Johann Wilhelm Jankowski (1825-1870) was an Austrian painter.
We have sections
- One-to-One Functions and Inverses
- The Logarithm Function
- The Exponential Function
- Integration of the Trigonometric Functions
- The Inverse Trigonometric Functions
- The Hyperbolic Functions
The real numbers that satisfy polynomial equations with integer coefficients are called algebraic. For example, \(\frac{3}{5}\) satisfies the equation \(5x-3=0\) and \(\sqrt{2}\) satisfies the equation \(x^{2}-2=0\). The numbers that are not algebraic are called transcendental. For example the real number \(\pi\). The functions \(f\) that satisfy polynomial equations with polynomial coefficients are called algebraic. For example, \(f(x)=x/(\pi x+\sqrt{2})\) satisfies the equation \((\pi x+\sqrt{2})f(x)-x=0\) and \(f(x)=2\sqrt{x}-3x^{2}\) satisfies the equation \([f(x)]^{2}+6x^{2}f(x)+(9x^{4}-4x)=0\). The functions that are not algebraic are called transcendental. The functions that we study here, the logarithm, the exponential, and the trigonometric functions and their inverses, are all transcendental.
\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}
One-to-One Functions and Inverses.
A function \(f\) with domain \(D\) is said to be one-to-one if and only if no two distinct points in \(D\) have the same image under \(f\). In other words, \(f(x_{1})\neq f(x_{2})\) whenever \(x_{1}\neq x_{2}\) for \(x_{1},x_{2}\in D\). Equivalently, \(f\) is one-to-one if and only if
\[f(x_{1})=f(x_{2})\mbox{ implies }x_{1}=x_{2}\]
for any pair of points \(x_{1},x_{2}\) in the domain of \(f\).
Theorem. Suppose that \(f\) is a one-to-one function. Then, there is one and only one function \(g\) that is defined on the range of \(f\) and satisfies the equation \(f(g(x))=x\) for all \(x\) in the range of \(f\). \(\sharp\)
The function that we have named \(g\) in the theorem is called the inverse of \(f\) and is denoted by \(f^{-1}\). Let \(f\) be a one-to-one function. The inverse of \(f\), denoted by \(f^{-1}\), is the unique function and is defined on the range of \(f\) and satisfies the equation \(f(f^{-1}(x))=x\) for all \(x\) in the range of \(f\).
\begin{equation}{\label{t11}}\tag{1}\mbox{}\end{equation}
Theorem \ref{t11}. Suppose that \(f\) is either an increasing function or a decreasing function. Then \(f\) is one-to-one and hence has an inverse. \(\sharp\)
Suppose that the function \(f\) has an inverse. Then, by definition, \(f^{-1}\) satisfies the equation \(f(f^{-1}(x))=x\) for all \(x\) in the range of \(f\). We also have \(f^{-1}(f(x))=x\) for all \(x\) in the domain of \(f\).
Theorem. Let \(f\) be a one-to-one function defined on an interval \(I\). Suppose that \(f\) is continuous. Then, its inverse \(f^{-1}\) is also continuous. \(\sharp\)
Theorem. Suppose that \(f\) has an inverse and is differentiable. Let \(a\) be a point in the domain of \(f\), and let \(b=f(a)\). If \(f'(a)\neq 0\), then \((f^{-1})'(b)\) exists and is given by
\[(f^{-1})'(b)=\frac{1}{f'(a)}.\]
Be sure to notice that \((f^{-1})'(b)\) is obtained by evaluating \(f’\) at \(a\), not at \(b\). Since \(a=f^{-1}(b)\), it follows
\[(f^{-1})'(b)=\frac{1}{f'(a)}=\frac{1}{f'[f^{-1}(b)]}.\]
Thus we can write
\[(f^{-1})'(x)=\frac{1}{f'[f^{-1}(x)]}.\]
Let \(y=f(x)\) so that \(x=f^{-1}(y)\). Then
\[\frac{dx}{dy}=(f^{-1})'(y)\]
and
\[\frac{dy}{dx}=f'(x)=f'(f^{-1}(y)).\]
Thus we also can write
\[\frac{dx}{dy}=\frac{1}{dy/dx}.\]
Theorem. Suppose that \(f\) has an inverse and is differentiable, and set \(y=f(x)\). Let \(a\) be a number in the domain of \(f\) and let \(b=f(a)\). If \({\displaystyle \frac{dy}{dx}(a)\neq 0}\), then \({\displaystyle \frac{dx}{dy}(b)}\) exists and is given by
\[\frac{dx}{dy}(b)=\frac{1}{{\displaystyle \frac{dy}{dx}(a)}}.\]
Example. The function defined by \(y=x-\cos x\) for \(-\pi\leq x\leq\pi\) is differentiable. (a) Show that this function has an inverse. (b) Calculate the derivative of the inverse function. (c) Evaluate the derivative of the inverse function at \(y=-1\).
(a) Since \(dy/dx=1+\sin x\geq 0\) for \(-\pi <x<\pi\) and \(1+\sin x=0\) only at \(x=-\pi /2\), it follows that this function has an inverse by Theorem \ref{t11}.
(b) We have
\[\frac{dx}{dy}=\frac{1}{dy/dx}=\frac{1}{1+\sin x}.\]
(c) Note that if we let \(x=0\), then \(y=-1\). Thus
\[\frac{dx}{dy}(-1)=\frac{1}{{\displaystyle \frac{dy}{dx}(0)}}=\frac{1}{1+\sin 0}=1.\]
\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}
The Logarithm Function.
Let \(b\) be a positive number different from \(1\). The logarithm to the base \(b\) is defined by setting
\[c=\log_{b} a\mbox{ if and only if }b^{c}=a.\]
The fundamental property of logarithms is that they transform multiplication into addition given by
\[\log_{b} xy=\log_{b} x+\log_{b} y.\]
Definition. A logarithm function is a non-constant differentiable function \(f\) defined on the set of positive numbers satisfying \(f(xy)=f(x)+f(y)\) for all \(x>0\) and \(y>0\). \(\sharp\)
Theorem. Suppose that \(f\) is a logarithm function and \(x\) is any positive number. Then, we have
\[f(1)=0\mbox{ and }f'(x)=\frac{1}{x}f'(1).\]
The most natural choice, the one that will keep calculations as simple as possible, is to set \(f'(1)=1\). The derivative is then \(1/x\). This function, which takes on the value \(0\) at \(1\) and has derivative \(1/x\) for \(x>0\), must, by the fundamental theorem of calculus, take the form
\[\int_{1}^{x} \frac{dt}{t}\]
since
\begin{align*} f(x) & =f(x)-f(1)\\ & =\int_{1}^{x} f'(t)dt=\int_{1}^{x}\frac{dt}{t}.\end{align*}
Definition. The function
\[L(x)=\int_{1}^{x} \frac{dt}{t}\mbox{ for }x>0\]
is called the natural logarithm function. We also write \(L(x)=\ln x\). \(\sharp\)
There is one and only one number at which the function \(\ln x\) takes on the value \(1\). This unique number is denoted by the letter \(e\). That is, \(\ln e=1\). We also call \(\ln x\) the logarithm to the base \(e\), i.e., \(\ln x=\log_{e} x\).
