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5.5Logarithmic, Exponential and Hyperbolic Functions

With the Fundamental Theorems of Calculus it is possible to rigorously develop the logarithmic, exponential and hyperbolic functions.

5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS231

Definition 5.5.1 For each x > 0 we define the natural logarithm of x, denoted ln x, by the equation

The natural logarithm is an increasing, continuous and di erentiable function on (0, ∞).

(ii)If a > 0 and b > 0, then ln(ab) = ln(a) + ln(b).

(iii)If a > 0 and b > 0, then ln(a/b) = ln(a) + ln(b).

(iv)If a > 0 and n is a natural number, then ln(an) = n ln a.

(v)The range of ln x is (−∞, ∞).

(vi)ln x is one-to-one and has a unique inverse, denoted ex.

Proof.

(i)Since 1/t is continuous on (0, ∞), (i) follows from the Fundamental Theorem of Calculus, Second Form.

(ii)Suppose that a > 0 and b > 0. Then

= ln a + ln b.

5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS233

Hence, ln 4 > 1. ln(4n) = n ln 4 > n and ln 4−n = −n ln 4 < −n. By the intermediate value theorem, every interval (−n, n) is contained in the range of ln x. Therefore, the range of ln x is (−∞, ∞), since the derivative of ln x is always positive, ln x is increasing and hence one-to- one. The inverse of ln x exists.

(vi)Let e denote the number such that ln(e) = 1. Then we define y = ex if and only if x = ln(y) for x (−∞, ∞), y > 0.

This completes the proof.

Definition 5.5.2 If x is any real number, we define y = ex if and only if x = ln y.

Theorem 5.5.2 (Exponential Function) The function y = ex has the following properties:

(i)e0 = 1, ln(ex) = x for every real x and dxd (ex) = ex.

(ii)ea · eb = ea+b for all real numbers a and b.

(iii)ea = ea−b for all real numbers a and b. eb

(iv)(ea)n = ena for all real numbers a and natural numbers n.

Proof.

(i)Since ln(1) = 0, e0 = 1. By definition y = ex if and only if x = ln(y) = ln(ex). Suppose y = ex. Then x = ln y. By implicit di erentiation, we get

1 = y1 dxdy , dxdy = y = ex.

Therefore,

dxd (ex) = ex.

(iii)ea = ea−b ↔

eb

ea

ln eb = ln(ea−b) ↔

ln(ea) − ln(eb) = a − b ↔

a − b = a − b.

It follows that for all real numbers a and b,

ea = ea−b. eb

(iv)(ea)n = ena ↔

ln((ea)n) = ln(ena) ↔

n ln(ea) = na ↔

na = na.

Therefore, for all real numbers a and natural numbers n, we have

(ea)n = ena.

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Definition 5.5.3 Suppose b > 0 and b 6= 1. Then we define the following:

(i)ln(bx) = x ln b, for all real numbers x.

(ii)dxd (bx) = bx ln b, for all real numbers x.

(iii)bx1 · bx2 = bx1+x2 , for all real numbers x1 and x2.

(iv)bx1 = bx1−x2 , for all real numbers x1 and x2. bx2

(v)(bx1 )x2 = bx1x2 , for all real numbers x1 and x2.

=bx ln b.

(iii)bx1 · bx2 = ex1 ln b · ex2 ln b

=e(x1 ln b+x2 ln b)

=e(x1+x2) ln b

=b(x1+x2)

=ex1 ln b−x2 ln b

=e(x1−x2) ln b

=b(x1−x2).

(v)By Definition 5.5.3 (i), we get

=ex2 ln(ex1 ln b)

=ex2·x1 ln b

=e(x1x2) ln b

=bx1x2 .

where D is some constant. This completes the proof.

Theorem 5.5.4 If u(x) > 0 for all x, and u(x) and v(x) are di erentiable functions, then we define

y = (u(x))v(x) = ev(x) ln(u(x)).

5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS237

Proof. This theorem follows by the chain rule and the product rule as follows

Theorem 5.5.5 The following di erentiation formulas for the hyperbolic functions are valid.

Proof. We use the definitions and properties of hyperbolic functions given in Chapter 1 and the di erentiation formulas of this chapter.

11

=−sinh2 x · cosh2 x = −sinh2 x

=−csch2x.

(v)dxd (sech x) = dxd (cosh x)−1 = −1(cosh x)−2 · sinh x

=− sech x tanh x.

(vi)dxd (csch x) = dxd (sinh x)−1 = −1(sinh x)−2 · cosh x

=− coth x csch x.cosh2 x

This completes the proof.

Theorem 5.5.6 The following integration formulas are valid:

Proof. Each formula can be easily verified by di erentiating the right-hand side to get the integrands on the left-hand side. This proof is left as an exercise.

Theorem 5.5.7 The following di erentiation and integration formulas are valid:

5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS239

−

The proof is left as an exercise.

Exercises 5.5

1.Prove Theorem 5.5.6.

2.Prove Theorem 5.5.7.

3.Show that sinh mx and cosh mx are linearly independent if m 6= 0. (Hint:

7.Determine the relation between c1 and c2 in problem 5 with A and B in problem 6.

8.Prove the basic identities for hyperbolic functions:

(i)sinh(x + y) = sinh x cosh y + cosh x sinh y.

(ii)sinh(x − y) = sinh x cosh y − cosh x sinh y.

(iii)cosh(x + y) = cosh x cosh y + sinh x sinh y.

(iv)cosh(x − y) = cosh x cosh y − sinh x sinh y.

(v)sinh 2x = 2 sinh x cosh x.

(vi)cosh2 x + sinh2 x = 2 cosh2 x − 1 = 1 + 2 sinh2 x = cosh 2x.

(vii) cosh2 x − sinh2 x = 1, 1 − tanh2 x = sech2x, coth2 x − 1 = csch2x.

√

(iii)x2 − a2, x = a cosh t.

10.Compute y0 in each of the following:

12. Compute y0 in each of the following:

5.5. LOGARITHMIC, EXPONENTIAL AND HYPERBOLIC FUNCTIONS241

15.Logarithmic Di erentiation is a process of computing derivatives by first taking logarithms and then using implicit di erentiation. Find y0 in each of the following, using logarithmic di erentiation.

In problems 16–30, compute f0(x) each f(x).