Chapter 3 Diﬀerentiationmathstat.sci.tu.ac.th/.../MA111-119/stnote111-119-ch3.pdf · 2019. 8. 22. · Chapter 3 Diﬀerentiation 3.1 Tangent Lines and Rates of Change 3.1.1 Tangent

Chapter 3

Differentiation

3.1 Tangent Lines and Rates of Change

3.1.1 Tangent Lines

If a curve C has equation y = f(x) and we want to find the tangent to C at the pointP(

a, f(a))

, then we consider the nearby point Q(

x, f(x))

, where x 6= a, and computethe slope of the secant line PQ:

mPQ =f(x)− f(a)

x− a

y

x

P (a, f(a))

Q(x, f(x))

0 a x

x− a

f(x)− f(a)

Secant Line PQ

y

x0

bb

b

a x

Q

L

P

Q approach P

Then we let Q approach P along the curve C by letting x approach a. If mPQ

approaches a number m, then we define the tangent L to be the line through P withslope m.

The tangent line to the curve y = f(x) at the point P(

a, f(a))

is the linethrough P with slope

m = limx→a

f(x)− f(a)

x− a

provided that this limit exists.

Definition 3.1.

33

MA111: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 34

Find an equation of the tangent line to y = x3 − x at the point P (2, 6).

Example 3.1.

Solution

There is another expression for the slope of a tangent line that is sometimes easierto use. Let h = x− a. Then x = a+ h, so the slope of the secant line PQ is

mPQ =f(a+ h)− f(a)

h

Hence the expression for the slope of the tangent line in Definition 3.1 becomes

m = limh→0

f(a+ h)− f(a)

h

Find an equation of the tangent line to f(x) =√5x+ 1 at the point (3, 4).

Example 3.2.

Solution


3.1.2 Rate of Change

Velocities

Suppose an object moves along a straight line according to an equation of motion s =f(t), where s is the displacement of the object from the origin at time t. The functionf that describes the motion is called the position function of the object. In thetime interval from t = a to t = a + h the change in position is f(a + h) − f(a)

(

seeFigure 3.1 (a)

)

. The average velocity over this time interval is

average velocity =displacement

time=

f(a+ h)− f(a)

h

which is the same as the slope of the secant line PQ in Figure 3.1 (b)

b

f(a+ h)− f(a)

position at

time t = a

position at

time t = a+ h

0s

f(a)

f(a+ h)

(a)

h

a a+ h

P (a, f(a))

Q(

a+ h, f(a+ h))

s

t

s = f(t)

0

(b)

Figure 3.1:

Now suppose we compute the average velocities over shorter and shorter time intervals[a, a+h]. In other words, we let h approach 0. We define the velocity of instantaneousvelocity v(a) at time t = a to be the limit of these average velocities:

v(a) = limh→0

f(a+ h)− f(a)

h

This mean that the velocity at time t = a is equal to the slope of the tangent line at P .

Exercises 3.1

1. A curve has equation y = f(x).

(a) Write an equation for the slope of the secant line through the points P(

3, f(3))

and Q(

x, f(x))

.

(b) Write an equation for the slope of the tangent line at P .

2. Suppose an object moves with position function s = f(t).

(a) Write an equation for the average velocity of the object in the time intervalfrom t = a to t = a+ h.

(b) Write an equation for the instantaneous velocity at time t = a.


3. Find an equation of the tangent line to the curve at the given point.

(a) y = x2 − 2, (1,−1) (b) y = x2 − 3x, (−2, 10)

(c) y = 1− 2x− 3x2, (−2,−7) (c) y =1

x2, (−2, 1

4)

(e) y =2

x+ 1, (1, 1) (f) y =

√x+ 3, (−2, 1)

4. (a) Find the slope of the tangent to the curve y =2

x+ 3at the point where

x = a.(b) Find the slopes of the tangent lines at the points whose x-coordinates are

(i) −1 (ii) 0 and (iii) 1.

5. (a) Find the slope of the tangent to the curve y = x3 − 4x+ 1 at the point wherex = a.

(b) Find equations of the tangent lines at the point (1,−2) and (2, 1).

6. The following functions represent the position of an object at time t seconds. Findthe average velocity between (i) t = 0 and t = 2 (ii) t = 1 and t = 2 (iii) t = 1.9and t = 2 and (iv) t = 1.99 and t = 2 and (v) estimate the instantaneous velocityat t = 2.

(a) s = f(t) = 16t2 + 10

(b) s = f(t) =√t2 + 8t

7. Use the position function f(t) to find the velocity at time t = a.

(a) f(t) = −16t2 + 5, a = 1

(b) f(t) =√t + 16 , a = 0

8. The displacement (in meters) of the particle moving in a straight line is given bythe equation of motion s = f(t) = 4t3 + 6t + 2, where t is measured in seconds.Find the velocity of the particle at times t = a, t = 1, t = 2, and t = 3.

Answer to Exercise 3.1

1. (a)f(x)− f(3)

x− 3(b) lim

x→3

f(x)− f(3)

x− 3

2. (a)f(a+ h)− f(a)

h(b) lim

h→0

f(a+ h)− f(a)

h

3. (a) y = 2x−3 (b) y = −7x−4 (c) y = 10x+13 (d) y = 14x+ 3

4(e) y = −1

2x+ 3

2

(f) y = 12x+ 2

4. (a) −2/(a+ 3)2 (b) (i) −12

(ii) −29

(iii) −18

5. (a) 3a2 − 4 (b) y = −x− 1, y = 8x− 15

6. (a) (i) 32 (ii) 48 (iii) 62.4 (iv) 63.84 (v) 64

(b) (i) 2.236 (ii) 1.472 (iii) 1.351 (iv) 1.343 (v) 1.342

7. (a) −32 (b) 18

8. 12a2 + 6, 18 m/s, 54 m/s, 114 m/s


3.2 Derivatives

In previous section we define the tangent line to the curve with equation y = f(x) wherex = a to be

m = limh→0

f(a + h)− f(a)

h.

