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Chapter 3: Transcendental Functions

Winter 2016

Department of MathematicsHong Kong Baptist University

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Except for the power functions, the other basic elementaryfunctions are also called the transcendental functions.Transcendental functions cannot be constructed from realnumbers and a single real variable x by using finitely manyarithmetic operations (addition, subtraction, multiplication,division, and fractional powers).

We have studied the trigonometric functions in Section 2.5.This chapter is devoted to the study of other transcendentalfunctions, including

(i) exponential functions,

(ii) logarithmic functions,

(iii) inverse trigonometric functions (not required).

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Recall that the elementary functions are built from the basicelementary functions and constants through composition(f ◦ g) and combinations using operations (+, −, × and /).

By the rules of differentiation, we conclude that the derivativeof any elementary function can be expressed in terms of thebasic elementary functions and their derivatives.

In this chapter, we will study the differentiation rules for (i)exponential functions and (ii) logarithmic functions,respectively.

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§3.2 Exponential and Logarithmic Functions

An exponential function is a function of the form f (x) = ax ,where the base a is a positive constant and the exponent x is thevariable.

Lemma

Let a > 0 and b > 0. Then for any pair of real numbers x and y ,we have

(i) a0 = 1 (ii) ax+y = axay

(iii) a−x =1

ax(iv) ax−y =

ax

ay

(v) (ax)y = axy (vi) (ab)x = axbx

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If a = 1, then ax = 1x = 1 for every x , i.e., it is a constantfunction.

If a > 1, then ax is an increasing function of x with

limx→−∞

ax = 0 and limx→∞

ax =∞.

If 0 < a < 1, then ax is a decreasing function of x with

limx→−∞

ax =∞ and limx→∞

ax = 0.

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Definition

If a > 0 and a 6= 1, then function loga(x), called the logarithm ofx to the base a, is the inverse of the one-to-one function ax :

y = loga(x) ⇐⇒ x = ay .

loga(x) has domain (0,∞) and range (−∞,∞).

Remark: Since ax and loga(x) are inverse functions, the followingcancellation identities hold:

loga(ax) = x for all real x ,

aloga(x) = x for all x > 0.

In particular, loga(a) = 1 when x = 1 and loga(1) = 0 when x = 0.

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Laws of Logarithms

Lemma

If x > 0, y > 0, a > 0, b > 0, a 6= 1, and b 6= 1, then

(i) loga(1) = 0 (ii) loga(xy) = loga(x) + loga(y)

(iii) loga

(1

x

)= − loga(x) (iv) loga

(x

y

)= loga(x)− loga(y)

(v) loga(xy ) = y loga(x) (vi) loga(x) =logb(x)

logb(a)

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If a > 1, then loga(x) is an increasing function of x with

limx→0+

loga(x) = −∞ and limx→∞

loga(x) =∞.

If 0 < a < 1, then loga(x) is a decreasing function of x with

limx→0+

loga(x) =∞ and limx→∞

loga(x) = −∞.

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Example 1:

(a) Simplify 3log9(4); (b) Solve the equation 3x−1 = 2x .

Solution:

(a) Note that

log9(4) =log3(4)

log3(9)=

1

2log3(4).

So 3log9(4) = 3log3(4)·12 = 41/2 = 2.

(b) Let a = 2 and take logarithms of both sides,

(x − 1) log2(3) = x .

This leads to

x =log2(3)

log2(3)− 1≈ 2.71,

where log2(3) ≈ 1.585.

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Example 1:

(a) Simplify 3log9(4); (b) Solve the equation 3x−1 = 2x .

Solution:

(a) Note that

log9(4) =log3(4)

log3(9)=

1

2log3(4).

So 3log9(4) = 3log3(4)·12 = 41/2 = 2.

(b) Let a = 2 and take logarithms of both sides,

(x − 1) log2(3) = x .

This leads to

x =log2(3)

log2(3)− 1≈ 2.71,

where log2(3) ≈ 1.585.