Theorem. We have the following properties.
(i) \(\ln x\) is defined on \((0,+\infty )\) with derivative
\[(\ln x)’=\frac{1}{x}\mbox{ for all }x>0.\]
$(\ln x)’$ is positive on \((0,+\infty )\), and therefore \(\ln x\) is an increasing function.
(ii) \(\ln x\) is continuous on \((0,+\infty )\) since it is differentiable on \((0,+\infty )\).
(iii) \(\ln x\) is negative if \(0<x<1\), \(\ln 1=0\), \(\ln x\) is positive if \(x>1\).
(iv) If \(x\) and \(y\) are positive, then \(\ln (1/x)=-\ln x\), \(\ln xy=\ln x+\ln y\) and \(\ln (x/y)=\ln x-\ln y\).
(v) If \(x\) is positive and \(p/q\) is rational, then
\[\ln x^{\frac{p}{q}}=\frac{p}{q}\ln x.\]
(vi) The range of \(\ln x\) is \((-\infty ,+\infty )\). \(\sharp\)
Let \(u\) be a positive differentiable function of \(x\). Then, using the chain rule, we have
\[\frac{d}{dx}(\ln u)=\frac{d}{du}(\ln u)\frac{du}{dx}=\frac{1}{u}\frac{du}{dx}.\]
For example, we have
\[\frac{d}{dx}[\ln (1+x^{2})]=\frac{2x}{1+x^{2}}\mbox{ for all real }x\]
and
\[\frac{d}{dx}[\ln (1+3x)]=\frac{3}{1+3x}\mbox{ for all }x>-\frac{1}{3}.\]
Example. Find the domain of \(f\) and find \(f'(x)\) for \(f(x)=\ln (x\sqrt{1+3x})\). We must have \(x\sqrt{1+3x}>0\), which implies \(x>0\). The domain is the set of positive numbers. Since
\begin{align*} f(x) & =\ln (x\sqrt{1+3x})\\ & =\ln x+\ln [(1+3x)^{1/2}]\\ & =\ln x+\frac{1}{2}\ln (1+3x),\end{align*}
we have
\begin{align*} f'(x) & =\frac{1}{x}+\frac{1}{2}\cdot\frac{3}{1+3x}\\ & =\frac{9x+2}{6x^{2}+2x}.\end{align*}
Example. Prove
\[\frac{d}{dx}(\ln |x|)=\frac{1}{x}\mbox{ for all }x\neq 0.\]
For \(x>0\), we have
\[\frac{d}{dx}(\ln |x|)=\frac{d}{dx}(\ln x)=\frac{1}{x}.\]
For \(x<0\), we have
\begin{align*} \frac{d}{dx}(\ln |x|) & =\frac{d}{dx}[\ln (-x)]\\ & =\frac{1}{-x}\frac{d}{dx}(-x)=\frac{1}{x}.\end{align*}
Similarly, if \(u\) is a differentiable function of \(x\), then, for \(u\neq 0\), we have
\[\frac{d}{dx}(\ln |u|)=\frac{1}{u}\frac{du}{dx}.\]
Example. Find the domain of \(f\) and find \(f'(x)\) for \(f(x)=(\ln x^{2})^{3}\). We must have \(x^{2}>0\). Thus, the domain of \(f\) consists of all \(x\neq 0\). We first have
\[f(x)=(\ln x^{2})^{3}=(2\ln |x|)^{3}=8(\ln |x|)^{3}.\]
It follows
\[f'(x)=24(\ln |x|)^{2}\frac{d}{dx}(\ln |x|)=\frac{24}{x}(\ln |x|)^{2}.\]
Example. Given
\[f(x)=\ln\left (\frac{x^{4}}{x-1}\right ).\]
Find the domain of \(f\) and \(f'(x)\). The domain of \(f\) is the open interval \((1,+\infty )\) since the logarithm function is defined only for positive numbers. Since
\[f(x)=\ln x^{4}-\ln (x-1)=4\ln x-\ln (x-1),\]
we have
\[f'(x)=\frac{4}{x}-\frac{1}{x-1}=\frac{3x-4}{x(x-1)}.\]
Now, we have the indefinite integral
\[\int \frac{dx}{x}=\ln |x|+C\mbox{ for }x\neq 0.\]
In practice, we have
\[\int \frac{g'(x)}{g(x)}dx\mbox{ is reduced to }\int \frac{du}{u}\]
by setting \(u=g(x)\) and \(du=g'(x)dx\). Therefore, we obtain
\[\int \frac{g'(x)}{g(x)}dx=\ln |g(x)|+C\mbox{ for }g(x)\neq 0.\]
Example. Calculate
\[\int \frac{x^{2}}{1-4x^{3}}dx.\]
Set \(u=1-4x^{3}\). Then \(du=-12x^{2}\). Therefore, we obtain
\begin{align*} \int \frac{x^{2}}{1-4x^{3}}dx & =-\frac{1}{12}\int \frac{du}{u}\\ & =-\frac{1}{12}\ln |u|+C\\ & =-\frac{1}{12}\ln |1-4x^{3}|+C.\end{align*}
Example. Calculate
\[\int \frac{\ln x}{x}dx.\]
Set \(u=\ln x\). Then \(du=\frac{1}{x}dx\). Therefore, we obtain
\begin{align*} \int \frac{\ln x}{x}dx & =\int udu=\frac{1}{2}u^{2}+C\\ & =\frac{1}{2}(\ln x)^{2}+C.\end{align*}
Example. Evaluate
\[\int_{1}^{2} \frac{6x^{2}-2}{x^{3}-x+1}dx.\]
Set \(u=x^{3}-x+1\),. Then \(du=(3x^{2}-1)dx\). At \(x=1\), we have \(u=1\), and at \(x=2\), we have \(u=7\). Therefore, we obtain
\begin{align*} \int_{1}^{2} \frac{6x^{2}-2}{x^{3}-x+1}dx & =2\int_{1}^{7} \frac{du}{u}\\ & =2\left [\ln |u|\right ]_{1}^{7}\\ & =2(\ln 7-\ln 1)=2\ln 7.\end{align*}
Consider the product
\[g(x)=g_{1}(x)g_{2}(x)\cdots g_{n}(x).\]
Then, we have
\[\ln |g(x)|=\ln |g_{1}(x)|+\ln |g_{2}(x)|+\cdots +\ln |g_{n}(x)|\]
Taking differentiation, we obtain
\[\frac{g'(x)}{g(x)}=\frac{g’_{1}(x)}{g_{1}(x)}+\frac{g’_{2}(x)}{g_{2}(x)}+\cdots \frac{g’_{n}(x)}{g_{n}(x)},\]
which says
\[g'(x)=g(x)\left (\frac{g’_{1}(x)}{g_{1}(x)}+\frac{g’_{2}(x)}{g_{2}(x)}+\cdots \frac{g’_{n}(x)}{g_{n}(x)}\right ).\]
The process by which \(g'(x)\) was obtained is called logarithmic differentiation. Logarithmic differentiation is valid at all points where \(g(x)\neq 0\).