We also saw that the velocity of an object with position function s = f(t) at time t = ais

v(a) = limh→0

f(a+ h)− f(a)

h

In fact, limits of the form

limh→0

f(a+ h)− f(a)

h

arise whenever we calculate a rate of change in any of the sciences or engineering, suchas a rate of reaction in chemistry or a marginal cost in economics. Since this type oflimit occurs so widely, it is given a special name and notation.

The derivative of a function f(x) at x = a is defined by

f ′(a) = limh→0

f(a+ h)− f(a)

h

provided the limit exists. If the limit exists, we say that f is differentiable atx = a.

Definition 3.2.

Compute the derivative of f(x) = x3 + 7x at x = 1.

Example 3.3.

Solution


The derivative of f(x) is the function f ′(x) given by

f ′(x) = limh→0

f(x+ h)− f(x)

h

provided the limit exists. The process of computing a derivative is called dif-

ferentiation.

Definition 3.3.

Find the derivative of the function f(x) =2x− 3

5− xusing the definition of deriva-

tive.

Example 3.4.

Solution

Other Notations

If we use the notation y = f(x) to indicate that the independent variable is x and thedependent variable is y, then some common alternative notations for the derivative areas follows:

f ′(x) = y′ =dy

dx=

df

dx=

d

dxf(x) = Df(x) = Dxf(x)

The symbols D and d/dx are called differentiation operators.

If we want to indicate the value of a derivative dy/dx at a specific number a, we usethe notation

f ′(a) =dy

dx

∣

∣

∣

∣

x=a

=d

dxf(x)

∣

∣

∣

∣

x=a


A function f is differentiable at a if f ′(a) exists. It is differentiable on an

open interval (a, b) or (a,∞) or (−∞, a) or (−∞,∞) if it is differentiable atevery number in the interval.

Definition 3.4.

If f is differentiable at a, then f is continuous at x = a.

Theorem 3.1.

Exercise 3.2

1. If f(x) = 3x2 − 5x, find f ′(2) and use it to find an equation of the tangent line tothe parabola y = 3x2 − 5x at the point (2, 2).

2. If g(x) = x3 − 5x+ 1, find g′(1) and use it to find an equation of the tangent lineto the curve y = x3 − 5x+ 1 at the point (1,−3).

3. Compute f ′(a) using Definition 2.2.

(a) f(x) = 3x+ 1, a = 1 (b) f(x) =√3x+ 1 , a = 1

(c) f(x) = x2 + 2x, a = 0 (d) f(x) = x3 + 4, a = −1

(e) f(x) =x

2x− 1, a = 1 (f) f(x) =

2√3− x

, a = −1

4. Compute the derivative function f ′(x) using Definition 2.3.

(a) f(x) = 3x2 + 1 (b) f(x) =3

x+ 1

(c) f(x) =√3x+ 1 (d) f(x) = x3 + 2x− 1

5. Determine whether or not f ′(0) exists.

(a) f(x) =

{

2x+ 1 if x < 03x+ 1 if x ≥ 0

(b) f(x) =

{

x sin1

xif x 6= 0

0 if x = 0


1. 7, y = 7x− 12 2. −2, y = −2x− 1

3. (a) 3 (b) 34

(c) 2 (d) 3 (e) −1 (f) 18

4. (a) 6x (b)−3

(x+ 1)2(c)

3

2√3x+ 1

(d) 3x2 + 2

5. (a) Does not exist (b) Does not exist


3.3 Techniques of Differentiation

We have now computed numerous derivative using the limit definition. In this sectionwe develop rules for finding derivatives without having to use the definition directly.These differentiation rules enable us to calculate the derivatives of polynomials, rationalfunctions, algebraic functions, exponential and logarithmic functions, and trigonometricand inverse trigonometric functions.

The Power Rule

For any constant c, d

dx(c) = 0.

Theorem 3.2.

d

dx(x) = 1

Theorem 3.3.

If n in any real number, then

d

dx(xn) = nxn−1

Theorem 3.4.

Differentiate:

(a) f(x) = x2018 (b) f(x) =1

x−3/4

(c) g(x) =3√x7 (d) h(x) = xπ

Example 3.5.

Solution


General Derivative Rules

If f and g are differentiable at x and c is any constant, then

(i)d

dx

[

cf(x)]

= cd

dxf(x)

(ii)d

dx

[

f(x) + g(x)]

=d

dxf(x) +

d

dxg(x)

(iii)d

dx

[

f(x)− g(x)]

=d

dxf(x)− d

dxg(x)

Theorem 3.5.

Find the derivative of f(x) = 5x5/2 + 3√x− 2

√x+ 6x4 − 5.

Example 3.6.

Solution

Find the derivative of f(x) =5x3 − 2x+ 4

√x

x2.

Example 3.7.

Solution


The Product and Quotient Rules

If f and g are both differentiable, then

d

dx

[

f(x)g(x)]

= f(x)d

dx

[

g(x)]

+ g(x)d

dx

[

f(x)]

Theorem 3.6 (The Product Rules).

Use the product rule to find the derivative of

f(x) = (2− x− 3x3)(7 + x5).

Example 3.8.

Solution

If f(x) = x2/3g(x), where g(3) = 4 and g′(3) = 2, find f ′(3).

Example 3.9.

Solution


Suppose that f and g are differentiable at x and g(x) 6= 0. Then

d

dx

[

f(x)

g(x)

]

=g(x)

d

dx[f(x)]− f(x)

d

dx[g(x)]

[

g(x)]2 .

Theorem 3.7 (The Quotient Rule).

Compute the derivative of f(x) =x2 + x+ 1

x3 + 4.

Example 3.10.

Solution

Note. Don’t use the Quotient Rule every time you see the quotient. Sometimes it’seasier to rewrite a quotient first to put it in a form that is simpler for the purpose ofdifferentiation.

For instance, although it is possible to differentiate the function

F (x) =2x3 + x2 + 3

√x

x

using the Quotient Rule, it is much easier to perform the division first and then writethe function as

F (x) = 2x2 + x+ 3x− 1

2

before differentiating.