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§3.3 The Natural Logarithm and Exponential

This section considers the base a = e, where the e constant iscalled the Euler’s number. Specifically, we write

loge(x) = ln(x) and ex = exp(x).

The e constant was defined by Bernoulli,

e = limn→∞

(1 +

1

n

)n

= limr→0

(1 + r)1/r .

The number e is irrational (in fact, transcendental) with valuee ≈ 2.71828. In §4.10 we will also learn that

e = 1 +1

1!+

1

2!+

1

3!+

1

4!+ · · ·

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Theorem

(1) For any x > 0,d

dxln(x) =

1

x.

(2) For any x ∈ (−∞,∞),

d

dxex = ex .

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Proof: (1) By definition, we have

d

dxln(x) = lim

h→0

ln(x + h)− ln(x)

h

= limh→0

1

hln

(1 +

h

x

)=

1

x· limr→0

ln(1 + r)

r︸ ︷︷ ︸=:L

,

where we let h = rx . But taking the natural logarithm of thedefinition of e gives

1 = ln(e) = ln(

limr→0

(1 + r)1/r)

= limr→0

1

rln(1 + r) = L.

Thus, we have ddx ln(x) = 1

x , as required.15 / 32

(2) The derivative of y = ex can be calculated by implicitdifferentiation. Note that

y = ex =⇒ x = ln(y).

By the Chain rule and take derivatives of both sides on x ,

1 =1

y

dy

dx.

This leads tody

dx= y = ex .

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Example 2:

Find the derivatives of (a) ln(cos2(x)), and (b) ln(x +√x2 + 1).

Solution:

(a) By the Chain rule,

d

dxln(cos2(x)) =

1

cos2(x)· ddx

cos2(x) = −2 tan(x).

(b) By the Chain rule,

d

dxln(x +

√x2 + 1) =

1

x +√x2 + 1

· ddx

(x +√

x2 + 1)

=1

x +√x2 + 1

(1 +

2x

2√x2 + 1

)=

1√x2 + 1

.

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Example 2:

Find the derivatives of (a) ln(cos2(x)), and (b) ln(x +√x2 + 1).

Solution:

(a) By the Chain rule,

d

dxln(cos2(x)) =

1

cos2(x)· ddx

cos2(x) = −2 tan(x).

(b) By the Chain rule,

d

dxln(x +

√x2 + 1) =

1

x +√x2 + 1

· ddx

(x +√x2 + 1)

=1

x +√x2 + 1

(1 +

2x

2√x2 + 1

)=

1√x2 + 1

.

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Example 3:

Find the derivatives of (a) ex2−3x , and (b)

√1 + e2x .

Solution:

(a) By the Chain rule,

d

dxex

2−3x = ex2−3x · d

dx(x2 − 3x) = (2x − 3)ex

2−3x .

(b) By the Chain rule,

d

dx

√1 + e2x =

1

2√

1 + e2x· ddx

(1 + e2x) =e2x√

1 + e2x.

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Example 3:

Find the derivatives of (a) ex2−3x , and (b)

√1 + e2x .

Solution:

(a) By the Chain rule,

d

dxex

2−3x = ex2−3x · d

dx(x2 − 3x) = (2x − 3)ex

2−3x .

(b) By the Chain rule,

d

dx

√1 + e2x =

1

2√

1 + e2x· ddx

(1 + e2x) =e2x√

1 + e2x.

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General Exponentials and Logarithms

Theorem

(1) For any x ∈ (−∞,∞),

d

dxax =

d

dxex ln(a) = ex ln(a) ln(a) = ax ln(a).

(2) For any x > 0,

d

dxloga(x) =

1

x ln(a).

Remark: The derivative of loga(x) can be derived (i) by implicit

differentiation on x = ay , or (ii) by noting that loga(x) =ln(x)

ln(a).