Example. Given
\[g(x)=\frac{(x^{2}+1)^{3}(2x-5)^{2}}{(x^{2}+5)^{2}},\]
find \(g'(x)\) for \(x\neq\frac{5}{2}\). We can write
\[g(x)=(x^{2}+1)^{3}(2x-5)^{2}(x^{2}+5)^{-2}.\]
Then, according to the above formula, we have
\begin{align*}
g'(x) & =\frac{(x^{2}+1)^{3}(2x-5)^{2}}{(x^{2}+5)^{2}}\left [\frac{3(x^{2}+1)^{2}(2x)}{(x^{2}+1)^{3}}+
\frac{2(2x-5)(2)}{(2x-5)^{2}}+\frac{(-2)(x^{2}+5)^{-3}(2x)}{(x^{2}+5)^{-2}}\right ]\\
& =\frac{(x^{2}+1)^{3}(2x-5)^{2}}{(x^{2}+5)^{2}}\left (\frac{6x}{x^{2}+1}+\frac{4}{2x-5}-\frac{4x}{x^{2}+5}\right ).
\end{align*}
Equivalently, using the basic properties, we have
\begin{align*}
\ln |g(x)| & =\ln (x^{2}+1)^{3}+\ln (2x-5)^{2}-\ln (x^{2}+5)^{2}\\
& =3\ln (x^{2}+1)+2\ln |2x-5|-2\ln (x^{2}+5).
\end{align*}
Therefore, we obtain
\[\frac{g'(x)}{g(x)}=\frac{3(2x)}{x^{2}+1}+\frac{2(2)}{2x-5}-\frac{2(2x)}{x^{2}+5},\]
which implies
\[g'(x)=g(x)\left [\frac{6x}{x^{2}+1}+\frac{4}{2x-5}-\frac{4x}{x^{2}+5}\right ].\]
\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}
The Exponential Function.
The function \(E(x)=e^{x}\) for all real \(x\) is called the exponential function.
Theorem. We have the following properties.
(i) We have \(\ln e^{x}=x\) for all real \(x\). This says that the exponential function is the inverse of the logarithm function.
(ii) We have \(e^{x}>0\) for all real \(x\).
(iii) \(\ln 1=0\) gives \(e^{0}=1\).
(iv) We have \(e^{x}\rightarrow 0\) as \(x\rightarrow -\infty\).
(v) We have \(e^{\ln x}=x\) for all \(x>0\).
(vi) We have \(e^{x+y}=e^{x}\cdot e^{y}\) for all real \(x\) and \(y\).
(vii) We have \({\displaystyle e^{-y}=\frac{1}{e^{y}}}\).
(viii) We have \({\displaystyle e^{x-y}=\frac{e^{x}}{e^{y}}}\). \(\sharp\)
Theorem. The exponential function is its own derivative for all real \(x\), i.e.,
\[\frac{d}{dx}(e^{x})=e^{x}.\]
If \(u\) is a differentiable function of \(x\), then the chain rule gives
\[\frac{d}{dx} (e^{u})=e^{u}\frac{du}{dx}.\]
For example, we have
\begin{align*} \frac{d}{dx}(e^{kx}) & =Ke^{kx}\\
\frac{d}{dx}(e^{\sqrt{x}}) & =e^{\sqrt{x}}\frac{d}{dx}(\sqrt{x})=\frac{e^{\sqrt{x}}}{2\sqrt{x}}\\
\frac{d}{dx}(e^{-x^{2}}) & =e^{-x^{2}}\frac{d}{dx}(-x^{2})=-2xe^{-x^{2}}
\end{align*}
Now, we have the indefinite integral
\[\int e^{x}dx=e^{x}+C.\]
In practice, we have
\[\int e^{g(x)}g'(x)dx\mbox{ is reduced to }\int e^{u}du\]
by setting \(u=g(x)\) and \(du=g'(x)dx\). Then, we have
\[\int e^{g(x)}g'(x)dx=e^{g(x)}+C.\]
Example. Find
\[\int 9e^{3x}dx.\]
Set \(u=3x\). Then \(du=3dx\). Therefore, we obtain
\[\int 9e^{3x}dx=3\int e^{u}du=3e^{u}+C=3e^{3x}+C.\]
Example. Find
\[\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx.\]
Set \(u=\sqrt{x}\). Then \(du=dx/2\sqrt{x}\). Therefore, we obtain
\begin{align*} \int \frac{e^{\sqrt{x}}}{\sqrt{x}} & =2\int e^{u}du\\ & =2e^{u}+C=2e^{\sqrt{x}}+C.\end{align*}
Example. Find
\[\int \frac{e^{3x}}{e^{3x}+1}dx.\]
Set \(u=e^{3x}+1\). Then \(du=3e^{3x}dx\). Therefore, we obtain
\begin{align*} \int \frac{e^{3x}}{e^{3x}+1}dx & =\frac{1}{3}\int \frac{du}{u}\\ & =\frac{1}{3}\ln |u|+C\\ & =\frac{1}{3}\ln (e^{3x}+1)+C.\end{align*}
Example. Evaluate
\[\int_{0}^{\sqrt{2\ln 3}} xe^{-x^{2}/2}dx.\]
Set \(u=-\frac{1}{2}x^{2}\). Then \(du=-xdx\). At \(x=0\), we have \(u=0\), and, at \(x=\sqrt{2\ln 3}\), we have \(u=-\ln 3\). Therefore, we obtain
\begin{align*} \int_{0}^{\sqrt{2\ln 3}} xe^{-x^{2}/2}dx & =-\int_{0}^{-\ln 3} e^{u}du\\ & =
-\left [e^{u}\right ]_{0}^{-\ln 3}=1-e^{-\ln 3}\\ & =1-\frac{1}{3}=\frac{2}{3}.\end{align*}
Example. Evaluate
\[\int_{0}^{1} e^{x}(e^{x}+1)^{1/5}dx.\]
Set \(u=e^{x}+1\). Then \(du=e^{x}dx\). At \(x=0\), we have \(u=2\), and, at \(x=1\), we have \(u=e+1\). Therefore, we obtain
\begin{align*}\int_{0}^{1} e^{x}(e^{x}+1)^{1/5}dx & =\int_{2}^{e+1} u^{1/5}du\\ & =
\left [\frac{5}{6}e^{\frac{6}{5}}\right ]_{2}^{e+1}\\ & =\frac{5}{6}[(e+1)^{6/5}-2^{6/5}].\end{align*}
We have shown
\[\frac{d}{dx}(x^{p})=px^{p-1}\mbox{ for any rational number \(p\).}\]
We can now extend this result to arbitrary powers. First of all, we define \(x^{z}\) for irrational \(z\) by setting \(x^{z}=e^{z\ln x}\). Then, for \(x>0\), we have \(x^{r}=e^{r\ln x}\) for all real numbers \(r\). Given any real number \(r\), we also have
\begin{align*} \frac{d}{dx}(x^{r}) & =\frac{d}{dx}(e^{r\ln x})=e^{r\ln x}\frac{d}{dx}(r\ln x)\\ & =x^{r}\cdot\frac{r}{x}=rx^{r-1}.\end{align*}
Let \(u\) be a positive differentiable function of \(x\), and let \(r\) be any real number. Then, using the chain rule, we have
\[\frac{d}{dx}(u^{r})=ru^{r-1}\frac{du}{dx}.\]
We also have the indefinite integral
\[\int x^{r}dx=\frac{x^{r+1}}{r+1}+C\mbox{ for }r\neq 1.\]
Example. Find
\[\int \frac{x^{3}}{(2x^{4}+1)^{\pi}}dx.\]
Set \(u=2x^{4}+1\). Then \(du=8x^{3}dx\). Therefore, we obtain
\begin{align*} \int \frac{x^{3}}{(2x^{4}+1)^{\pi}}dx & =\frac{1}{8}\int \frac{du}{u^{\pi}}
\\ & =\frac{1}{8}\left (\frac{u^{1-\pi}}{1-\pi}\right )+C\\ & =\frac{(2x^{4}+1)^{1-\pi}}{8(1-\pi )}+C.\end{align*}
Example. Find
\[\frac{d}{dx}[(x^{2}+1)^{3x}].\]
Note that \((x^{2}+1)^{3x}=e^{3x\ln (x^{2}+1)}\). Then, we have
\begin{align*}
\frac{d}{dx}[(x^{2}+1)^{3x}] & =\frac{d}{dx}[e^{3x\ln (x^{2}+1)}]\\
& =e^{3x\ln (x^{2}+1)}\left [3x\cdot\frac{2x}{x^{2}+1}+3\ln (x^{2}+1)\right ]\\
& =(x^{2}+1)^{3x}\left [\frac{6x^{2}}{x^{2}+1}+3\ln (x^{2}+1)\right ].