Exercise 3.3

1. Differentiate the function.

(a) f(x) = x3 − 2x+ 1 (b) f(x) = 3x2 − 4

(c) f(x) = 4 (d) f(x) = 3x3 − 2√x

(e) f(x) =3

x− 8x+ 1 (f) f(x) =

10√x− 2x

(g) f(x) = 2x3/2 − 3x−1/3 (h) f(x) = 2 3√x+ 3

(i) f(x) = x (3x2 −√x) (j) f(x) =

3x2 − 3x+ 1

2x

(k) f(x) = x+5√x2 (l) f(x) = x

√x+

1

x2√x


(a) y = x+4

x, (2, 4) (b) y = x+

√x, (1, 2)

3. Find the points on the curve y = x3 − x2 − x+ 1 where the tangent is horizontal.

4. Show that the curve y = 6x3 + 5x− 3 has no tangent line with slope 4.

5. Let

f(x) =

{

2− x if x ≤ 1x2 − 2x+ 2 if x > 1

Is f differentiable at x = 1?

6. For what values of x is the function f(x) = |x2 − 9| differentiable?

7. For what values of a and b is the line 2x+ y = b tangent to the parabola y = ax2

when x = 2?

8. Find a cubic function y = ax3 + bx2 + cx+ d whose graph has horizontal tangentsat the points (−2, 6) and (2, 0).

9. Differentiate.

(a) f(x) = (x2 + 3)(x3 − 3x+ 1)

(b) f(x) = (3x+ 4)(x3 − 2x2 + x)

(c) f(x) = (√x+ 3x)

(

5x2 − 3

x

)

(d) f(x) =3x− 2

5x+ 1(e) f(x) =

x− 2

x2 + x+ 1

(f) f(x) =3x− 6

√x

5x2 − 2(g) f(x) =

(x+ 1)(x− 2)

x2 − 5x+ 1

(h) f(x) =x2 + 3x− 2√

x(i) f(x) = x( 3

√x+ 3)

(j) f(x) = (x2 + 1)x3 + 3x2

x2 + 2(k) f(x) =

x

x+c

x



(a) y =2x

x+ 1, (1, 1) (b) y =

1

1 + x2, (−1, 1

2)

11. Write put the product rule for the function f(x)g(x)h(x).

12. Find the derivative of each function using the general product rule developed inexercise 12.

(a) f(x) = x2/3(x2 − 2)(x3 − x+ 1)

(b) f(x) = (x+ 1)(x3 + 4x)(x5 − 3x2 + 1)

13. Suppose that f(5) = 1, f ′(5) = 6, g(5) = −3, and g′(5) = 2. Find the values of (a)(fg)′(5), (b) (f/g)′(5), and (c) (g/f)′(5).


1. (a) 3x2 − 2 (b) 6x (c) 0 (d) 9x2 − 1√x

(e) − 3x2 − 8 (f) −5x−3/2 − 2

(g) 3x1/2 + x−4/3 (h) 23x−2/3 (i) 9x2 − 3

2x1/2 (j) 3

2− 1

2x−2 (k) 1 + 2/(5

5√x3)

(l) 32

√x− 5/(2x3

√x)

2. (a) y = 4 (b) y = 32x+ 1

23. (1, 0), (−1

3, 3227) 5. No

6. Not differentiable at x = −3 or 3 7. a = −12, b = 2 8. y = 3

16x3 − 9

4x+ 3

9. (a) 2x(x3 − 3x+ 1) + (x2 + 3)(3x2 − 3)

(b) 3(x3 − 2x2 + x) + (3x+ 4)(3x2 − 4x+ 1)

(c)

(

1

2x−1/2 + 3

)(

5x2 − 3

x

)

+ (√x+ 3x)(10x+ 3x−2) (d)

13

(5x+ 1)2

(e)−x2 + 4x+ 3

(x2 + x+ 1)2(f)

(3− 3x−1/2)(5x2 − 2)− (3x− 6√x)(10x)

(5x− 2)2

(g)−4x2 + 6x− 11

(x2 − 5x+ 1)2(h) 3

2x1/2 + 3

2x−1/2 + x−3/2 (i) 4

3x1/3 + 3

(j) 2xx3 + 3x2

x2 + 2+ (x2 − 1)

(3x2 + 6x)(x2 + 2)− (x3 + 3x2)(2x)

(x2 + 2)2(k)

2cx

(x2 + c)2

10. (a) y = 12x+ 1

2(b) y = 1

2x+ 1

11. f ′(x)g(x)h(x) + f(x)g′(x)h(x) + f(x)g(x)h′(x)

12. (a) 23x−1/3(x2 − 2)(x3 − x+ 1) + 2x4/3(x3 − x+ 1) + x2/3(x2 − 2)(3x2 − 1)

(b) (x3+4x)(x5−3x2+1)+(x+1)(3x2+4)(x5−3x2+1)+(x+1)(x3+4x)(5x4−6x)

13. (a) −6 (b) −209

(c) 20


3.4 The Chain Rule

If g is differentiable at x and f is differentiable at g(x), then

d

dx

[

f(

g(x))]

= f ′(

g(x))

g′(x)

It is often to think of the chain rule in Leibniz notation. If y = f(u) andu = g(x), then y = f

(

g(x))

and the chain rule says that

dy

dx=

dy

du· dudx

Theorem 3.8 (Chain Rule).

If n is any real number and u = g(x) is differentiable, then

d

dx(un) = nun−1du

dx

ord

dx

[

g(x)]n

= n[

g(x)]n−1 · g′(x)

Theorem 3.9.

Differentiate:

(a) y = (x5 − 4x3 + πx)101 (b) g(x) =1

3√x2 + 5

(c) h(x) =

(

x− 2

2x+ 1

)9

Example 3.11.

Solution


Let y = u5 − 2u3 + 5u and u = 3x− 2. Finddy

dx

∣

∣

∣

∣

x=1

.

Example 3.12.

