\in I\) when the righthand limit
\[\lim_{x\rightarrow c+}\frac{f(x)-f(c)}{x-c}\]
exists as a finite value, or if the limit is \(+\infty\) or \(-\infty\). This limit is denoted by \(f’_{+}(c)\). The lefthand derivative is denoted and defined by
\[f’_{-}(c)=\lim_{x\rightarrow c-}\frac{f(x)-f(c)}{x-c}.\]
In addition, if \(c\) is an interior point of \(I\), we say that \(f\) has the derivative \(f'(c)=+\infty\) when both the righthand and lefthand derivatives at \(c\) are \(+\infty\). The derivative \(f'(c)=-\infty\) can be similarly defined. \(\sharp\)
It is clear that \(f\) has a derivative (finite or infinite) at an interior point \(c\) if and only if \(f’_{+}(c)=f’_{-}(c)\). In this case, we also have
\[f’_{+}(c)=f’_{-}(c)=f'(c).\]
Proposition. Suppose that the real-valued functions \(f,g:D\rightarrow\mathbb{R}\) are defined on \(D\subseteq\mathbb{R}\) and differentiable at \(c\in D\). Then \(f+g\), \(f-g\) and \(f\cdot g\) are differentiable at \(c\). If \(g(c)\neq 0\), then \(f/g\) is also differentiable at \(c\). Moreover, we have the following rules for differentiation:
- \((f+g)'(c)=f'(c)+g'(c)\);
- \((f-g)'(c)=f'(c)-g'(c)\);
- \((f\cdot g)'(c)=f(c)\cdot g'(c)+g(c)\cdot f'(c)\);
- \({\displaystyle \left (\frac{f}{g}\right )'(c)=\frac{g(c)\cdot f'(c)-f(c)\cdot g'(c)}{g^{2}(c)}}\) provided that \(g(c)\neq 0\). \(\sharp\)
Proposition. (Chain Rule). Let \(I\) and \(J\) be open intervals in \(\mathbb{R}\). We consider the real-valued functions \(f:I\rightarrow J\) and \(g:J\rightarrow\mathbb{R}\). Suppose that \(f\) is differentiable at \(c\in I\) and \(g\) is differentiable at \(d=f(c)\in J\). Then, the composition function \(g\circ f\) is differentiable at \(c\) and
\[(g\circ f)'(c)=g'(f(c))\cdot f'(c).\]
Theorem. (Inverse Function Theorem). Let \(I\) be an open interval in \(\mathbb{R}\). Suppose that the real-valued function \(f:I\rightarrow\mathbb{R}\) is differentiable on \(I\) with \(f'(x)\neq 0\) for any \(x\in I\). Then, we have the following properties.
(i) \(f\) is one-to-one.
(ii) \(f^{-1}\) is continuous on \(f(I)\).
(iii) \(f^{-1}\) is differentiable on \(f(I)\).
(iv) \({\displaystyle (f^{-1})'(b)=\frac{1}{f'(c)}}\) with \(b=f(c)\). \(\sharp\)
Definition. Let \(f:I\rightarrow\mathbb{R}\) be a real-valued function defined on an open interval \(I\subseteq\mathbb{R}\). Then \(f\) is said to be uniformly differentiable on \(I\) when \(f\) is differentiable on \(I\) and, for each \(\epsilon >0\), there exists \(\delta >0\) such that
\[0<|x-p|<\delta\mbox{ with }x,p\in I\mbox{ implies }\left |\frac{f(x)-f(p)}{x-p}-f'(x)\right |<\epsilon.\]
It is clear to see that the real-valued functions \(\sin x\) and \(\cos x\) are uniformly differentiable on \(\mathbb{R}\).
Proposition. Suppose that \(f\) is uniformly differentiable on an open interval \(I\). Then \(f’\) is continuous on \(I\).
Proof. By definition, given any \(\epsilon >0\), there exists \(\delta >0\) such that
\[0<|x-p|<\delta\mbox{ with }x,p\in I\mbox{ implies }\left |\frac{f(x)-f(p)}{x-p}-f'(x)\right |<\frac{\epsilon}{2}.\]
Then, we obtain
\begin{align*} |f'(x)-f'(p)| & =\left|f'(x)-\frac{f(x)-f(p)}{x-p}\right |+\left |\frac{f(p)-f(x)}{p-x}-f'(p)\right |\\ & <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\end{align*}
This shows that \(f’\) is continuous at \(p\) for any \(p\in\mathbb{R}\). This completes the proof. \(\blacksquare\)
\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}
Local Optima.