\end{align*}
We can also use the logarithmic differentiation by setting \(f(x)=(x^{2}+1)^{3x}\). Then, we have \(\ln f(x)=3x\ln (x^{2}+1)\). Taking differentiation, we obtain
\begin{align*} \frac{f'(x)}{f(x)} & =3x\cdot\frac{2x}{x^{2}+1}+3\ln (x^{2}+1)\\ & =\frac{6x^{2}}{x^{2}+1}+3\ln (x^{2}+1).\end{align*}
We get the same result. \(\sharp\)
The function \(f(x)=p^{x}\) is called exponential function with base \(p\). We are going to consider the differentiation of \(p^{x}\). Now, we have
\begin{align*} \frac{d}{dx}(p^{x}) & =\frac{d}{dx}(e^{x\ln p})\\ & =e^{x\ln p}\ln p=p^{x}\ln p.\end{align*}
For example, we have
\[\frac{d}{dx} (2^{x})=2^{x}\ln 2\mbox{ and }\frac{d}{dx}(10^{x})=10^{x}\ln 10.\]
Let \(u\) be a differentiable function of \(x\). Then, using the chain rule, we have
\[\frac{d}{dx} (p^{u})=p^{u}\ln p\frac{du}{dx}.\]
For example, we have
\[\frac{d}{dx}(2^{3x^{2}})=2^{3x^{2}}(\ln 2)(6x)=6×2^{3x^{2}}\ln 2.\]
We also have the indefinite integral
\[\int p^{x}dx=\frac{p^{x}}{\ln p}+C\mbox{ for }p>0,p\neq 1.\]
Example. Find
\[\int x5^{-x^{2}}dx.\]
Set \(u=-x^{2}\). Then \(du=-2xdx\). Therefore, we obtain
\begin{align*} \int x5^{-x^{2}}dx & =-\frac{1}{2}\int 5^{u}du\\ & =-\frac{1}{2}\cdot\frac{5^{u}}{\ln 5}+C\\ & =-\frac{5^{-x^{2}}}{2\ln 5}+C.\end{align*}
Example. Evaluate
\[\int_{1}^{2} 3^{2x-1}dx.\]
Set \(u=2x-1\). Then \(du=2dx\). At \(x=1\), we have \(u=1\), and, at \(x=2\), we have \(u=3\). Therefore, we obtain
\begin{align*} \int_{1}^{2} 3^{2x-1}dx & =\frac{1}{2}\int_{1}^{3} 3^{u}du\\ & =\frac{1}{2}\left [\frac{3^{u}}{\ln 3}\right ]_{1}^{3}\\ & =\frac{12}{\ln 3}.\end{align*}
The logarithm of \(x\) to the base \(p\) is defined by
\[\log_{p} x=\frac{\ln x}{\ln p}\mbox{ for }p>0,p\neq 1,x>0.\]
For example, we have
\[\log_{2} 32=\frac{\ln 32}{\ln 2}=\frac{\ln 2^{5}}{\ln 2}=\frac{5\ln 2}{\ln 2}=5.\]
The differentiation of \(\log_{p} x\) is given by
\[\frac{d}{dx} (\log_{p} x)=\frac{1}{x\ln p}.\]
Let \(u\) be a positive differentiable function of \(x\). Then, using the chain rule, we have
\[\frac{d}{dx}(\log_{p} u)=\frac{1}{u\ln p}\cdot\frac{du}{dx}.\]
For example, we have
\begin{align*} \frac{d}{dx}[\log_{2} (3x^{2}+1)] & =\frac{1}{(3x^{2}+1)\ln 2}\cdot\frac{d}{dx}(3x^{2}+1)\\ & =\frac{6x}{3x^{2}+1)\ln 2}.\end{align*}
and
\begin{align*}\frac{d}{dx}(\log_{5} |x|) & =\frac{d}{dx}\left [\frac{\ln |x|}{\ln 5}\right ]\\ & =\frac{1}{x\ln 5}.\end{align*}
Next we are going to estimate the number \(e\).
Theorem. For each positive integer \(n\), we have
\[\left (1+\frac{1}{n}\right )^{n}\leq e\leq\left (1+\frac{1}{n}\right )^{n+1}.\]
Let \(n=1,000,000\). Then, wee have
\[\left (1+\frac{1}{1,000,000}\right )^{1,000,000}\approx 2.7182805\]
and
\[\left (1+\frac{1}{1,000,000}\right )^{1,000,001}\approx 2.7182832.\]
Therefore, for the five decimal places, \(e\approx 2.71828\). We can also show \((1+1/n)^{n}\rightarrow e\) as \(n\rightarrow\infty\).
\begin{equation}{\label{d}}\tag{D}\mbox{}\end{equation}
Integration of the Trigonometric Functions.
Example. Find
\[\int \tan xdx.\]
Since \(\tan x=\sin x/\cos x\), we set \(u=\cos x\). Then \(du=-\sin x\). Therefore, we obtain
\begin{align*}
\int \tan xdx & =\int \frac{\sin x}{\cos x}dx = -\int \frac{du}{u}\\
& =-\ln |u|+C \\ & = -\ln |\cos x| +C\\ & = \ln\left |\frac{1}{\cos x}\right |+C\\ & =\ln |\sec x|+C.
