The function $f(t)=1/t$ is continuous on $(0, \infty ) $. By the fundamental theorem of calculus, $f$ has an antiderivative on on the interval with end points $x$ and $1$ whenever $x>0 $. This observation allows us to make the following definition.

Definition 9.2.1 The natural logarithm $\ln(x)$ is an antiderivative of $1/x$, given by $$\ln x = \int _1 ^x {1\over t}\,dt.$$

Figure 9.2.1 gives a geometric interpretation of $\ln$. Note that when $x< 1$, $\ln x$ is negative.

Figure 9.2.1. $\ln(x)$ is an area.

Some properties of this function $\ln x$ are now easy to see.

Theorem 9.2.2 Suppose that $x,y >0 $ and $q\in \Q$.

    a. $\ds{d\over dx} \ln x = {1\over x}.$

    b. $\ln(1) = 0$.

    c. $\ln (xy) = \ln x+ \ln y $

    d. $\ln(x/y) = \ln x - \ln y $

    e. $\ln x^q = q\ln x $.

Proof.
Part (a) is simply the Fundamental Theorem of Calculus (7.2.2). Part (b) follows directly from the definition, since $$\ln(1)=\int_1^1 {1\over t}\,dt.$$

Part (c) is a bit more involved; start with: $$\ln(xy)=\int_1^{xy} {1\over t}\,dt= \int_1^{x} {1\over t}\,dt+\int_x^{xy} {1\over t}\,dt =\ln(x)+\int_x^{xy} {1\over t}\,dt.$$

In the remaining integral, use the substitution $u=t/x$ to get $$\int_x^{xy} {1\over t}\,dt=\int_1^{y} {1\over xu}x\,du =\int_1^{y} {1\over u}\,du=\ln(y).$$

Parts (d) and (e) are left as exercises.

$\square$

Part (e) is in fact true for any real number $q$ (not just rationals) but one of the points of our approach here is to give a rigorous definition of real powers which so far we have not done.

We now turn to the task of sketching the graph of $\ln x$.

Theorem 9.2.3 $\ln x $ is increasing and its graph is concave down everywhere.

Proof.
Since ${d\over dx}\ln x =1/x$ is positive for $x>0$, the Mean Value Theorem (6.5.2) implies that $\ln x$ is increasing. The second derivative of $\ln x$ is then $-1/x^2$ which is negative, so the graph is concave down.

$\square$

Notice that this theorem implies that $\ln x$ is injective.

Theorem 9.2.4 $\ds\lim_{x\to\infty} \ln x =\infty$

Proof.
Note that $\ln 2 >0 $ and for $n\in\N$, $\ln 2^n =n\ln 2$. Since $\ln x$ is increasing, when $x> 2^n$, $\ln(x)>n\ln2$. Since $\ds\lim_{n\to\infty}n\ln2=\infty$, also $\ds\lim_{x\to\infty} \ln x =\infty$.

$\square$

Corollary 9.2.5 $\lim_{x\to 0^+} \ln x =-\infty$

Proof.
If $0< x< 1 $, then $(1/x)>1$ and $\lim_{x\to 0^+} (1/x) = \infty $. Let $y=1/x$; then $\lim _{x\to 0^+} \ln x= \lim_{y\to \infty}\ln(1/y) =\lim_{y\to \infty}\ln(1)-\ln(y)=\lim_{y\to \infty}-\ln(y) =-\infty$.

$\square$

Thus, the domain of $\ln $ is $(0, \infty ) $ and the range is $\R$; $\ln(x)$ is shown in figure 9.2.2.

Figure 9.2.2. The graph of $\ln(x)$.

By the intermediate value theorem (2.5.6) there is a number $e$ such that $\ln e = 1 $. The number $e$ is also known as Napier's constant.

It turns out that $e$ is not rational. In fact, $e$ is not the root of a polynomial with rational coefficients which means that $e$ is a transcendental number. We will not prove these assertions here. The value of $e$ is approximately $2.718$.

Example 9.2.6 Let $f(x) =\ln (x^5 + 7x +12) $. Compute $f'(x) $.

Using the chain rule: $\ds f'(x) = {1\over x^5 +7x+ 12}(5x+7)$.