\begin{equation}{\label{mad254}}\tag{3}\mbox{}\end{equation}
Definition \ref{mad254}. Let \(f\) be a real-valued function defined on a subset \(S\) of an interval \(I\). Given \(a\in S\), we say that \(f\) has a local maximum at \(a\) when there exists \(\delta>0\) satisfying
\[f(x)\leq f(a)\mbox{ for all }x\in (a-\delta ,a+\delta )\cap S.\]
We say that \(f\) has a local minimum at \(a\) when
\[f(x)\geq f(a)\mbox{ for all }x\in (a-\delta ,a+\delta )\cap S.\]
\begin{equation}{\label{map3}}\tag{4}\mbox{}\end{equation}
Proposition \ref{map3}. Let \(f\) be a real-valued function defined on an open interval \(I\). Then, we have the following properties.
(i) Suppose that, for some \(c\in I\), we have \(f'(c)>0\) or \(f'(c)=+\infty\). Then, there exists an one-dimensional ball \(B(c)\subseteq I\) satisfying
\begin{equation}{\label{ma98}}\tag{5}
f(x)>f(c)\mbox{ for }x>c\mbox{ and }f(x)<f(c)\mbox{ for }x<c.
\end{equation}
(ii) Suppose that, for some \(c\in I\), we have \(f'(c)<0\) or \(f'(c)=-\infty\). Then, there exists an one-dimensional ball \(B(c)\subseteq I\) satisfying
\[f(x)<f(c)\mbox{ for }x>c\mbox{ and }f(x)>f(c)\mbox{ for }x<c.\]
Proof. It suffices to prove part (i). Suppose that \(f'(c)>0\). We define a function \(f^{*}\) on \([a,b]\) by
\[f^{*}(x)=\left\{\begin{array}{ll}{\displaystyle \frac{f(x)-f(c)}{x-c}} & \mbox{if \(x\neq c\)}\\f'(c) & \mbox{if \(x=c\)}. \end{array}\right .\]
Since
\begin{align*} \lim_{x\rightarrow c}f^{*}(x) & =\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}\\ & =f'(c)=f^{*}(c),\end{align*}
it follows that \(f^{*}\) is continuous at \(c\). Proposition 9a in page Continuity of Functions says that there exists a one-dimensional open ball \(B(c)\) such that \(f^{*}(x)\) has the same sign as \(f^{*}(c)\) in \(B(c)\). Since
\[f^{*}(x)=\frac{f(x)-f(c)}{x-c}\mbox{ for }x\neq c\]
and \(f^{*}(c)=f'(c)>0\), it follows that \(f(x)-f(c)\) has the same sign as \(x-c\), which proves (\ref{ma98}). Suppose that \(f'(c)=+\infty\). Then, there exists a one-dimensional open ball \(B(c)\) satisfying
\[\frac{f(x)-f(c)}{x-c}>1\mbox{ for }x\neq c.\]
Since this quotient is positive, it follows that \(f(x)-f(c)\) and \(x-c\) must have the same sign. This completes the proof. \(\blacksquare\)
\begin{equation}{\label{map4}}\tag{6}\mbox{}\end{equation}
Proposition \ref{map4}. Let \(f\) be a real-valued function defined on an open interval \(I\). Suppose that \(f\) has a local maximum or minimum at an interior point \(c\) of \(I\). Then \(f'(c)=0\).
Proof. Suppose that \(f'(c)\) is positive or \(+\infty\). Then, part (i) of Proposition \ref{map3} says that \(f\) cannot have a local extremum at \(c\). Suppose that \(f'(c)\) is negative or \(-\infty\). Then, part (ii) of Proposition \ref{map3} says that \(f\) cannot have a local extremum at \(c\). Therefore, we must have \(f'(c)=0\). This completes the proof. \(\blacksquare\)
The converse of Proposition\ref{map4} is not true in general. For example, the function \(f(x)=x^{3}\) satisfies \(f'(0)=0\), but it is increasing in every neighborhood of \(0\). We also remark that \(f\) can have a local maximum or minimum at \(c\) without \(f'(c)=0\). For example, the function \(f(x)=|x|\) has a minimum at \(0\). However, the derivative \(f'(0)\) does not exist.
\begin{equation}{\label{mat9}}\tag{7}\mbox{}\end{equation}
Lemma \ref{mat9}. Let \(f\) be a real-valued function defined on a metric space \((M,d)\). Suppose that \(f\) is continuous on a compact subset \(X\) of \((M,d)\). Then, there exist \(p,q\in X\) satisfying
\[f(p)=\inf_{x\in X}f(x)=\min_{x\in X}f(x)\]
and
\[f(q)=\sup_{x\in X}f(x)=\max_{x\in X}f(x).\]
The above Lemma \ref{mat9} can refer to the page of Continuity of Functions.