\end{align*}
Similarly, we can show
\[\int \cot xdx=\ln |\sin x|+C.\]
Example. Find
\[\int \sec xdx.\]
We have
\begin{align*}
\int \sec xdx & =\int \sec x\frac{\sec x+\tan x}{\sec x+\tan x}dx\\
& =\int \frac{\sec x\tan x+\sec^{2}x}{\sec x+\tan x}dx\\ & \quad\mbox{ (set \(u=\sec x+\tan x\), \(du=(\sec x\tan x+\sec^{2} x)dx\))}\\
& =\int \frac{du}{u} = \ln {u}+C = \ln |\sec x+\tan x|+C.
\end{align*}
Similaly, we can show
\[\int \csc xdx=\ln |\csc x-\cot x|+C.\]
Example. Calculate
\[\int \frac{\sec^{2} 3x}{1+\tan 3x}dx.\]
Set \(u=1+\tan 3x\). Then \(du=3\sec^{2} 3xdx\). Therefore, we obtain
\begin{align*} \int \frac{\sec^{2} 3x}{1+\tan 3x}dx & =\frac{1}{3}\int \frac{du}{u}\\ & =\frac{1}{3}\ln |u|+C\\ & =\frac{1}{3}\ln |1+\tan 3x|+C.\end{align*}
Example. Calculate the area of the region bounded by the graph of
\[f(x)=\frac{\sin x}{2+\cos x}\]
and the \(x\)-axis for \(x\in [0,2\pi /3]\). The area is given by
\[\int_{0}^{2\pi /3} \frac{\sin x}{2+\cos x}dx.\]
Set \(u=2+\cos x\). Then \(du=-\sin xdx\). At \(x=0\), we have \(u=3\), and, at \(x=2\pi /3\), we have \(u=3/2\). Therefore, we obtain
\begin{align*} \int_{0}^{2\pi /3} \frac{\sin x}{2+\cos x}dx & =-\int_{3}^{3/2} \frac{du}{u}\\ & =\int_{3/2}^{3} \frac{du}{u}\\ & =\ln 3-\ln (3/2)=\ln 2.\end{align*}
\begin{equation}{\label{e}}\tag{E}\mbox{}\end{equation}
The Inverse Trigonometric Functions.
The inverse sine function \(y=\sin^{-1} x\) with domain \([-1,1]\) and range \([-\frac{\pi}{2},\frac{\pi}{2}]\). We also have
\[\sin (\sin^{-1} x)=x\mbox{for all \(x\in [-1,1]\) }\]
and
\[\sin^{-1}(\sin x)=x\mbox{for all \(x\in [-\frac{\pi}{2},\frac{\pi}{2}]\) }.\]
We can see
\begin{align*} \sin^{-1}(\sin \frac{8}{3}\pi ) & =\sin^{-1}(\sin (2\pi +\frac{2}{3}\pi ))\\ & =\sin^{-1}(\sin \frac{2}{3}\pi ))\\ & =\sin^{-1}(\sin \frac{pi}{3})=\frac{\pi}{3}.\end{align*}
$\sin (\sin^{-1} 2)$ makes no sense since \(2\) is not in the domain of \(\sin^{-1} x\). Reading from the figure, we have
\begin{align*}
& \sin (\sin^{-1} x)=x\\ & \cos (\sin^{-1} x)=\sqrt{1-x^{2}}\\
& \tan (\sin^{-1} x)=\frac{x}{\sqrt{1-x^{2}}}\\ & \cot (\sin^{-1} x)=\frac{\sqrt{1-x^{2}}}{x}\\
& \sec (\sin^{-1} x)=\frac{1}{\sqrt{1-x^{2}}}\\ & \csc (\sin^{-1} x)=\frac{1}{x}.
\end{align*}
For example, let \(\sin^{-1} x=\theta\). Then \(\sin\theta =x\). We have
\[\cos (\sin^{-1} x)=\cos\theta =\sqrt{1-x^{2}}.\]
Since the derivative of the sine function,
\[\frac{d}{dx}(\sin x)=\cos x,\]
does not take on the value \(0\) on the open inerval \((-\frac{\pi}{2},\frac{\pi}{2})\) (recall that \(\cos x>0\) for \(-\frac{\pi}{2}<x<\frac{\pi}{2}\)), the inverse sine function is differentiable on the open interval \((-1,1)\). We are going to find the derivative. Since \(y=\sin^{-1} x\), we have \(\sin y=x\). Then, we obtain \(\cos y\frac{dy}{dx}=1\) using the implicit differentiation. That is, we have
\begin{align*} \frac{dy}{dx} & =\frac{1}{\cos y}=\frac{1}{\cos (\sin^{-1} x)}\\ & =\frac{1}{\sqrt{1-x^{2}}},\end{align*}
which implies
\[\frac{d}{dx}(\sin^{-1} x)=\frac{1}{\sqrt{1-x^{2}}}.\]
We also have the following integral
\[\int \frac{dx}{\sqrt{1-x^{2}}}=\sin^{-1} x+C.\]
Example. Find
\[\frac{d}{dx}[\sin^{-1}(3x^{2})].\]
In this case, the domain is the set of all real numbers \(x\) satisfying \(-1\leq 3x^{2}\leq 1\), or \(|x|\leq 1/\sqrt{3}\). Using the chain rule, we have
\begin{align*} \frac{d}{dx}[\sin^{-1}(3x^{2}) & =\frac{1}{\sqrt{1-(3x^{2})^{2}}}\frac{d}{dx}(3x^{2})\\ & =\frac{6x}{\sqrt{1-9x^{4}}}.\end{align*}
Example. For \(a>0\), prove
\[\int \frac{dx}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}\left (\frac{x}{a}\right )+C.\]
Let \(au=x\). Then \(adu=dx\). Therefore, we have
\begin{align*} \int \frac{dx}{\sqrt{a^{2}-x^{2}}} & =\int \frac{adu}{\sqrt{a^{2}-a^{2}u^{2}}}\\ & =\int \frac{du}{\sqrt{1-u^{2}}}\\ & =\sin^{-1} u+C\\ & =\sin^{-1}\left (\frac{x}{a}\right )+C.\end{align*}
Now, we can evaluate
\[\int_{0}^{\sqrt{3}} \frac{dx}{\sqrt{4-x^{2}}}\]
It follows
\begin{align*} \int_{0}^{\sqrt{3}} \frac{dx}{\sqrt{4-x^{2}}} & =\left [\sin^{-1}\left (\frac{x}{2}\right )\right ]_{0}^{\sqrt{3}}\\ & =\sin^{-1}\left (\frac{\sqrt{3}}{2}\right )-\sin^{-1} 0\\ & =\frac{\pi}{3}.\end{align*}
The inverse tangent function \(y=\tan^{-1} x\) with domain \((-\infty ,\infty)\) and range \((-\frac{\pi}{2},\frac{\pi}{2})\). We also have
\[\tan (\tan^{-1} x)=x\mbox{for all real numbers \(x\)}\]
and
\[\tan^{-1}(\tan x)=x\mbox{for all \(x\in (-\frac{\pi}{2},\frac{\pi}{2})\),}.\]
Reading from the figure, we have
\begin{align*}
\sin (\tan^{-1} x) & =\frac{x}{\sqrt{1+x^{2}}}\\ \cos (\tan^{-1} x) & =\frac{1}{\sqrt{1+x^{2}}}\\
\tan (\tan^{-1} x) & =x \\ \cot (\tan^{-1} x) & =\frac{1}{x}\\
\sec (\tan^{-1} x) & =\sqrt{1+x^{2}}\\ \csc (\tan^{-1} x) & =\frac{\sqrt{1+x^{2}}}{x}.
