math 151 engineering mathematics iroquesol/math_151_week_4.pdf · math 151 engineering mathematics...

Chapter 3. Derivatives

MATH 151 Engineering Mathematics ISpring 2017, WEEK 4

JoungDong Kim

02/14/2017

JoungDong Kim MATH 151 Engineering Mathematics I Spring 2017, WEEK 4

Chapter 3. Derivatives Section 3.1 Derivatives

Derivatives

Definition. DerivativeThe Derivative of a function f at a number a, denoted by f ′(a), is

f ′(a) = limh→0

f (a + h)− f (a)

hor

f ′(a) = limx→a

f (x)− f (a)

x − a

if this limit exists.



Derivatives

Ex10) Find the derivative of the following functions at the numbera given. (use definition)

a) f (x) = x2 − 8x + 9, a = −2

f ′(a) = limh→0f (a + h)− f (a)

h

f ′(−2) = limh→0f (−2 + h)− f (−2)

h

f ′(−2) = limh→0

((−2 + h)2 − 8(−2 + h) + 9

)−((−2)2 − 8(−2) + 9

)h

f ′(−2) = limh→04− 4h + h2 + 16− 8h + 9− 29

h

f ′(−2) = limh→0h2 − 12h

h= limh→0

h(h − 12)

h= limh→0(h − 12) = −12



Derivatives

b) f (x) =x

x + 1, a = 2

f ′(a) = limh→0f (a + h)− f (a)

h

f ′(2) = limh→0f (2 + h)− f (2)

h

f ′(2) = limh→0

((2+h)

(2+h)+1

)−(

(2)(2)+1

)h

f ′(2) = limh→0

(2+h3+h

)−(23

)h

f ′(2) = limh→0

(2+h)3−2(3+h)3(3+h)

h



Derivatives

f ′(2) = limh→0(6 + 3h)− (6 + 2h)

3h(3 + h)

f ′(2) = limh→0h

3h(3 + h)

f ′(2) = limh→01

3(3 + h)=

1

9



Derivatives

Interpretations of the Derivatives

Geometrically the tangent line to y = f (x) at (a, f (a)) is the linethrough (a, f (a)) whose slope is equal to f ′(a), the derivative of fat a.

1 The slope of the tangent line to the graph of f (x) atx = a.

2 The instantaneous rate of change of f (x) at x = a.

3 The instantaneous velocity at x = a.

All use

limh→0

f (a + h)− f (a)

h= f ′(a)



Derivatives

Ex11) Recall the surface area of a sphere is given by A = 4πr2.Find the average rate of change of the area from r = 1 to r = 2.Find the instantaneous rate of change of the area at r = 1.

Ave. rate of change =A(rfinal )− A(rinitial )

rfinal − rinitial

=A(2)− A(1)

2− 1=

4π(2)2 − 4π(1)2

1= 16π − 6π = 12π

A′(a) = limh→0A(a + h)− A(a)

h

A′(1) = limh→0A(1 + h)− A(1)

h

A′(1) = limh→04π(1 + h)2 − 4π(1)2

h



Derivatives

A′(1) = limh→04π((1 + h)2 − 1

)h

A′(1) = limh→04π(1 + 2h + h2)− 4π

h

A′(1) = limh→08πh + 4πh2

h=

h(8π + 4πh)

h

A′(1) = limh→0 (8π + 4πh) = 8π

or



Derivatives

A′(a) = limx→aA(x)− A(a)

x − a

A′(1) = limx→1A(x)− A(1)

x − 1

A′(1) = limx→14π(x)2 − 4π(1)2

x − 1

A′(1) = limx→14π((x)2 − 1

)x − 1

A′(1) = limx→14π(x − 1)(x + 1)

x − 1

A′(1) = limx→1 4π(x + 1) = 8π



Derivatives

Definition. Differentiable A function f is differentiable at a iff ′(a) exists.

Theorem. If f is differentiable at a, then f is continuous at a.

How can a function FAIL to be differentiable

1 If the graph of a function f has a “corner” or “kink” in it,then the graph of f has no tangent at that point and f is notdifferentiable there.

2 If f is not continuous at a, then f is not differentiable at a.

3 The curve has a vertical tangent line when x = a, f is notdifferentiable at a.



Derivatives



Derivatives

Ex12) The graph of f is given. State, with reasons, the numbers atwhich f is not differentiable.

1.- The function is not differentiable at x = −2 because thefunction f has a discontinuity at that point.

2.- The function is not differentiable at x = −1 because thefunction f has a corner at that point.



Derivatives

3.- The function is not differentiable at x = 4 because the functionf has a discontinuity at that point.

4.- The function is not differentiable at x = 8 because the functionf has a corner at that point.

5.- The function is not differentiable at x = 11.2 because thefunction f has a vertical tangent line at that point.