Theorem. (Rolle’s Theorem). Suppose that the real-valued function \(f:[a,b]\rightarrow\mathbb{R}\) is defined on the closed interval \([a,b]\) such that the following conditions are satisfied.
- \(f\) is continuous at both endpoints \(a\) and \(b\).
- \(f\) is differentiable on \((a,b)\) in which each derivative can be finite or infinite.
- \(f(a)=f(b)\).
Then, there exists some point \(c\in (a,b)\) satisfying \(f'(c)=0\).
Proof. From Proposition \ref{map5}, we see that \(f\) is continuous on \([a,b]\). Suppose that \(f'(c)\neq 0\) for all \(c\in (a,b)\). Since \(f\) is continuous on a compact set \([a,b]\), it attains its maximum \(M\) and its minimum \(m\) at some point in \([a,b]\) by Lemma \ref{mat9}. Using Proposition \ref{map4}, since we assume that \(f'(c)\neq 0\) for each \(c\in (a,b)\), we conclude that \(f\) must attain its maximum and minimum at both endpoints \(a\) and \(b\). Since \(f(a)=f(b)\), we have \(m=M\), which says that \(f\) is constant on \([a,b]\). This contradicts \(f'(c)\neq 0\) for each \(c\in (a,b)\), and the proof is complete. \(\blacksquare\)
\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}
Mean-Value Theorem.
We are going to present many types of mean-value theorem for differentiation and its application to obtain the Intermediate-Value Theorem for derivatives.
\begin{equation}{\label{ma99}}\tag{8}\mbox{}\end{equation}
Theorem \ref{ma99}. (Mean-Value Theorem). Suppose that the real-valued function \(f:[a,b]\rightarrow\mathbb{R}\) is defined on the closed interval \([a,b]\) such that the following conditions are satisfied.
- \(f\) is continuous at both endpoints \(a\) and \(b\).
- $latexf$ is differentiable on \((a,b)\) in which each derivative can be finite or infinite.
Then, there exists some point \(c\in (a,b)\) satisfying
\[f'(c)=\frac{f(a)-f(b)}{a-b}=\frac{f(b)-f(a)}{b-a}.\]
The proof will be given in the following more general theorem.
\begin{equation}{\label{map6}}\tag{9}\mbox{}\end{equation}
Theorem \ref{map6}. (Generalized Mean-Value Theorem). Suppose that the real-valued functions \(f,g:[a,b]\rightarrow\mathbb{R}\) are defined on the closed interval \([a,b]\) such that the following conditions are satisfied.
- \(f\) and \(g\) are continuous at both endpoints \(a\) and \(b\).
- \(f\) and \(g\) are differentiable on \((a,b)\) in which each derivative can be finite or infinite.
- There is no \(c\in (a,b)\) at which \(f'(c)\) and \(g'(c)\) are infinite.
Then, there exists some point \(c\in (a,b)\) satisfying
\[f'(c)\cdot [g(b)-g(a)]=g'(c)\cdot [f(b)-f(a)].\]
Proof. We define
\[h(x)=f(x)\cdot [g(b)-g(a)]-g(x)\cdot [f(b)-f(a)].\]
Then, we see that \(h'(x)\) is finite when \(f'(x)\) and \(g'(x)\) are finite. The third assumption also says that \(h'(x)\) is infinite when exactly \(f'(x)\) or \(g'(x)\) is infinite. We also see that \(h\) is continuous at the endpoints satisfying
\[h(a)=h(b)=f(a)g(b)-g(a)f(b).\]
Using the Rolle’s theorem, we have \(h'(c)=0\) for some \(c\in (a,b)\). This completes the proof. \(\blacksquare\)
Suppose that we take \(g(x)=x\) in Theorem \ref{map6}. Then, we obtain
\[f'(c)\cdot (b-a)=f(b)-f(a),\]
which leads to Theorem \ref{ma99}. There is also an extension which does not require continuity at the endpoints.