\end{align*}
Since the derivative of the tangent function,
\[\frac{d}{dx}(\tan x)=\sec^{2} x=\frac{1}{\cos^{2} x},\]
is never \(0\), the inverse tangent function is everywhere differentiable. We are going to find the derivative . Since \(y=\tan^{-1} x\), we have \(\tan y=x\). Therefore, we obtain \(\sec^{2} y\frac{dy}{dx}=1\). That is, we have
\begin{align*} \frac{dy}{dx} & =\frac{1}{\sec^{2} y}=\cos^{2} y\\ & =\cos^{2}(\tan^{-1} x)=\frac{1}{1+x^{2}},\end{align*}
which implies
\[\frac{d}{dx}(\tan^{-1} x)=\frac{1}{1+x^{2}}.\]
We also have the following integral
\[\int \frac{dx}{1+x^{2}}=\tan^{-1} x+C.\]
Example. Calculate
\[\frac{d}{dx}[\tan^{-1}(ax^{2}+bx+c)].\]
Using the chain rule, we have
\begin{align*} \frac{d}{dx}[\tan^{-1}(ax^{2}+bx+c) & =\frac{1}{1+(ax^{2}+bx+c)^{2}}
\frac{d}{dx}(ax^{2}+bx+c)\\ & =\frac{2ax+b}{1+(ax^{2}+bx+c)^{2}}.\end{align*}
Example. For \(a\neq 0\), prove
\[\int \frac{dx}{a^{2}+x^{2}}=\frac{1}{a}\tan^{-1}\left (\frac{x}{a}\right )+C.\]
Let \(au=x\). Then \(adu=dx\). Therefore, we obtain
\begin{align*} \int \frac{dx}{a^{2}+x^{2}} & =\int \frac{adu}{a^{2}+a^{2}u^{2}}\\ & =\frac{1}{a}\int \frac{du}{1+u^{2}}\\ & =\frac{1}{a}\tan^{-1} u+C\\ & =\frac{1}{a}\tan^{-1}\left (\frac{x}{a}\right )+C.\end{align*}
Now we can evaluate
\[\int_{0}^{2} \frac{dx}{4+x^{2}}\]
It follows
\begin{align*} \int_{0}^{2} \frac{dx}{4+x^{2}} & =\left [\frac{1}{2}\tan^{-1}\left (\frac{x}{2}\right )\right ]_{0}^{2}\\ & =\frac{1}{2}\tan^{-1} 1-
\frac{1}{2}\tan^{-1} 0\\ & =\frac{\pi}{8}.\end{align*}
The inverse secant function \(y=\sec^{-1} x\) with domain \((-\infty ,-1]\cup [1,\infty)\) and range \([0,\frac{\pi}{2})\cup (\frac{\pi}{2},\pi ]\). We also have
\[\sec (\sec^{-1} x)=x\mbox{for all \(x\) such that \(|x|\geq 1\),}\]
and
\[\sec^{-1}(\sec x)=x\mbox{for all \(x\in [0,\frac{\pi}{2} )\cup (\frac{\pi}{2},\pi ]\) }.\]
Reading from the figure, for \(x>1\), we have
\begin{align*}
\sin (\sec^{-1} x) & =\frac{\sqrt{x^{2}-1}}{x}\\ \cos (\sec^{-1} x) & =\frac{1}{x}\\
\tan (\sec^{-1} x) & =\sqrt{x^{2}-1}\\ \cot (\sec^{-1} x) & =\frac{1}{\sqrt{x^{2}-1}}\\
\sec (\sec^{-1} x) & =x\\ \csc (\sec^{-1} x) & =\frac{x}{\sqrt{x^{2}-1}}.
\end{align*}
The derivative of the secant function,
\[\frac{d}{dx}(\sec x)=\sec x\tan x,\]
is nonzero on \((0,\frac{\pi}{2})\cup (\frac{\pi}{2},2\pi )\). Therefore, the inverse secant function is differentiable for \(|x|>1\). We are going to find the derivative. Since \(y=\sec^{-1} x\), we have \(\sec y=x\). Then \(\sec y\tan y\frac{dy}{dx}=1\). Therefore, we obtain
\[\frac{dy}{dx}=\frac{1}{\sec y\tan y}.\]
Clearly, we see that \(\sec y=x\) and \(\tan y=\pm\sqrt{x^{2}-1}\). Now if \(x>1\) then \(0<y<\frac{\pi}{2}\) and \(\tan y=\sqrt{x^{2}-1}\), and if \(x<-1\) then \(\frac{\pi}{2}<y<\pi\) and \(\tan y=-\sqrt{x^{2}-1}\). In either case, the product \(\sec y\tan y\) is positive. Therefore, we have
\[\frac{d}{dx}(\sec^{-1} x)=\frac{1}{|x|\sqrt{x^{2}-1}}.\]
Example. Calculate
\[\frac{d}{dx}[\sec^{-1}(2\ln x)].\]
Using the chain rule, we have
\begin{align*} \frac{d}{dx}[\sec^{-1}(2\ln x)] & =\frac{1}{|2\ln x|\sqrt{(2\ln x)^{2}-1}}\frac{d}{dx}(2\ln x)\\ & =\frac{1}{x|\ln x|\sqrt{4(\ln x)^{2}-1}}.\end{align*}
Example. Prove
\[\int \frac{dx}{x\sqrt{x^{2}-1}}=\sec^{-1}|x|+C.\]
We need \(x^{2}-1>0\). Thus \(|x|>1\). For \(x>1\), we have \(\sec^{-1}|x|=\sec^{-1} x\) and
\[\frac{d}{dx}(\sec^{-1} x)=\frac{1}{|x|\sqrt{x^{2}-1}}=\frac{1}{x\sqrt{x^{2}-1}}.\]
For \(x<-1\), we have \(\sec^{-1}|x|=\sec^{-1}(-x)\) and
\begin{align*} \frac{d}{dx}[\sec^{-1}(-x)] & =\frac{1}{|-x|\sqrt{x^{2}-1}}(-1)\\ & =\frac{-1}{-x\sqrt{x^{2}-1}}=\frac{1}{x\sqrt{x^{2}-1}}.\end{align*}
Therefore, it follows that \(F(x)=\sec^{-1}|x|\) is an anti-derivative for \(f(x)=(1/x)\sqrt{x^{2}-1}\). Similarly, we can show that if \(a>0\) is a constant, then (let \(au=x\))
\[\int\frac{dx}{x\sqrt{x^{2}-a^{2}}}=\frac{1}{a}\sec^{-1}\frac{|x|}{a}+C.\]
Now we can evaluate
\begin{align*}
\int_{2\sqrt{2}}^{4} \frac{dx}{x\sqrt{x^{2}-4}} & =\left [\frac{1}{2}\sec^{-1}\frac{|x|}{2}\right ]_{2\sqrt{2}}^{4}\\
& =\frac{1}{2}\sec^{-1}2-\frac{1}{2}\sec^{-1}\sqrt{2}\\
& =\frac{1}{2}(\frac{\pi}{3}-\frac{\pi}{4})=\frac{\pi}{24}.