Derivatives

Ex13) Where is f (x) = |x2 − 4| not differentiable.

Solution

-4 -3 -2 -1 0 1 2 3 4

x

-4

-2

0

2

4

6

8

10

12

f

Function f(x)= x2

- 4

-4 -3 -2 -1 0 1 2 3 4

x

0

2

4

6

8

10

12

f

Function f(x)= | x2

- 4 |

Ex14) Where is f (x) =|x + 1|x + 1

not differentiable.

SolutionJoungDong Kim MATH 151 Engineering Mathematics I Spring 2017, WEEK 4


Derivatives

The function f is defined by

f (x) =|x + 1|x + 1

=

{−1 x < −1

1 x > −1

and its graph is given by

-6 -4 -2 0 2 4

-5

-4

-3

-2

-1

0

1

2

3

4

5Function f(x)= abs(x + 1)/(x+1)

Therefore the function f is not differentiable at x = −1 because fis not continuous there.



Derivatives

Ex15) Given the graph of f (x) below, sketch the graph of thederivative.

a)



Derivatives

The graph of the derivative is

-6 -4 -2 0 2 4

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2Function df/dx

2/3

-5/9



Derivatives

b)



Derivatives

The graph of the derivative is

-5 0 5

-15

-10

-5

0

5

10

15Function df/dx

eb

c



Derivatives

Definition. Derivative of functionThe derivative of f is defined as

f ′(x) = limh→0

f (x + h)− f (x)

h

Ex16) Find the derivative of the following functions as well as thedomain of the derivative.

a) f (x) = x2 − 8x + 9

f ′(x) = limh→0f (x + h)− f (x)

h



Derivatives

f ′(x) = limh→0

((x + h)2 − 8(x + h) + 9

)−(x2 − 8x + 9

)h

f ′(x) = limh→0

((x2 + 2xh + h2)− (8x + 8h) + 9

)−(x2 − 8x + 9

)h

f ′(x) = limh→0

(x2 + 2xh + h2 − 8x − 8h + 9

)−(x2 − 8x + 9

)h

f ′(x) = limh→0

(2xh + h2 − 8h

)h

= limh→0h(2x + h − 8h)

h

f ′(x) = 2x − 8



Derivatives

b) f (x) =3x + 1

x − 2

f ′(x) = limh→0f (x + h)− f (x)

h

f ′(x) = limh→0

3(x+h)+1(x+h)−2 −

3x+1x−2

h

f ′(x) = limh→0

[3(x+h)+1][x−2]−[3x+1][(x+h)−2][(x+h)−2][x−2]

h

f ′(x) = limh→0

[3x+3h+1][x−2]−[3x+1][x+h−2][x+h−2][x−2]

h



Derivatives

f ′(x) = limh→0

[3x2+3hx+x−6x−6h−2]−[3x2+3xh−6x+x+h−2][x+h−2][x−2]

h

f ′(x) = limh→0

−7h[x+h−2][x−2]

h

f ′(x) = limh→0−7h

h[x + h − 2][x − 2]=

−7

(x − 2)2



Derivatives

c) f (x) =4√x

f ′(x) = limh→0f (x + h)− f (x)

h

f ′(x) = limh→0

4√x+h− 4√

x

h

f ′(x) = limh→0

4√

x−4√

x+h√x+h√

x

h

f ′(x) = limh→04[√x −√x + h]

h√x + h

√x

[√x +√x + h

√x +√x + h

]



Derivatives

f ′(x) = limh→04x − 4(x + h)

h√x + h

√x [√x +√x + h]

f ′(x) = limh→0−4h

h√x + h

√x [√x +√x + h]

f ′(x) = limh→0−4√

x + h√x [√x +√x + h]

f ′(x) =−4√

x√x [√x +√x ]

f ′(x) =−4

x [2√x ]

=−2

x√x

=−2

x3/2



Derivatives

Ex17) The limit below represent the derivative of some functionf (x) at some number a. Identify f (x) and a for each limit.

a) limh→0

(2 + h)5 − 32

h

= limh→0

(2 + h)5 − 25

h

= limh→0

f (2 + h)− f (2)

h; f (x) = x5

= limh→0

f (a + h)− f (a)

h; f (x) = x5; a = 2

= limh→0

f (a + h)− f (a)

h= f ′(a); f (x) = x5; a = 2



Derivatives

b) limx→3π

cos x + 1

x − 3π

= limx→3π

cos x − cos(3π)

x − 3π

= limx→3π

f (x)− f (3π)

x − 3π; f (x) = cos(x)

= limx→3π

f (x)− f (a)

x − a; f (x) = cos(x); a = 3π

= limx→3π

f (x)− f (a)

x − a= f ′(a); f (x) = cos(x); a = 3π


math 151 engineering mathematics iroquesol/math_151_week_4.pdf · math 151 engineering mathematics...

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