Given a function \(f\), the right-limit \(f(a+)\) and left-limit \(f(a-)\) can refer to the page Discontinuity and Semi-continuity of Functions.
\begin{equation}{\label{ma8}}\tag{10}\mbox{}\end{equation}
Theorem \ref{ma8}. (Generalized Mean-Value Theorem). Suppose that the real-valued functions \(f,g:[a,b]\rightarrow\mathbb{R}\) are defined on the
closed interval \([a,b]\) such that the following conditions are satisfied.
- The limits \(f(a+)\), \(g(a-)\), \(f(b+)\) and \(g(b-)\) exist as finite values.
- \(f\) and \(g\) are differentiable on \((a,b)\) in which each derivative can be finite or infinite.
- There is no \(c\in (a,b)\) at which \(f'(c)\) and \(g'(c)\) are infinite.
Then, there exists some point \(c\in (a,b)\) satisfying
\[f'(c)\cdot [g(b-)-g(a+)]=g'(c)\cdot [f(b-)-f(a+)].\]
Proof. We define new functions \(F\) and \(G\) on \([a,b]\) as follows:
- \(F(x)=f(x)\) and \(G(x)=g(x)\) for \(x\in (a,b)\);
- \(F(a)=f(a+)\), \(G(a)=g(a+)\), \(F(b)=f(b-)\) and \(G(b)=g(b-)\).
Then, we see that \(F\) and \(G\) are continuous on \([a,b]\). Therefore, the desired result follows immediately from Theorem \ref{map6}. This completes the proof. \(\blacjsquare\)
Example. Let \(f:[0,\infty)\rightarrow\mathbb{R}\) be differentiable on \([0,\infty)\). Suppose that
\begin{equation}{\label{ma154}}\tag{11}
\lim_{x\rightarrow\infty}\left [f(x)+f'(x)\right ]=0.
\end{equation}
We are going to claim that \(\lim_{x\rightarrow\infty}f(x)=0\). Let \(g(x)=e^{x}f(x)\). Then, we have
\begin{equation}{\label{ma155}}\tag{12}
g'(x)=e^{x}\left [f(x)+f'(x)\right ].
\end{equation}
From (\ref{ma154}), given any \(\epsilon>0\), there exists \(N_{1}>0\) such that
\begin{equation}{\label{ma156}}\tag{13}
x>N_{1}\mbox{ implies }\left |f(x)+f'(x)\right |<\frac{\epsilon}{2}.
\end{equation}
Using the generalized mean-value Theorem \ref{map6} by considering functions \(g\) and \(e^{x}\), there exists \(c\) satisfying \(N_{1}<c<x\) satisfying
\[e^{c}\left [g(x)-g(N_{1})\right ]=g'(c)\left (e^{x}-e^{N_{1}}\right ).\]
Using (\ref{ma155}), we obtain
\[e^{c}\left [g(x)-g(N_{1})\right ]=e^{c}\left [f(c)+f'(c)\right ]\left (e^{x}-e^{N_{1}}\right ),\]
which implies
\[f(c)+f'(c)=\frac{g(x)-g(N_{1})}{e^{x}-e^{N_{1}}}.\]
Using (\ref{ma156}), we have
\[\left |g(x)-g(N_{1})\right |<\frac{\epsilon}{2}\left |e^{x}-e^{N_{1}}\right |,\]
which implies
\[|g(x)|<\frac{\epsilon}{2}\left |e^{x}-e^{N_{1}}\right |+\left |g(N_{1})\right |.\]
Since \(g(x)=e^{x}f(x)\), we obtain
\[\left |e^{x}f(x)\right |<\frac{\epsilon}{2}\left |e^{x}-e^{N_{1}}\right |+\left |e^{N_{1}}f(N_{1})\right |,\]
which implies
\begin{align*} |f(x)| & <\frac{\epsilon}{2}\left |1-e^{N_{1}-x}\right |+\left |e^{N_{1}-x}f(N_{1})\right |\\ & <\frac{\epsilon}{2}+\left |e^{N_{1}-x}f(N_{1})\right |\mbox{ (since \(N_{1}-x<0\))}.\end{align*}
Therefore, we obtain
\begin{equation}{\label{ma157}}\tag{14}
x>N_{1}\mbox{ implies }|f(x)|<\frac{\epsilon}{2}+\left |e^{N_{1}-x}f(N_{1})\right |
\end{equation}
Since
\[\lim_{x\rightarrow\infty}\left |e^{N_{1}-x}f(N_{1})\right |=0,\]
there exists \(N_{2}>0\) such that
\[x>N_{2}\mbox{ implies }\left |e^{N_{1}-x}f(N_{1})\right |<\frac{\epsilon}{2}.\]
Let \(N=\max\{N_{1},N_{2}\}\). Then, for any \(x>N\), using (\ref{ma157}), we have
\[|f(x)|<\frac{\epsilon}{2}+\left |e^{N_{1}-x}f(N_{1})\right |<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\]
This shows \(\lim_{x\rightarrow\infty}f(x)=0\). \(\sharp\)
Proposition. Suppose that the real-valued function \(f\) defined on the open interval \((a,b)\) is differentiable on \((a,b)\) and continuous at the endpoints \(a\) and \(b\) in which each derivative can be finite or infinite. Then, we have the following properties.