Solution

Exercise 3.4

1. Find the derivative of each function.

(a) f(x) = (x3 + 4x)7 (b) f(x) = (x3 + x− 1)3

(c) f(x) =√x2 − 7x (d) f(x) =

√x2 + 4

(e) f(x) =

(

x− 1

x

)3/2

(f) f(x) = (3x− 2)10(5x2 − x+ 1)12

(g) f(x) = (2x− 5)4(8x2 − 5)−3 (h) f(x) =

(

x− 6

x+ 7

)3

(i) f(x) =1

5√2x− 1

(j) f(x) =√

x+√x

2. Find an equation of the tangent line to y = f(x) at x = a.

(a) y =√x2 + 16 , a = 3 (b) y =

8√4 + 3x

, a = 4

3. Suppose that F (x) = f(

g(x))

and g(3) = 6, g′(3) = 4, f ′(3) = 2, and f ′(6) = 7.Find F ′(3).

4. A table of values for f , g, f ′, and g′ is given.

x f(x) g(x) f ′(x) g′(x)1 3 2 4 62 1 8 5 73 7 2 7 9

(a) If h(x) = f(

g(x))

, find h′(1).

(b) If H(x) = g(

f(x))

, find H ′(1).



1. (a) 7(x3 + 4x)6(3x2 + 4) (b) 3(3x2 + 1)(x3 + x− 1)2 (c)2x− 7

2√x2 − 7x

(d) x(x2 + 4)−1/2 (e) 32(x− 1/x)1/2(1 + 1/x2)

(f) 6(3x−2)9(5x2−x+1)11(85x2−51x+9) (g) 8(2x−5)3(8x2−5)−4(−4x2+30x−5)

(h)39(x− 6)2

(x+ 7)4(i) −2

5(2x− 1)−6/5 (j) [1 + 1/(2

√x)]/(2

√

x+√x)

2. (a) y = 35x+ 16

5(b) y = − 3

16x+ 11

43. 28

4. (a) 30 (b) 36

3.5 Implicit Differentiation

Compare the following two equations:

y =√x2 + 3 and x2 + y2 = 9.

The first equation defines y as a function of x explicitly, since for each x, the equationgives an explicit formula y = f(x) for finding the corresponding value of y. On theother hand, the second equation does not define a function. However, we can solve fory and find at least two functions

(

y =√9− x2 and y = −

√9− x2

)

that are definedimplicitly by the equation x2 + y2 = 9.

To find the derivatives of functions defined implicitly, we can use the method ofimplicit differentiation. This consists of differentiating both sides of the equation with

respect to x and then solving the resulting equation for y′. In the examples and exercisesof this section it is always assumed that the given equation determines y implicitly as adifferentiable function of x so that the method of implicit differentiation can be applied.

Finddy

dxif x3 + y2 − 3y = 5.

Example 3.13.

Solution


Find y′(x) if x3y2 − 4x = 4− 2y.

Example 3.14.

Solution

Exercise 3.5

1. Find the derivative y′(x) implicitly.

(a) x2 + y2 = 1 (b) x3 + x2y + 4y2 = 6

(c) x2y + xy2 = 3x (d) x2y2 + 3y = 4x

(e)√xy − 4y2 = 12 (f)

x+ 3

y= 4x+ y2

(g)√x+ y − 4x2 = y (h)

y

x− y= x2 + 1

(i)√xy = 1 + xy


(a) x2 − 4y2 = 0, (2, 1) (b) x2 − 4y3 = 0, (2, 1)

(c) x2y2 = 4y, (2, 1) (d) x3y3 = 9y, (1, 3)

(e)x2

16− y2

9= 1, (−5, 9

4) (f) y2 = x3(2− x), (1, 1)

(g) 2(x2 + y2)2 = 25(x2 − y2), (3, 1) (h) y2 = x3 + 3x2, (1, 2)

(i) y(y2 − 1)(y − 2) = x(x− 1)(x− 2), (0, 1)

3. Find an equation of the tangent line to the hyperbola

x2

a2− y2

b2= 1

at the point (x0, y0).


4. Find all points on the curve x2y2 + xy = 2 where the slope of the tangent line is−1.

5. If x[

f(x)]3

+ xf(x) = 6 and f(3) = 1, then find f ′(3).


1. (a)−x

y(b)

−x(3x + 2y)

x2 + 8y(c)

3− 2xy − y2

x2 + 2xy(d)

4− 2xy2

3 + 2x2y(e)

y

16y√xy − x

(f)y − 4y2

x+ 3 + 2y3(g)

16x√x+ y − 1

1− 2√x+ y

(h)2x(x− y)2 + y

x(i)

2y√xy − y

x− 2x√xy

2. (a) y = 12x (b) y = 1

3x+ 1

3(c) y = −x+ 3 (d) y = −9

2x+ 15

2(e) y = −5

4x− 4

(f) y = x (g) y = − 913x+ 40

13(h) y = 9

2x− 5

2(i) y = −x+ 1

3.x0x

a2− y0y

b2= 1 4. (−1,−1), (1, 1) 5. −1

6

3.6 Derivatives of Trigonometric Functions

We collect all the differentiation formulas for trigonometric functions in the followingtable.

d

dx(sin x) = cosx

d

dx(cosx) = − sin x

d

dx(tanx) = sec2 x

d

dx(cot x) = − csc2 x

d

dx(sec x) = sec x tanx

d

dx(csc x) = − csc x cot x

Differentiate.

(a) f(x) = x4 sin x (b) f(x) = 3 tanx− 2 csc x

(c) f(x) =sec x

1 + tanx(d) f(x) =

√cotx− cos 3x

Example 3.15.

Solution


Find y′(x) if x2 cos y + sin 2y = xy.

Example 3.16.