\end{align*}
There are three other trigonometric inverses given below.
- The inverse cosine \(y=\cos^{-1} x\) is the inverse of \(y=\cos x\) for \(x\in [0,\pi ]\).
- The inverse cotangent \(y=\cot^{-1} x\) is the inverse of \(y=\cot x\) for \(x\in (0,\pi )\).
- The inverse cosecant \(y=\csc^{-1} x\) is the inverse of \(y=\csc x\) for \(x\in [-\frac{\pi}{2},0)\cup (0,\frac{\pi}{2}]\).
The differentiation formulas for these functions are given by
\begin{align*}
&\frac{d}{dx}(\cos^{-1} x)=\frac{-1}{\sqrt{1-x^{2}}}=-\frac{d}{dx}(\sin^{-1} x)\\
&\frac{d}{dx}(\cot^{-1} x)=\frac{-1}{1+x^{2}}=-\frac{d}{dx}(\tan^{-1} x)\\
&\frac{d}{dx}(\csc^{-1} x)=\frac{-1}{|x|\sqrt{x^{2}-1}}=-\frac{d}{dx}(\sec^{-1} x)
\end{align*}
\begin{equation}{\label{f}}\tag{F}\mbox{}\end{equation}
The Hyperbolic Functions.
The hyperbolic sine \(\sinh\), and the hyperbolic cosine \(\cosh\) are the functions defined by
\[\sinh x=\frac{1}{2}(e^{x}-e^{-x})\mbox{ and }\cosh x=\frac{1}{2}(e^{x}+e^{-x}).\]
The hyperbolic sine and cosine functions satisfy identities similar to those satisfied by the sine and cosine, which are give by
\begin{align*}
\cosh^{2} x-\sinh^{2} x & =1\\
\sinh (x+y) & =\sinh x\cosh y+\cosh x\sinh y\\
\cosh (x+y) & =\cosh x\cosh y+\sinh x\sinh y\\
\sinh 2x & =2\sinh x\cosh x\\
\cosh 2x & =\cosh^{2} x+\sinh^{2} x.
\end{align*}
Now, we have
\begin{align*} \frac{d}{dx}(\sinh x) & =\frac{d}{dx}\left [\frac{1}{2}(e^{x}-e^{-x})\right ]\\ & =\frac{1}{2}(e^{x}+e^{-x})=\cosh x\end{align*}
and
\begin{align*} \frac{d}{dx}(\cosh x) & =\frac{d}{dx}\left [\frac{1}{2}(e^{x}+e^{-x})\right ]\\ & =\frac{1}{2}(e^{x}-e^{-x})=\sinh x.\end{align*}
Example. Let
\[A(t)=\frac{1}{2}\cosh t\sinh t-\int_{1}^{\cosh t}\sqrt{x^{2}-1}dx.\]
We wish to show \(A(t)=\frac{t}{2}\) for all \(t\geq 0\). Differentiating \(A(t)\), we have
\[A'(t)=\frac{1}{2}\left [\cosh t\cdot\frac{d}{dt}(\sinh t)+\sinh t\cdot
\frac{d}{dt}(\cosh t)\right ]-\frac{d}{dt}\left (\int_{1}^{\cosh t}\sqrt{x^{2}-1}dx\right ),\]
which implies
\begin{align*}
A'(t) & =\frac{1}{2}(\cosh^{2} t+\sinh^{2} t)-\sqrt{\cosh^{2} t-1}\cdot\frac{d}{dt}(\cosh t)\\
& =\frac{1}{2}(\cosh^{2} t+\sinh^{2} t)-\sinh t\sinh t\\
& =\frac{1}{2}(\cosh^{2} t-\sinh^{2} t)=\frac{1}{2}.
\end{align*}
Therefore, we obtain \(A(t)=\frac{t}{2}+C\). Now
\begin{align*} A(0) & =\frac{1}{2}\sinh 0\cosh 0-\int_{1}^{\cosh 0}\sqrt{x^{2}-1}dx\\ & =\frac{1}{2}\cdot 1\cdot 0-\int_{1}^{1}\sqrt{x^{2}-1}dx=0,\end{align*}
which says \(C=0\). Therefore, we obtain \(A(t)=\frac{t}{2}\). \(\sharp\)
The hyperbolic tangent, hyperbolic cotangent, hyperbolic secant and hyperbolic cosecant are defined by setting
\begin{align*} & \tanh x=\frac{\sinh x}{\cosh x}\\ & \coth x=\frac{\cosh x}{\sinh x}\\ &\mbox{sech} x=\frac{1}{\cosh x}\\ & \mbox{csch} x=\frac{1}{\sinh x}.\end{align*}
Their derivatives are given below.
\begin{align*}
& \frac{d}{dx}(\tanh x)=\mbox{sech}^{2} x\\ & \frac{d}{dx}(\coth x)=-\mbox{csch}^{2} x\\
& \frac{d}{dx}(\mbox{sech }x)=-\mbox{sech }x\tanh x\\ &\frac{d}{dx}(\mbox{csch }x)=-\mbox{csch }x\coth x.
\end{align*}
These formulas are easy to verify. For instance, we have
\begin{align*}
\frac{d}{dx}(\tanh x) & =\frac{d}{dx}\left (\frac{\sinh x}{\cosh x}\right )\\
& =\frac{\cosh x\frac{d}{dx}(\sinh x)-\sinh x\frac{d}{dx}(\cosh x)}{\cosh^{2} x}\\
& =\frac{\cosh^{2} x-\sinh^{2} x}{\cosh^{2} x}\\
& =\frac{1}{\cosh^{2} x}=\mbox{sech}^{2} x.
\end{align*}
Of the six hyperbolic functions, only the hyperbolic cosine and hyperbolic secant are not one-to-one. Thus the hyperbolic sine, hyperbolic tangent, hyperbolic cotangent and hyperbolic cosecant functions all have inverses. If we restrict the domains of the hyperbolic cosine and hyperbolic secant functions to \(x\geq 0\), then the functions will also have inverses. The hyperbolic inverses that are important to us are the inverse hyperbolic sine, the inverse hyperbolic cosine, and the inverse hyperbolic tangent.
Theorem. We have the following properties.
(i) We have \(\sinh^{-1} x=\ln (x+\sqrt{x^{2}+1})\) for \(x\in\mathbb{R}\).