(i) Suppose that \(f'(c)>0\) or \(f'(c)=+\infty\) for each \(c\in (a,b)\). Then \(f\) is strictly increasing on \([a,b]\).
(ii) Suppose that \(f'(c)<0\) or \(f'(c)=-\infty\) for each \(c\in (a,b)\). Then \(f\) is strictly decreasing on \([a,b]\).
(iii) Suppose that \(f'(c)=0\) for each \(c\in (a,b)\). Then \(f\) is constant on \([a,b]\).
Proof. For any \(x<y\) with \([x,y]\subseteq [a,b]\), using the mean-value theorem on \([x,y]\), we have
\begin{equation}{\label{maeq7}}\tag{15}
f'(c)=\frac{f(y)-f(x)}{y-x}
\end{equation}
for some \(c\in (x,y)\). Therefore, the desired results follow from (\ref{maeq7}) immediately. \(\blacksquare\)
\begin{equation}{\label{mat8}}\tag{16}\mbox{}\end{equation}
Lemma \ref{mat8}. (Intermediate-Value Theorem). Let \(f\) be a real-valued and continuous function defined on a closed interval \(I\) in \(\mathbb{R}\). Suppose that \(\alpha <\beta\) in \(I\) satisfying \(f(\alpha )\neq f(\beta )\). Then \(f\) takes every value between \(f(\alpha )\) and \(f(\beta )\) in the interval \((\alpha ,\beta )\).
The above Lemma \ref{mat8} can refer to the page Continuity of Functions. Lemmas \ref{mat9} and \ref{mat8} say that a continuous real-valued function \(f\) defined on a closed interval \([a,b]\) assumes every value between its maximum and minimum on \([a,b]\). In particular, \(f\) assumes every value between \(f(a)\) and \(f(b)\). The similar result for the functions with derivatives will be shown below.
\begin{equation}{\label{mat10}}\tag{16}\mbox{}\end{equation}
Theorem \ref{mat10}. (Intermediate-Value Theorem for Derivatives). Suppose that the real-valued function \(f\) defined on the closed interval \([a,b]\) is differentiable on the open interval \((a,b)\) in which each derivative can be finite or infinite. We also assume that \(f\) has finite one-sided derivatives \(f’_{+}(a)\) and \(f’_{-}(b)\) at the endpoints with \(f’_{+}(a)\neq f’_{-}(b)\). Suppose that \(c\) is a real number between \(f’_{+}(a)\) and \(f’_{-}(b)\). Then, there exists \(x_{0}\in (a,b)\) satisfying \(f'(x_{0})=c\).
Proof. We define a new function
\[g(x)=\frac{f(x)-f(a)}{x-a}\mbox{ and }g(a)=f’_{+}(a)\]
for \(x\in (a,b]\) Then, we see that \(g\) is continuous at the endpoints \(a\) and \(b\). Since \(f\) is continuous on \((a,b)\) by Proposition \ref{map5}, it follows that \(g\) is continuous on \([a,b]\). Using Theorem \ref{mat8}, the real-valued function \(g\) takes every value between
\[g(a)=f’_{+}(a)\mbox{ and }g(b)=\frac{f(b)-f(a)}{b-a}\]
on \((a,b)\). Using the mean-value theorem, we also have \(g(x)=f'(\eta)\) for some \(\eta\in (a,x)\) whenever \(x\in (a,b)\). This shows that \(f’\) takes every value between \(g(a)\) and \(g(b)\) on \((a,b)\).