Solution

Exercise 3.6

1. Find the derivative of each function.

(a) f(x) = 4 sin x− x (b) f(x) = tan x− csc x

(c) f(x) = x cosx (d) f(x) = 4√x− 2 sin x

(e) f(x) =sin x

x(f) f(x) =

tanx

x

(g) f(x) =cosx− 1

x2(h) f(x) =

x

sin x+ cosx

(i) f(x) = csc x cot x (j) f(x) = sin x sec x

(k) f(x) = 2 sinx cos x (l) f(x) = 4x2 tanx

(m) f(x) = 4 sin2 x+ 4 cos2 x (n) f(x) = sin(2x2 + 3)

(o) f(x) = cos(a3 + x3) (p) f(x) = tan2 x

(q) f(x) = tan(cosx) (r) f(x) = x2 sin 4x

(s) f(x) = sec3 4x (t) f(x) = sin(

tan√sin x

)

2. Prove, using the definition of derivative, that if f(x) = cosx, then f ′(x) = − sin x.


3. Find an equation of the tangent line to the given curve at the specified point.

(a) y = sin x, (π2, 1) (b) y = cosx, (π

2, 0)

(c) y = tanx, (π4, 1) (d) y = x+ cosx, (0, 1)

(e) y = x cosx, (π,−π) (f) y = 3 tanx− 2 csc x at x = π3

(g) y = sin(sin x) at x = π

4. For what values of x does the graph of f(x) = x+2 sin x have a horizontal tangent.

5. Find all points on the graph of the function f(x) = 2 sinx + sin2 x at which thetangent line is horizontal.

6. Find the derivative y′(x) implicitly.

(a) 4 cosx sin y = 1 (b) xy = cot(xy)

7. Using the identities sin 2x = 2 sin x cosx and cos 2x = cos2 x− sin2 x, prove that iff(x) = sin 2x, then f ′(x) = 2 cos 2x.

8. Using the identities sin 2x = 2 sin x cosx and cos 2x = cos2 x− sin2 x, prove that iff(x) = cos 2x, then f ′(x) = −2 sin 2x.


1. (a) 4 cosx− 1 (b) sec2 x+ csc x cot x (c) cosx− x sin x (d) 2x−1/2 − 2 cosx

(e)x cos x− sin x

x2(f)

x sec2 x− tan x

x(g)

−x2 sin x− (cosx− 1)2x

x4

(h)sin x+ cosx+ x sin x− x cosx

1 + sin 2x(i) − csc x cot2 x− csc3 x (j) sec2 x

(k) 2 cos2 x− 2 sin2 x (l) 8x tanx+ 4x2 sec2 x (m) 0 (n) 4x cos(2x2 + 3)

(o) −3x2 sin(a3 + x3) (p) 2 tanx sec2 x (q) − sin x sec2(cos x)

(r) 2x sin 4x+4x2 cos 4x (s) 12 sec3 4x tan 4x (t)cos(

tan√sin x

)

(sec2√sin x)(cosx)

2√sin x

3. (a) y = 1 (b) y = −x+ π2

(c) y = 2x+ 1− π2

(d) y = x+ 1 (e) y = −x

(f) y = 403

(

x− π3

)

+ 3√3− 4√

3(g) y = −x+ π

4. (2n+ 1)π ± π3, n an integer 5.

(

π2+ 2nπ, 3

)

,(

3π2+ 2nπ,−1

)

, n an integer

6. (a) tanx tan y (b)−y

x


3.7 Derivatives of Exponential Functions

For any constant a > 0,d

dx(ax) = ax ln a

Theorem 3.10.

The most commonly used base is the irrational number e. Since ln e = 1, the deriva-tive of f(x) = ex is simply:

d

dx(ex) = ex ln e = ex

We now have the following result.

d

dx(ex) = ex and

d

dx(e−x) = −e−x.

Theorem 3.11.

Differentiate

(a) f(x) = 4x cosx (b) g(x) =ex − 5x+ 1

x3 − ln 2

(c) h(x) = 2secx +1

3√etan x

Example 3.17.

Solution


Exercise 3.7

Find the derivative of each function.

1. f(x) = 4ex − x 2. f(x) = xex

3. f(x) = x− 2x 4. f(x) =ex − e−x

ex + e−x

5. f(x) = (1/3)x 6. f(x) = x2 + 2ex

7. f(x) =√xex 8. f(x) = x2e−x

9. f(x) =ex

x10. f(x) =

ex

x+ ex

11. f(x) = 2ex+1 + 4−x+1

12. Find an equation of the tangent line to y =2

1 + e−xat x = 0.


1. 4ex − 1 2. ex + xex 3. 1 + (ln 2)2x 4.4

(ex + e−x)25.(

ln 13

) (

13

)x6. 2x+ 2ex

7.√xex+

ex

2√x

8. 2xe−x−x2e−x 9.xex − ex

x210.

xex − ex

(x+ ex)211. 2ex+1−(ln 4)4−x+1

12. y = 12x+ 1

3.8 Derivatives of Inverse Trigonometric Functions

We collect all the differentiation formulas for inverse trigonometric functions in the fol-lowing table.

d

dx

(

sin−1 x)

=1√

1− x2when −1 < x < 1

d

dx(cos−1 x) =

−1√1− x2

when −1 < x < 1

d

dx(tan−1 x) =

1

1 + x2

d

dx(cot−1 x) =

−1

1 + x2

d

dx(sec−1 x) =

1

|x|√x2 − 1

when |x| > 1

d

dx(csc−1 x) =

−1

|x|√x2 − 1

when |x| > 1


Differentiate

(a) y = cos−1(5x2) (b) y = x tan−1√x

(c) y = cot−1 (sin√x)

Example 3.18.

Solution


Exercise 3.8


1. y = sin−1(x2) 2. y = tan−1(ex)

3. y = cos−1(x3) 4. y = sec−1(x2)

5. H(x) = (1 + x2) tan−1 x 6. g(t) = sin−1(4/t)

7. y = x2 cot−1(3x) 8. f(x) = ex − x2 tan−1 x

9. y = tan−1(cos 2x) 10. y = x cos−1(2x)

11. y = cos−1(sin x) 12. y = tan−1(sec x)


1.2x√1− x4

2.ex

1 + e2x3.

