1 Chapter 6 Differential Calculus The two basic forms of calculus are differential calculus and integral calculus. This chapter will be devoted to the.

1

Chapter 6Differential Calculus

The two basic forms of calculus are differential calculus and integral calculus. This chapter will be devoted to the former and Chapter 7 will be devoted to the latter. Finally, Chapter 8 will be devoted to a study of how MATLAB can be used for calculus operations.

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Differentiation and the Derivative

The study of calculus usually begins with the basic definition of a derivative. A derivative is obtained through the process of differentiation, and the study of all forms of differentiation is collectively referred to as differential calculus.If we begin with a function and determine its derivative, we arrive at a new function called the first derivative. If we differentiate the first derivative, we arrive at a new function called the second derivative, and so on.

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The derivative of a function is the slope at a given point.

x

y

( )y f x

x

y

4

Various Symbols for the Derivative

( ) or '( ) or

dy df xf x

dx dx

0Definition: lim

x

dy y

dx x

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Figure 6-2(a). Piecewise Linear Function (Continuous).

(a) ( )y f x

x

Continuous Function

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Figure 6-2(b). Piecewise Linear Function (Finite Discontinuities).

(b)

Discontinuities

( )y f x

x1x 2x 3x

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Piecewise Linear Segment

1 1( , )x y

2 2( , )x y

2 1y y

2 1x x

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Slope of a Piecewise Linear Segment

2 1

2 1

slopey ydy

dx x x

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Example 6-1. Plot the first derivative of the function shown below.

2 4 6 8 10 x

( )y f x12

-12

(a)

10

x

'( )dy

f xdx

3

-12

6

2 4 6 8 10 x

( )y f x12

-12

(a)

(b)

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Development of a Simple Derivative

2y x2( )y y x x

2 22 ( )y y x x x x

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Development of a Simple DerivativeContinuation

22 ( )y x x x

2y

x xx

0lim 2x

dy yx

dx x

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Chain Rule

( )y f u ( )u u x( )

'( )dy df u du du

f udx du dx dx

( )'( )

df uf u

duwhere

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Example 6-2. Approximate the derivative of y=x2 at x=1 by forming small changes.

2(1) (1) 1y 2(1.01) (1.01) 1.0201y

1.0201 1 0.0201y 0.0201

2.010.01

dy y

dx x

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Example 6-3. The derivative of sin u with respect to u is given below.

sin cosd

u udu

Use the chain rule to find the derivative with respect to x of

24siny x

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Example 6-3. Continuation.

2u x

2du

xdx

2

'( )

4(cos )(2 ) 8 cos

dy du dy duf u

dx dx du dx

u x x x

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Table 6-1. Derivatives( )f x '( )f x Derivative Number

( )af x '( )af x D-1

( ) ( )ux vx '( ) '( )u x v x D-2

( )f u ( )'( )du df u du

f udx du dx

D-3

a 0 D-4

( 0)nx n 1nnx D-5

( 0)nu n 1n dunudx

D-6

uv dv duu vdx dx D-7

u

v2

du dvv udx dxv

D-8

ue uduedx

D-9

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Table 6-1. Derivatives (Continued)ua l n u d u

a ad x

D - 1 0

l n u 1 d u

u d xD - 1 1

l o g a u 1l o g a

d ue

u d xD - 1 2

s i n uc o s

d uu

d x

D - 1 3

c o s us i n

d uu

d x D - 1 4

t a n u 2s e cd u

ud x

D - 1 5

1s i n u1

2

1 s i n

2 21

d uu

d xu

D - 1 6

1c o s u

2

1

1

d u

d xu

1 0 c o s u D - 1 7

1t a n u1

2

1 t a n

1 2 2

d uu

u d x

D - 1 8

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Example 6-4. Determine dy/dx for the function shown below.

2 siny x x

2

2 sinsin

dy dv duu v

dx dx dx

d xd xx x

dx dx

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2

2

cos sin 2

cos 2 sin

dyx x x x

dx

x x x x

21


sin xy

x

2 2

2

sinsin

cos sin

d x d xdu dvv u x xdy dx dx dx dx

dx v xx x x

x

22


2

2

x

y e

2

2

xu

2

2 12

2

xd

dux x

dx dx

2 2

2 2

x xdye x xe

dx

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Higher-Order Derivatives

( )y f x( )

'( )dy df x

f xdx dx

2 2

2 2

( )''( )

d y d f x d dyf x

dx dx dx dx

3 3 2(3)

3 3 2

( )( )

d y d f x d d yf x

dx dx dx dx

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Example 6-7. Determine the 2nd derivative with respect to x of the function below.

5sin 4y x

5(cos 4 ) (4 ) 20cos 4dy d

x x xdx dx

2

220 sin 4 (4 ) 80sin 4

d y dx x x

dx dx

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Applications: Maxima and Minima

1. Determine the derivative.2. Set the derivative to 0 and solve for

values that satisfy the equation.3. Determine the second derivative.

(a) If second derivative > 0, point is a minimum.

(b) If second derivative < 0, point is a maximum.

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Displacement

Velocity

Acceleration

dyv

dt

Displacement, Velocity, and Acceleration

y

2

2

dv d ya

dt dt

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Example 6-8. Determine local maxima or minima of function below.

3 2( ) 6 9 2y f x x x x 23 12 9

dyx x

dx

23 12 9 0x x

1 and 3x x

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23 12 9dy

x xdx

For x = 1, f”(1) = -6. Point is a maximum and ymax= 6.

For x = 3, f”(3) = 6. Point is a minimum andymin = 2.

2

26 12

d yx

dx

1 Chapter 6 Differential Calculus The two basic forms of calculus are differential calculus and integral calculus. This chapter will be devoted to the.

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