Now, we define another function
\[h(x)=\frac{f(x)-f(b)}{x-b}\mbox{ and }h(b)=f’_{-}(b)\]
for \(x\in [a,b)\). Then, the above similar argument can show that \(f’\) takes every value between \(h(a)=g(b)\) and$h(b)$ on \((a,b)\). Combining these results, we see that \(f’\) takes every value between \(g(a)=f’_{+}(a)\) and \(h(b)=f’_{-}(b)\) on \((a,b)\). This completes the proof. \(\blacksquare\)
We also remark that Theorem \ref{mat10} is still valid if one or both of the one-sided derivatives \(f’_{+}(a)\) and \(f’_{-}(b)\) is infinite. In this case, the proof can consider the function \(g\) defined by \(g(x)=f(x)-cx\) for \(x\in [a,b]\).
Proposition. Suppose that the real-valued function \(f\) defined on the open interval \((a,b)\) is differentiable on \((a,b)\) and \(f’\) is monotonic on \((a,b)\). Then \(f’\) is continuous on \((a,b)\).
Proof. Suppose that there exits \(c\in (a,b)\) such that \(f’\) is discontinuous at \(c\). We take a closed interval \([\alpha ,\beta ]\subset (a,b)\) satisfying \(c\in (\alpha ,\beta )\). Since \(f’\) is monotonic on \([\alpha ,\beta ]\), the discontinuity at \(c\) must be a jump discontinuity. This says that \(f’\) shall skip some values between \(f'(\alpha )\) and \(f'(\beta )\), which contradicts Theorem \ref{mat10}. This completes the proof. \(\blacksquare\)
Taylor’s Theorem.
Using the Mean-Value Theorem \ref{ma99}, we have
\[f(b)=f(a)+f'(c)(b-a)\]
for some \(c\) that is between \(a\) and \(b\). Suppose that the real-valued function \(f\) is differentiable at \(c\). Then \(f\) can be approximated by a linear function near \(c\) given by the following form
\[f(x)\approx f(c)+f'(c)(x-c).\]
This approximation is correct when \(x-c\) is sufficiently small. In general, the Taylor’s theorem tells us that \(f\) can be approximated by a polynomial of degree \(n\) when \(f\) has a derivative of order \(n+1\).
\begin{equation}{\label{mat187}}\tag{17}\mbox{}\end{equation}
Theorem \ref{mat187}. (Taylor’s Theorem). Given any integer \(n\in\mathbb{N}\), suppose that the real-valued function \(f\) has finite derivatives of order \(n+1\) everywhere on an open interval \((a,b)\), and that \(f^{(n)}\) is continuous on the closed interval \([a,b]\). Given \(c\in [a,b]\), for every \(x\in [a,b]\) with \(x\neq c\), we have
\[f(x)=f(c)+f'(c)(x-c)+\frac{f”(c)}{2!}(x-c)^{2}+\cdots +\frac{f^{(n)}(c)}{n!}(x-c)^{n}+R_{n}(x),\]
where the remainder term \(R_{n}(x)\) is given by
\[R_{n}(x)=\frac{f^{(n+1)}(x_{0})}{(n+1)!}(x-c)^{n+1}\]
for some \(x_{0}\) in the interior of the interval joining \(c\) and \(x\). \(\blacksquare\)
A more general results will be given below, which can obtain the Taylor’s theorem.
\begin{equation}{\label{mat11}}\tag{18}\mbox{}\end{equation}
Theorem \refl{mat11}. Given any integer \(n\in\mathbb{N}\), suppose that the real-valued functions \(f\) and \(g\) have finite derivatives of order \(n+1\) everywhere on an open interval \((a,b)\), and that \(f^{(n)}\) and \(g^{(n)}\) are continuous on the closed interval \([a,b]\). Given \(c\in [a,b]\), for every \(x\in [a,b]\) with \(x\neq c\), we have
\[\left [f(x)-f(c)-\sum_{k=1}^{n}\frac{f^{(k)}(c)}{k!}(x-c)^{k}\right ]g^{(n+1)}(x_{0})
=f^{(n+1)}(x_{0})\left [g(x)-g(c)-\sum_{k=1}^{n}\frac{g^{(k)}(c)}{k!}(x-c)^{k}\right ],\]
where \(x_{0}\) is in the interior of the interval joining \(c\) and \(x\).
Proof. Without loss of generality, we assume \(c<b\) and \(x>c\). For each \(t\in [c,x]\), we define two functions
\[F(t)=f(t)+\sum_{k=1}^{n}\frac{f^{(k)}(t)}{k!}(x-t)^{k}\]
and
\[G(t)=g(t)+\sum_{k=1}^{n}\frac{g^{(k)}(t)}{k!}(x-t)^{k}\]
Then, the real-valued functions \(F\) and \(G\) are continuous on the closed interval \([c,x]\) and have finite derivatives everywhere on the open interval \((c,x)\).