−3x2

√1− x6

4.2

x√x4 − 1

5. 1+2x tan−1 x 6.−4√

t4 − 16t2

7. 2x cot−1(3x)− 3x2

1 + 9x28. ex − x2

1 + x2− 2x tan−1 x 9.

−2 sin 2x

1 + cos2 2x

10. cos−1 2x− 2x√1− 4x2

11.− cosx

√

1− sin2 x= ±1 12.

sec x tan x

1 + sec2 x

3.9 Derivatives of Logarithmic Function

d

dx(loga x) =

1

x ln a(3.1)

Theorem 3.12.

If we put a = e in Formula (3.1), then the factor ln a on the right side becomesln e = 1 and we get the formula for the derivative of the natural logarithmic functionloge x = ln x:

d

dx(ln x) =

1

x(3.2)

In general, if we combine Formula (3.2) with the Chain Rule, we get

d

dx(ln u) =

1

u

du

dxor

d

dx

[

ln g(x)]

=g′(x)

g(x)

Differentiate

(a) y = ln(2x3 + 5x) (b) f(x) = log5(3 + tanx)

(c) f(x) = ln[

cos(

x2 − sec x)]

Example 3.19.

Solution


Finddy

dxif y = sec(π2) + x tan−1 (x) + (ln 4)(52x).

Example 3.20.

Solution

Logarithmic Differentiation

The calculation of derivatives of complicated functions involving products, quotients, orpowers can often be simplified by taking logarithms. The method used in the followingexample is called logarithmic differentiation.

Use logarithmic differentiation to find the derivative of y =x3/4

√x2 + 1

(3x+ 2)5.

Example 3.21.

Solution


Steps in Logarithmic Differentiation

1. Take natural logarithms of both sides of an equationy = f(x) and use the Laws of Logarithms to simplify.

2. Differentiate implicitly with respect to x.3. Solve the resulting equation for y′.

Differentiate y = (lnx)cos x.

Example 3.22.

Solution

Finddy

dxif yx = sin y.

Example 3.23.

Solution


Exercise 3.9

1. Differentiate the function.

(a) f(x) = ln(2x) (b) f(x) = ln(x3)

(c) f(θ) = ln(cos θ) (d) f(x) = log3(x2 − 4)

(e) f(x) = ln√x (f) f(x) =

√x ln x

(g) f(x) = ln

(

a− x

a+ x

)

(h) f(x) = ex ln x

(i) f(x) =ln x

1 + x(j) f(x) = |x3 − x2|

(k) f(x) = ln(e−x + xe−x) (l) f(x) = x2 ln(1− x2)


(a) y = x2 lnx, (1, 0) (b) y = ln(ln x), (e, 0)

3. Use logarithmic differentiation to find the derivative of the function.

(a) y = (2x+ 1)5(x4 − 3)6 (b) y =sin2 x tan4 x

(x2 + 1)2

(c) y = xx (d) y = xsinx

(e) y = (ln x)x (f) y = xex


1. (a) f ′(x) =1

x(b) f ′(x) =

3

x(c) f ′(θ) = − tan θ (d) f ′(x) =

2x

(x2 − 4) ln 3

(e) f ′(x) =1

2x(f) f ′(x) =

2 + ln x

2√x

(g) f ′(x) =−2a

a2 − x2

(h) f ′(x) = ex(

ln x+1

x

)

(i) f ′(x) =1 + x− x ln x

x(1 + x)2(j) f ′(x) =

3x− 2

x(x− 1)

(k) f ′(x) =−x

1 + x(l) f ′(x) = 2x ln(1− x2)− 2x3

1− x2

2. (a) y = x− 1 (b) x− ey = e

3. (a) y′ = (2x+ 1)5(x4 − 3)6(

10

2x+ 1+

24x3

x4 − 3

)

(b) y′ =sin2 x tan4 x

(x2 + 1)2

(

2 cotx+4 sec2 x

tan x− 4x

x2 + 1

)

(c) y′ = xx(ln x+ 1)

(d) y′ = xsinx

[

cosx ln x+sin x

x

]

(e) y′ = (ln x)x(

ln ln x+1

ln x

)

(f) y′ = exxex(

ln x+1

x

)


3.10 Derivatives of Hyperbolic Functions

In this section we shall study certain combinations of ex and e−x, called hyperbolic

functions. These functions have numerous engineering applications and arise naturallyin many mathematical problems.

The hyperbolic sine and hyperbolic cosine functions, denoted by sinh andcosh, respectively, are defined by

sinh x =ex − e−x

2and cosh x =

ex + e−x

2

Definition 3.5.

The remaining hyperbolic functions, hyperbolic tangent, hyperbolic cotangent, hy-perbolic secant, and hyperbolic cosecant are defined in terms of sinh and cosh asfollows.

tanh x =sinh x

cosh x=

ex − e−x

ex + e−x

coth x =cosh x

sinh x=

ex + e−x

ex − e−x

sech x =1

cosh x=

2

ex + e−x

csch x =1

sinh x=

2

ex − e−x

The graphs of these functions are shown in Figure below.

y = sinh xy = cosh x

−1

1

y = tanhx

y = cschxy = sechx y = coth x

The hyperbolic functions satisfy a number of identities that are similar to well-knowntrigonometric identities. We list some of them here.


Hyperbolic Identities

sinh(−x) = − sinh x cosh(−x) = cosh x

tanh(−x) = − tanh x cosh2 x− sinh2 x = 1

1− tanh2 x = sech 2x coth2 x− 1 = csch 2x

cosh x+ sinh x = ex cosh x− sinh x = e−x

sinh 2x = 2 sinh x cosh x cosh 2x = cosh2 x+ sinh2 x

sinh(x± y) = sinh x cosh y ± cosh x sinh y

cosh(x± y) = cosh x cosh y ± sinh x sinh y

tanh(x± y) =tanh x± tanh y

1± tanh x tanh y

The derivatives of the hyperbolic functions are easily computed. For example,

d

dx(sinh x) =

d

dx

(

ex − e−x

2

)

=ex + e−x

2= cosh x.