(ii) We have \(\cosh^{-1} x=\ln (x+\sqrt{x^{2}-1})\) for \(x\geq 1\).
(iii) We have
\[\tanh^{-1} x=\frac{1}{2}\ln \left (\frac{1+x}{1-x}\right )\]
for \(-1<x<1\).
Proof. To prove (i), we set \(y=\sinh^{-1} x\), i.e., \(\sinh y=x\), which says
\[\frac{1}{2}(e^{y}-e^{-y})=x, e^{y}-e^{-y}=2x, e^{2y}-2xe^{y}-1=0.\]
We find \(e^{y}=x\pm\sqrt{x^{2}+1}\). Since \(e^{y}>0\), we have \(e^{y}=x+\sqrt{x^{2}+1}\). Taking the natural log of both sides, we obtain \(y=\ln (x+\sqrt{x^{2}+1})\). \(\blacksquare\)
Theorem. We have the following properties.
(i) We have
\[\mbox{csch}^{-1}x=\sinh^{-1}\frac{1}{x}=\ln (\frac{1}{x}+\frac{\sqrt{x^{2}+1}}{|x|})\]
for \(x\neq 0\).
(ii) We have
\[\mbox{sech}^{-1}x=\cosh^{-1}\frac{1}{x}=\ln (\frac{1}{x}+\frac{\sqrt{x^{2}-1}}{x})\]
for \(0<x\leq 1\).
(iii) We have
\[\coth^{-1}x=\tanh^{-1}\frac{1}{x}=\frac{1}{2}\ln \left (\frac{x+1}{x-1}\right )\]
for \(x<-1\) or \(x>1\). \(\sharp\)
We can also prove
\begin{align*}
& \frac{d}{dx}(\sinh^{-1} x)=\frac{1}{\sqrt{x^{2}+1}}\mbox{ for \(x\in\mathbb{R}\)}\\
& \frac{d}{dx}(\cosh^{-1} x)=\frac{1}{\sqrt{x^{2}-1}}\mbox{ for \(x>1\)}\\
& \frac{d}{dx}(\tanh^{-1} x)=\frac{1}{1-x^{2}}\mbox{ for \(-1<x<1\)}\\
& \frac{d}{dx}(\coth^{-1} x)=\frac{1}{1-x^{2}}\mbox{ for \(|x|>1\)}\\
& \frac{d}{dx}(\mbox{sech}^{-1} x)=\frac{-1}{x\sqrt{1-x^{2}}}\mbox{ for \(0<x<1\)}\\
& \frac{d}{dx}(\mbox{csch}^{-1} x)=\frac{-1}{|x|\sqrt{1+x^{2}}}\mbox{ for \(x\neq 0\)}.
\end{align*}
For example, we have
\begin{align*} \frac{d}{dx}(\sinh^{-1} x) & =\frac{1+x/\sqrt{x^{2}+1}}{x+\sqrt{x^{2}+1}}\\ & =\frac{1}{\sqrt{x^{2}+1}}.\end{align*}
On the other hand, we can use the implicit differentiation to obtain \({\displaystyle \frac{d}{dx}(\mbox{csch}^{-1} x)}\) without knowing the exact formula of \(\mbox{csch}^{-1} x\). Let \(y=\mbox{csch}^{-1} x\). Then, we have \(\mbox{csch }y=x\) and \(\sinh y=\frac{1}{x}\). Using the implicit differentiation, we obtain \(\cosh y\cdot\frac{dy}{dx}=-\frac{1}{x^{2}}\) and
\begin{align*} \frac{dy}{dx} & =\frac{-1}{x^{2}\cosh y}\\ & =\frac{-1}{x^{2}\sqrt{1+1/x^{2}}}
\mbox{(since \(\cosh^{2} y-\sinh^{2} y=1\))}\\ & =\frac{-1}{|x|\sqrt{1+x^{2}}}.\end{align*}
Example. Find
\[\int\frac{dx}{\sqrt{a^{2}+x^{2}}}.\]
Let \(x=a\sinh u\). Then \(dx=a\cosh udu\).
\begin{align*} \int\frac{dx}{\sqrt{a^{2}+x^{2}}} & =\int\frac{a\cosh u}{a^{2}+a^{2}\sinh^{2}u}=\int du\\ & =\sinh^{-1}\left (\frac{x}{a}\right )+C\\ & =\ln\left (\frac{x}{a}+\sqrt{\frac{x^{2}}{a^{2}}+1}\right )+C.\end{align*}
Example. Prove that \(\tan^{-1}\frac{x+1}{x-1}+\tan^{-1}x\) is a constant \(C\), and find \(C\). Let
\[f(x)=\tan^{-1}\frac{x+1}{x-1}+\tan^{-1}x\]
for \(x\neq 1\). Then, we have
\begin{align*} f'(x) & =\frac{\frac{(x-1)-(x+1)}{(x-1)^{2}}}{1+\left (\frac{x+1}{x-1}
\right )^{2}}+\frac{1}{1+x^{2}}\\ & =\frac{\frac{-2}{(x-1)^{2}}}{\frac
{2(x^{2}+1)}{(x-1)^{2}}}+\frac{1}{1+x^{2}}\\ & =0,\end{align*}
which says \(f(x)=C\) a constant. Therefore, we obtain
\[C=f(0)=\tan^{-1}(-1)+\tan^{-1}(0)=-\tan^{-1}=-\frac{\pi}{4}.\]
Example. Find the extreme values of \(f(x)=\sin x+\cos x\). Let \(f'(x)=\cos x-\sin x=0\). Then, we have \(x=\frac{\pi}{4}+2k\pi\) or \(x=\frac{\pi}{4}+(2k+1)\pi\). Since \(f”(x)=-\sin x-\cos x\), we have \(f”(\frac{\pi}{4}+2k\pi)=-\sqrt{2}<0\) and \(f”(\frac{\pi}{4}+(2k+1)\pi)=\sqrt{2}>0\), which says that \(f\) has maximum \(\sqrt{2}\) at \(x=\frac{\pi}{4}+2k\pi\) and minimum \(-\sqrt{2}\) at \(x=\frac{\pi}{4}+(2k+1)\pi\). \(\sharp\)
Example. Find the derivative of \(f(x)=x^{a^{a}}+a^{x^{a}}+a^{a^{x}}\).
\begin{align*}
f'(x) & =\frac{d}{dx}\left (e^{\ln x^{a^{a}}}+e^{\ln a^{x^{a}}}+e^{\ln a^{a^{x}}}\right )\\
& =x^{a^{a}}\cdot a^{a}\cdot\frac{1}{x}+a^{x^{a}}\cdot\ln a\cdot a\cdotx^{a-1}+a^{a^{x}}\cdot\ln a\cdot\ln a\cdot a^{x}\\
& =a^{a}\cdot x^{a^{a}-1}+(a\ln a)x^{a-1}\cdot a^{x^{a}}+a^{x}\cdot
(\ln a)^{2}\cdot a^{a^{x}}
\end{align*}