Using the generalized mean-value Theorem \ref{map6}, we have
\[F'(x_{0})\cdot [G(x)-G(c)]=G'(x_{0})\cdot [F(x)-F(c)]\]
for some \(x_{0}\in (c,x)\). Since \(G(x)=g(x)\) and \(F(x)=f(x)\), we also obtain
\[F'(x_{0})\cdot [g(x)-G(c)]=G'(x_{0})\cdot [f(x)-F(c)];\]
that is, we have
\begin{equation}{\label{maeq12}}\tag{19}
F'(x_{0})\cdot\left [g(x)-g(c)-\sum_{k=1}^{n}\frac{g^{(k)}(c)}{k!}(x-c)^{k}\right ]
=G'(x_{0})\cdot\left [f(x)-f(c)-\sum_{k=1}^{n}\frac{f^{(k)}(c)}{k!}(x-c)^{k}\right ].
\end{equation}
We compute the derivatives with respect to the variable \(t\)
\begin{align*} F'(t) & =f'(t)+\sum_{k=1}^{n}\frac{f^{(k+1)}(t)}{k!}(x-t)^{k}
-\sum_{k=1}^{n}\frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}\\ & =\frac{(x-t)^{n}}{n!}f^{(n+1)}(t)\end{align*}
and
\begin{align*} G'(t) & =g'(t)+\sum_{k=1}^{n}\frac{g^{(k+1)}(t)}{k!}(x-t)^{k}
-\sum_{k=1}^{n}\frac{g^{(k)}(t)}{(k-1)!}(x-t)^{k-1}\\ & =\frac{(x-t)^{n}}{n!}g^{(n+1)}(t).\end{align*}
By taking \(t=x_{0}\), we obtain
\[F'(x_{0})=\frac{(x-x_{0})^{n}}{n!}f^{(n+1)}(x_{0})\]
and
\[G'(x_{0})=\frac{(x-x_{0})^{n}}{n!}g^{(n+1)}(x_{0}).\]
Substituting them into (\ref{maeq12}), we can obtain the desired formula, and the proof is complete. \(\blacksquare\)
Suppose that we take \(g(x)=(x-c)^{n+1}\) in Theorem \ref{mat11}. Then, we have \(g^{(k)}(c)=0\) for \(0\leq k\leq n\) and \(g^{(n+1)}(x)=(n+1)!\), which can obtain the Taylor’s theorem.
\begin{equation}{\label{d}}\tag{D}\mbox{}\end{equation}
Partial Derivatives.
Let \(S\) be an open subset of \(\mathbb{R}^{n}\), and let \(f:S\rightarrow\mathbb{R}\) be a real-valued function defined on \(S\). If the following limit exists
\[\frac{\partial}{\partial x_{k}}f({\bf c})=\lim_{x_{k}\rightarrow c_{k}}\frac{f(c_{1},\cdots ,
c_{k-1},x_{k},c_{k+1},\cdots ,c_{n})-f(c_{1},\cdots ,c_{n})}{x_{k}-c_{k}},\]
it is called the partial derivative of \(f\) at \({\bf c}\) with respect to the \(k\)th coordinate. In particular, if \(f(x,y,z)\) is a function of three variables, the partial derivatives of \(f\) with respect to \(x\), \(y\) and \(z\) are the functions \(f_{x}\), \(f_{y}\) and \(f_{z}\) defined by
\begin{align*}
f_{x}(x,y,z) & =\lim_{h\rightarrow 0}\frac{f(x+h,y,z)-f(x,y,z)}{h},\\
f_{y}(x,y,z) & =\lim_{h\rightarrow 0}\frac{f(x,y+h,z)-f(x,y,z)}{h},\\
f_{z}(x,y,z) & =\lim_{h\rightarrow 0}\frac{f(x,y,z+h)-f(x,y,z)}{h}
\end{align*}
provided these limits exist.
Example. We have the following examples.