We list the differentiation formulas for the hyperbolic functions as follows.

Derivatives of Hyperbolic Functions

d

dx(sinh x) = cosh x

d

dx(csch x) = −csch x coth x

d

dx(cosh x) = sinh x

d

dx(sech x) = −sech x tanh x

d

dx(tanh x) = sech 2x

d

dx(cothx) = −csch 2x

Note that any of these differentiation rules can be combined with the Chain Rule. Forinstance,

d

dx(cosh

√x) = sinh

√x · d

dx

√x =

sinh√x

2√x

.

Differentiate.

(a) f(x) = tanh 4x+ sinh2 x

(b) g(x) = ln(sinh x)− sinh(cosh x)

(c) h(x) = sinh x tanh x+ exsech x

Example 3.24.

Solution


Exercise 3.10


1. f(x) = x cosh x+ sinh(x2) 2. g(x) =1− cosh x

1 + cosh x

3. h(x) = coth√1 + x2 − tanh(ex) 4. y = ecosh 3x + ln(sinh x)

5. f(x) = (sech x)(1 − ln sech x) 6. g(x) = ln cosh x− 12tanh2 x


1. f ′(x) = x sinh x+ cosh x+ 2x cosh(x2) 2. g′(x) =−2 sinh x

(1 + cosh x)2

3. h′(x) =−x · csch 2

√1 + x2

√1 + x2

+ exsech 2(ex) 4. y′ = 3 ecosh 3x sinh x+ coth x

5. f ′(x) = (ln sech x)(sech x tanh x) 6. g′(x) = tanh3 x

3.11 Inverse Hyperbolic Functions

We can see from above Figures that sinh and tanh are one-to-one functions and so theyhave inverse functions denoted by sinh−1 and tanh−1. Moreover, the above Figure showthat cosh is not one-to-one, but when restricted to the domain [0,∞) it becomes one-to-one. The inverse hyperbolic cosine function is defined as the inverse of this restrictedfunction.

y = sinh−1 x ⇐⇒ sinh y = x

y = cosh−1 x ⇐⇒ cosh y = x if y ≥ 0

y = tanh−1 x ⇐⇒ tanh y = x

The remaining inverse hyperbolic functions are defined similarly. We can sketch thegraphs of sinh−1 x, cosh−1 x, and tanh−1 x in Figures below.

y = sinh−1 xy = cosh−1 x

1 −1 1

y = tanh−1 x

Since the hyperbolic functions are defined in terms of exponential functions, it’s notsurprising to learn that the inverse hyperbolic functions can be expressed in terms oflogarithms.


sinh−1 x = ln(x+√x2 + 1), x ∈ R csch −1x = ln

(

1

x+

√1 + x2

|x|

)

, x 6= 0

cosh−1 x = ln(x+√x2 − 1), x ≥ 1 sech −1x = ln

(

1 +√1− x2

x

)

, 0 < x ≤ 1

tanh−1 x = 12ln

(

1 + x

1− x

)

, |x| < 1 coth−1 x = 12ln

(

x+ 1

x− 1

)

, |x| > 1

The inverse hyperbolic functions are all differentiable because the hyperbolic func-tions are differentiable. We list the differentiation formulas for the inverse hyperbolicfunctions as follows.

Derivatives of Inverse Hyperbolic Functions

d

dx(sinh−1 x) =

1√1 + x2

d

dx(csch −1x) = − 1

|x|√x2 + 1

d

dx(cosh−1 x) =

1√x2 − 1

d

dx(sech −1x) = − 1

x√1− x2

d

dx(tanh−1 x) =

1

1− x2

d

dx(coth−1 x) =

1

1− x2

Differentiate.

(a) f(x) = sech −1√1− x2 , x > 0

(b) g(x) = x tanh−1 x+ ln√1− x2

(c) h(x) = x2 sinh−1 x+ tanh−1(sin x)

Example 3.25.

Solution


Exercise 3.11


1. f(x) = sinh−1√x 2. g(x) = (1− x) tanh−1 x

3. h(x) = (1− x) coth−1√x 4. y = cosh−1 x− xsech −1x

5. f(x) = x sinh−1(

x3

)

−√9 + x2 6. g(x) = coth−1

√x2 + 1


1. f ′(x) =1

2√

x(1 + x)2. g′(x) =

1

1 + x− tanh−1 x 3. h′(x) =

1

2√x− coth−1√x

4. y′ = −sech −1x 5. f ′(x) = sinh−1(

x3

)

6. g′(x) =−1

x√x2 + 1

3.12 Higher Derivatives

If the derivative f ′ of a function is itself differentiation, then the derivative of f ′ is denotedby f ′′ and is called the second derivative of f . As long as we have differentiability, wecan continue the process of differentiating derivatives to obtain third, fourth, fifth, andeven higher derivatives of f . The successive derivatives of f are denoted by

f ′ The first derivative of f

f ′′ = (f ′)′ The second derivative of f

f ′′′ = (f ′′)′ The third derivative of f

f (4) = (f ′′′)′ The fourth derivative of f

......

f (n) The n th derivative of f

If f(x) = 5x3 − 3x2 + 2, then

f ′(x) =

f ′′(x) =

f ′′′(x) =

f (4)(x) =...

f (n)(x) =

Example 3.26.


Successive derivatives can also be denoted as follows:

f ′(x) =d

dx[f(x)]

f ′′(x) =d

dx

[

d

dx[f(x)]

]

=d2

dx2[f(x)]

f ′′′(x) =d

dx

[

d2

dx2[f(x)]

]

=d3

dx3[f(x)]

...

In general, we write

f (n)(x) =dn

dxn[f(x)]

Let f(x) = 4x3/2. Find f ′(x), f ′′(x), f ′′′(x), and f (4)(x).

Example 3.27.

Solution

Find y′′ if x2 − xy + y2 = 6.

Example 3.28.