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Functions Derivatives

y = ex y ′ = ex

y = ax y ′ = ax ln(a)

y = ln(x) y ′ =1

x

y = loga(x) y ′ =1

x ln(a)

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Suppose we want to differentiate a function of the form

y = (f (x))g(x) (for f (x) > 0).

For instance, when the function is y = xx .

Since the variable appears in both the base and the exponent,neither the general power rule nor the exponential rule can bedirectly applied.

One method for finding the derivative of such a function is toexpress it in the form

y = eg(x) ln(f (x))

and then differentiate, using the Product rule to handle theexponent.

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Logarithmic Differentiation

Another method to obtain the derivative of such a function isto take the natural logarithms of both sides of the equationy = (f (x))g(x) and then to differentiate it implicitly:

ln(y) = g(x) ln f (x)

1

y

dy

dx= g ′(x) ln f (x) +

g(x)

f (x)f ′(x)

dy

dx= y

(g ′(x) ln f (x) +

g(x)

f (x)f ′(x)

)= (f (x))g(x)

(g ′(x) ln f (x) +

g(x)

f (x)f ′(x)

).

This technique is called logarithmic differentiation.

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Example 4: Find the derivative of y = xx .

Solution 1: Note that y = ex ln(x). By the Chain rule and theProduct rule,

dy

dx=

d

dxex ln(x) = ex ln(x) · d

dx(x ln(x)) = xx · (ln(x) + 1).

Solution 2: Let f (x) = x and g(x) = x . By logarithmicdifferentiation,

dy

dx= (f (x))g(x)

(g ′(x) ln f (x) +

g(x)

f (x)f ′(x)

)= xx · (ln(x) + 1).

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§3.5 Inverse Trigonometric Functions

The inverse trigonometric functions are inverse functions of sin,cos, tan, etc. For example, the arcsine function satisfies

y = arcsin(x) =⇒ x = sin(y).

Other inverse trigonometric functions are define similarly:

y = arccos(x) =⇒ x = cos(y),

y = arctan(x) =⇒ x = tan(y).

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y = arcsin(x) =⇒ x = sin(y).

Since −1 ≤ sin(y) ≤ 1 for all y , the arcsine function is onlydefined for x ∈ [−1, 1].

Since the equation x = sin(y) has many solutions for a givenx , we restrict the range of arcsine to the interval [−π/2, π/2].

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Lemma (Cancellation Identity)

For x ∈ [−π/2, π/2] and y ∈ [−1, 1], we have

arcsin(sin(x)) = x , sin(arcsin(y)) = y .

Examples:

(a) arcsin(12) = π/6;

(b) arcsin(2) is not defined;

(c) sin(arcsin(12)) = 12 ;

(d) arcsin(sin(7π6 )) = −π6 .

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Theorem (Derivative of arcsine)

For −1 < x < 1, we have

d

dxarcsin(x) =

1√1− x2

.

Proof: Let y = arcsin(x) =⇒ sin(y) = x , −π/2 < y < π/2.Then implicit differentiation gives

cos(y)dy

dx= 1 =⇒ dy

dx=

1

cos(y).

But cos(y) is positive for −π/2 < y < π/2, so that

cos2(y) = 1− sin2(y) = 1− x2 =⇒ cos(y) =√

1− x2.

Thus,dy

dx=

1√1− x2

.

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Example 5: Find the derivative of arcsin( xa ) and hence evaluate∫dx√

a2 − x2, where a > 0.

Solution: By the chain rule,

d

dxarcsin

(xa

)=

1√1− (x/a)2

· 1

a=

1√a2 − x2

.

Hence, ∫dx√

a2 − x2= arcsin

(xa

)+ C .

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The arccosine function is defined forx ∈ [−1, 1], with the range [0, π].

Theorem (Derivatives of arccos)

For −1 < x < 1,

d

dxarccos(x) = − 1√

1− x2.

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The arctangent function is defined for all x ∈ R, with range[−π/2, π/2].

Theorem (Derivatives of arctan)

d

dxarctan(x) =

1

1 + x2.

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