Example 9.2.7 Let $f(x) =\ln (-x) $ for $x< 0 $. Compute $f'(x) $. $$f'(x) ={1\over -x}(-1) ={1\over x}.$$ So the derivatives of $\ln(x)$ and $\ln(-x)$ are the same. Thus, you will often see $\ds\int{1\over x}\,dx=\ln|x|+C$ as the general antiderivative of $1/x$.

Example 9.2.8 Compute $\ds\int \tan x \,dx $.

Use $u=\cos x$: $$ \int\tan x\,dx=\int{\sin x\over\cos x}\,dx=\int -{1\over u}\,du =-\ln |u|+C = -\ln|\cos x| +C.$$ Using one of the properties of the logarithm, we could go further: $$-\ln|\cos x| +C=\ln|(\cos x)^{-1}|+C=\ln|\sec x|+C.$$

Example 9.2.9 Let $\ds f(x) = {x^{18} (x+2)^{6/7}\over (x^{10} + 4x^2 +1 )^6 }$. Compute $f'(x) $

Computing the derivative directly is straightforward but irritating. We therefore take an indirect approach. Note that $f(x) >0 $ for every $x$. Let $g(x) = \ln f(x) $. Then $g'(x) = f'(x)/f(x)$ and so $f'(x) =f(x)g'(x)$. Now $$\eqalign{ g(x) &=\ln\left({x^{18} (x+2)^{6/7}\over (x^{10} + 4x^2 +1 )^6 }\right)\cr &= 18\ln x +{6\over 7} \ln (x+2) - 6\ln (x^{10} +4x^2 +1)\cr}$$ Hence, $$g'(x) = {18\over x} +{6\over 7(x+2)} -{6(10x^9 +8x )\over x^{10} +4x^2 +1 }.$$ Therefore, $$f'(x) = {x^{18} (x+2)^{6/7}\over (x^{10} + 4x^2 +1 )^6 } \left({18\over x} +{6\over 7(x+2)} -{6(10x^9 +8x )\over x^{10} +4x^2 +1 }\right).$$

Exercises 9.2

Ex 9.2.1 Prove parts (d) and (e) of theorem 9.2.2.

In subsequent exercises, it is understood that the arguments in any logarithms are positive unless otherwise stated.

Ex 9.2.2 Expand $\ln ((x+45)^7 (x-2))$.

Ex 9.2.3 Expand $\ds\ln {x^3\over 3x-5 +(7/x)} $.

Ex 9.2.4 Sketch the graph of $y= \ln (x-7)^3 + 14 $.

Ex 9.2.5 Sketch the graph of $y=\ln |x| $ for $x\neq 0$.

Ex 9.2.6 Write $\ln 3x + 17 \ln (x-2) - 2\ln (x^2 + 4x + 1) $ as a single logarithm.

Ex 9.2.7 Differentiate $f(x) =x\ln x $.

Ex 9.2.8 Differentiate $f(x) =\ln (\ln (3x) )$.

Ex 9.2.9 Sketch the graph of $\ln (x^2 - 2x) $.

Ex 9.2.10 Differentiate $\ds f(x) ={1+\ln (3x^2 )\over 1+ \ln (4x)} $.

Ex 9.2.11 Differentiate $f(x) =\ln |\sec x +\tan x | $.

Ex 9.2.12 Find the second derivative of $f(x) =\sqrt{\ln (x^4 -2) } $.

Ex 9.2.13 Find the equation of the tangent line to $f(x) =\ln x $ at $x=a $.

Ex 9.2.14 Differentiate $\ds f(x) ={x^8 (x-23)^{1/2}\over 27 x^6(4x-6)^8 }$.

Ex 9.2.15 If $f(x) = \ln (x^3 + 2 ) $ compute $f'(e^{1/3} ) $.

Ex 9.2.16 Compute $\ds\int _1 ^e {1\over x}\,dx$.

Ex 9.2.17 Compute the derivative with respect to $x$ of $\ds\int _1 ^{\ln x} \ln t\, dt $. (Assume that $x>1$.)

Ex 9.2.18 Compute $\ds\int _0 ^{\pi/6} \tan (2x)\, dx$.

Ex 9.2.19 Compute $\ds\int {\ln x\over x}\,dx$.

Ex 9.2.20 Compute $\ds\int {\sin (2x)\over 1+\cos ^2 x }\,dx$.

Ex 9.2.21 Find the volume of the solid obtained by rotating the region under $y=1/\sqrt{x}$ from $1$ to $e$ about the $x$-axis.