Solution


Exercise 3.12

1. Find the first and second derivatives of the function.

(a) f(x) = x4 + 3x2 − 2 (b) f(x) = x5 + 6x2 − 7x

(c) f(x) = x6 +√x (d) f(x) =

√2x+ 1

(e) f(x) = e2x (f) y = cos 2θ

(g) h(x) =√x2 + 1 (h) F (s) = (3s+ 5)8

(i) y =x

1− x(j) y = (1− x2)3/4

(k) H(t) = tan 3t (l) g(t) = t3e5t

2. If f(x) = (2− 3x)−1/2, find f(0), f ′(0), f ′′(0), and f ′′′(0).

3. If f(θ) = cot θ, find f ′′′(π/6).

4. Find a formula for f (n)(x).

(a) f(x) =1

x(b) f(x) = e2x (c) f(x) =

1

3x3

5. Find a second-degree polynomial P such that P (2) = 5, P ′(2) = 3, and P ′′(2) = 2.

6. For what values of r does the function y = erx satisfy the equation y′′+5y′−6y = 0?


1. (a) f ′(x) = 4x3 + 6x, f ′′(x) = 12x2 + 6

(b) f ′(x) = 5x4 + 12x− 7, f ′′(x) = 20x3 + 12

(c) f ′(x) = 6x5 + 12x−1/2, f ′′(x) = 30x4 − 1

4x−3/2

(d) f ′(x) = (2x+ 1)−1/2, f ′′(x) = −(2x+ 1)−3/2

(e) f ′(x) = 2e2x, f ′′(x) = 4e2x (f) y′ = −2 sin 2θ, y′′ = −4 cos 2θ

(g) h′(x) =x√

x2 + 1, f ′′(x) =

1

(x2 + 1)3/2

(h) F ′(s) = 24(3s+ 5)7, F ′′(s) = 504(3s+ 5)6 (i) y′ =1

(1− x)2, y′′ =

2

(1− x)3

(j) y′ = −32x(1 − x2)−1/4, y′′ = 3

4(1− x2)−5/4(x2 − 2)

(k) H ′(t) = 3 sec2 3t, H ′′(t) = 18 sec2 3t tan 3t

(l) g′(t) = t2e5t(5t+ 3), g′′(t) = te5t(25t2 + 30t+ 6)

2.1√2,

3

4√2,

27

16√2,

405

64√2

3. −80 4. (a)(−1)nn!

xn+1(b) 2ne2x (c)

(−1)n(n+ 2)!

6xn+3

5. P (x) = x2 − x+ 3 6. r = 1,−6


3.13 Linear Approximations and Differentials

Linear Approximations

We have seen that a curve lies very close to its tangent near the point of tangency. Thisobservation is the basis for a method of finding approximate values of functions.

The idea is that it might be easy to calculate a value f(a) of a function, but difficult(or even impossible) to compute nearby values of f . So we settle for the easily computedvalues of the linear function L whose graph is the tangent line to f at (a, f(a)).

x

y

b

(a, f(a))

y = f(x)

y = L(x)

0

In other words, we use the tangent line at (a, f(a)) as an approximation to the curvey = f(x) when x is near a. An equation of this tangent line is

y = f(a) + f ′(a)(x− a)

and the approximation

f(x) ≈ f(a) + f ′(a)(x− a) (3.3)

is called linear approximation or tangent line approximation of f at a. Thelinear function whose graph is this tangent line, that is,

L(x) = f(a) + f ′(a)(x− a)

is called the linearization of f at x = a.

Find the linearization of f(x) =√x+ 3 at x = 1 and use it to approximate√

3.98 and√4.05.

Example 3.29.

Solution


Use a linearization to approximate 3√8.02 and 3

√25.2.

Example 3.30.

Solution

Differentials

The ideas behind linear approximations are sometimes formulated in the terminologyand notation of differentials. If y = f(x), where f is a differentiable function, then thedifferential dx is an independent variable; that is, dx can be given the value of anyreal number. The differential dy is then defined in terms of dx by the equation

dy = f ′(x)dx

So dy is a dependent variable; it depends on the values of x and dx.

Find the differential of the function.

(a) y = x3 − 3x2 − 4

(b) y = esec 3x

(c) y = sin(1 + 2x) ln(x2 + 1)

Example 3.31.

Solution


Exercise 3.13

1. Find the linearization L(x) of the function at a.

(a) f(x) = x3, a = 1 (b) f(x) = ln x, a = 1

(c) f(x) = e−2x, a = 0 (d) f(x) = 3√x, a = −8

2. Find the differential of the function.

(a) y = x2 + x− 3 (b) y = x4 + 5x

(c) y = (2x+ 3)−4 (d) y = x ln x

(e) y = (sin x+ cosx)3 (f) y =x+ 1

x− 1

(g) y = (7x2 + 3x− 1)−3/2 (h) y = 5

√

(x− 2)2

3. Use a linear approximation to estimate the given number.

(a)√402 (b)

√35.9

(c)√36.1 (d) 3

√1.02 + 4

√1.02

(e)1

10.1(f) sin 59◦

(g)13√9

(h) 3√8.01− 1

3√8.01


1. (a) L(x) = 3x− 2 (b) L(x) = x− 1 (c) L(x) = 1− 2x (d) L(x) = 112x− 4

3

2. (a) dy = (2x+ 1)dx (b) dy = (4x3 + 5)dx (c) dy = −8(2x+ 3)−5dx

(d) dy = (1 + ln x)dx (e) dy = 3(sin x+ cosx)2(cosx− sin x)dx

(f) dy = −2/(x− 1)2dx (g) dy = −32(7x2 + 3x− 1)−5/2(14x+ 3)dx

(h) dy =2

5 5

√

(x− 2)3dx

3. (a) 20.05 (b) 5.9917 (c) 6.0083 (d) 2.0117 (e) 0.099 (f) 0.857 (g) 0.479

(h) 1.501

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Chapter 3 Diﬀerentiationmathstat.sci.tu.ac.th/.../MA111-119/stnote111-119-ch3.pdf · 2019. 8. 22. · Chapter 3 Diﬀerentiation 3.1 Tangent Lines and Rates of Change 3.1.1 Tangent