(i) For \(f(x,y)=x\tan^{-1}xy\), the partial derivatives are
\[f_{x}(x,y)=\frac{xy}{1+(xy)^{2}}+\tan^{-1}xy\mbox{ and }f_{y}(x,y)=\frac{x^{2}}{1+x^{2}y^{2}}.\]
(ii) For \(f(x,y)=e^{xy}+\ln (x^{2}+y)\), the partial derivatives are
\[f_{x}(x,y)=ye^{xy}+\frac{2x}{x^{2}+y}\mbox{ and }f_{y}(x,y)=xe^{xy}+\frac{1}{x^{2}+y^{2}}.\]
(iii) For the function \(f(x,y,z)=xy^{2}z^{3}\), the partial derivatives are
\[f_{x}(x,y,z)=y^{2}z^{3}, f_{y}(x,y,z)=2xyz^{3}, f_{z}(x,y,z)=3xy^{2}z^{2}.\]
(iv) For the function \(g(x,y,z)=x^{2}e^{y/z}\), the partial derivatives are
\[g_{x}(x,y,z)=2xe^{y/z}, g_{y}(x,y,z)=\frac{x^{2}}{2}e^{y/z},
g_{z}(x,y,z)=-\frac{x^{2}y}{z^{2}}e^{y/z}.\]
(v) For a function of the form \(f(x,y,z)=F(x,y)G(y,z)\), the partial derivatives are
\begin{align*}
f_{x}(x,y,z) & =F_{x}(x,y)G(y,z),\\
f_{y}(x,y,z) & =F(x,y)G_{y}(y,z)+F_{y}(x,y)G(y,z),\\
f_{z}(x,y,z) & =F(x,y)G_{z}(y,z).
\end{align*}
We usually write the partial derivatives \(f_{x}\), \(f_{y}\) and \(f_{z}\) as
\[\frac{\partial f}{\partial x}=f_{x},\quad\frac{\partial f}{\partial y}=f_{y}\mbox{ and }\frac{\partial f}{\partial z}=f_{z}.\]
Let \(z=f(x,y)\) be a function of \(x\) and \(y\). Then, we have the following notations for second-order partials
\begin{align*}
(f_{x})_{x} & =f_{xx}=\frac{\partial}{\partial x}\left (\frac{\partial f}{\partial x}\right )=\frac{\partial^{2} f}{\partial x^{2}}=\frac{\partial^{2} z}{\partial x^{2}}\\
(f_{x})_{y} & =f_{xy}=\frac{\partial}{\partial y}\left (\frac{\partial f}{\partial x}\right )=\frac{\partial^{2} f}{\partial y\partial x}=\frac{\partial^{2} z}{\partial y\partial x}\\
(f_{y})_{x} & =f_{yx}=\frac{\partial}{\partial x}\left (\frac{\partial f}{\partial y}\right )=\frac{\partial^{2} f}{\partial x\partial y}=\frac{\partial^{2} z}{\partial x\partial y}\\
(f_{y})_{y} & =f_{yy}=\frac{\partial}{\partial y}\left (\frac{\partial f}{\partial y}\right )=\frac{\partial^{2} f}{\partial y^{2}}=\frac{\partial^{2} z}{\partial y^{2}}.
\end{align*}
Example. The function \(f(x,y)=\sin x^{2}y\) has first partial derivatives
\[\frac{\partial f}{\partial x}=2xy\cos x^{2}y\mbox{ and }\frac{\partial f}{\partial y}=x^{2}\cos x^{2}y.\]
The second-order partial derivatives are
\begin{align*}
\frac{\partial^{2}f}{\partial x^{2}} & =-4x^{2}y^{2}\sin x^{2}y+2x
\cos x^{2}y, \frac{\partial^{2}f}{\partial y\partial x}=-2x^{3}y\sin x^{2}y+2x\cos x^{2}y\\
\frac{\partial^{2}f}{\partial x\partial y} & =-2x^{3}y\sin x^{2}y+2x\cos x^{2}y,
\frac{\partial^{2}f}{\partial y^{2}}=-x^{4}\sin x^{2}y.
\end{align*}
In the case of a function of three variables, we can look for three first partial derivatives
\[\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z},\]
and nine second partial derivatives
\[\frac{\partial^{2} f}{\partial x^{2}},\frac{\partial^{2} f}{\partial x\partial y},
\frac{\partial^{2} f}{\partial x\partial z},\frac{\partial^{2} f}{\partial y^{2}},
\frac{\partial^{2} f}{\partial y\partial x},\frac{\partial^{2} f}{\partial y\partial z},
\frac{\partial^{2} f}{\partial z^{2}},\frac{\partial^{2} f}{\partial z\partial x},
\frac{\partial^{2} f}{\partial z\partial y}.\]
In general, the higher-order partial derivatives are given by
\[\frac{\partial^{2}f}{\partial x_{h}\partial x_{k}},\cdots ,\frac{\partial^{3}f}{\partial x_{p}\partial x_{q}\partial x_{r}}\mbox{ etc.}\]
