Fundamental Engineering Maths 3

8/9/2019 Fundamental Engineering Maths 3

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Chapter 3: DIFFERENTIATION RULES

Duong T. PHAM - EEIT2014

Fundamental Engineering Mathematics for EEIT2014

Vietnamese German UniversityBinh Duong Campus

Duong T. PHAM - EEIT2014 October 14, 2014 1 / 25


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Outline

1 Derivatives of polynomials and exponential functions

2 The product and quotient rules

3 Derivatives of trigonometric functions

4 The chain rule

5 Implicit differentiation

6 Linear approximation



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Derivative of a constant function

Let cbe a constant. Thend

dx(c) = 0

Proof: We have f(x) =c. Then

d

dx(c) = lim

h0f(x+h) f(x)

h = lim

h0c ch

= limh0

0

h = lim

h00

= 0.



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Derivatives of power functions

The power rule: Ifn is a positive integer, then

d

dx(xn) =nxn1

Proof: If f(x) =xn, then

d

dx(xn) = lim

h0f(x+h) f(x)

h = lim

h0(x+h)n xn

h

= limh

0

h

(x+h)n1 + (x+h)n2x+ . . .+ (x+h)xn2 +xn1

h

= limh0

(x+h)n1 + (x+h)n2x+ . . .+ (x+h)xn2 +xn1

=nxn1



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The power rule

The power rule (General version): If

is any real number, thend

dx(x) = x1

Ex: Find ddx

1x2

and ddx

xAns:

d

dx

1

x2

=

d

dx x2

= (2)x21 = 2x3 =2

x3

d

dx

x

= d

dx

x1/2

=

1

2x

121 =

1

2x

12 =

1

2x



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The constant multiple rule

The constant multiple rule: Ifc is a constant and f is a differentiable

function, thend

dx[cf(x)] =c

d

dxf(x)

Proof: Let g(x) =cf(x), then

ddx

g(x) = limh0

g(x+h) g(x)h

= limh0

cf(x+h) cf(x)h

= limh0

cf(x+h) f(x)

h

=c lim

h0f(x+h) f(x)

h

=c ddxf(x)

Ex: d

dx(3x4) = 3

d

dx(x4) = 3 4x3 = 12x3



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The sum rule

The sum rule: If f are gare both differentiable functions, then

d

dx[f(x) +g(x)] = d

dxf(x)+ d

dxg(x)

Proof: Let k(x) =f(x) +g(x), then

d

dxk(x) = lim

h0

k(x+h) k(x)h

= limh0

f(x+h) +g(x+h) [f(x) +g(x)]h

= limh0

f(x+h) f(x)

h +

g(x+h) g(x)h

= limh0

f(x+h) f(x)h

+ limh0

g(x+h) g(x)h

= d

dxf(x) + d

dxg(x)

Remark: The sum rule can be extended to sums of any number of functions. Forexample,

(f +g+k) =f +g +k



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Derivative of exponential functions

(ex) =ex

(ax) =ax ln a

(ln x) = 1

x

(logax) = 1

xln a

Exercises 3.1:

332, 454653, 57, 6770, 75



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The product rule

The product rule: If f and gare differentiable function, then

ddx

[f(x)g(x)] =g(x) ddx

f(x) +f(x) ddx

g(x)

Proof: Let k(x) =f(x)g(x), then

ddx

k(x) = limh0

k(x+h) k(x)h

= limh0

f(x+h)g(x+h) f(x)g(x)h

= limh0

[f(x+h) f(x)]g(x+h)

h +

f(x)[g(x+h) g(x)]h

= limh0

f(x+h) f(x)h

limh0

g(x+h) +f(x) limh0

g(x+h) g(x)h

=g(x)d

dxf(x) +f(x)

d

dxg(x)



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The quotient rule

The quotient rule: Ifuand vare differentiable function, thend

dxu(x)v(x)

=

u(x)v(x)

u(x)v(x)

v(x)2

Proof: Exercise

Ex: Let y= x2 x+ 3

x+ 2 . Findy.

Ans:

y =(x2 x+ 3)(x+ 2) (x2 x+ 3)(x+ 2)

(x+ 2)2

=(2x

1)(x+ 2)

(x2

x+ 3)

(x+ 2)2

= x2 + 5x 5

(x+ 2)2



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Differentiation formulae

ddx

(c) = 0 ddx

(x) = x1

ddx

(ex) =ex ddx

(ax) =ax ln a

(cf) =cf (f g) =f g

(fg) =fg+fgfg

= f

gfgg2



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Exercises

3.2:

126, 2730, 3132, 39, 42, 43, 46, 50,



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Derivatives of trigonometric functions

Recall that sec x = 1

cos x and csc x =

1

sin x. The following identities are

true:

(sin x)= cos x (cos x) = sin x

(tan x) = 1cos2 x

(cot x) = 1sin2 x

(csc x) = csc xcot x (sec x) = sec xtan x

Exercises: 3.3: 116



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The chain rule

The chain rule: Let fbe differentiable at a and let gbe differentiable at f(a)

= the composition g f is differentiable at a and(g f)(a) =g(f(a)) f (a)

Proof: We have

(g

f)(a) = limh0

(g f)(a+h) (g f)(a)h

= limh0

g(f(a+h)) g(f(a))h

= limh0

g(f(a+h)) g(f(a))

f(a+h) f(a) f(a+h) f(a)

h

= limh0

g(f(a+h)) g(f(a))f(a+h)

f(a)

limh0

f(a+h) f(a)h

=g(f(a)) f (a)

Note that in the last argument we use the fact that f is continuous at a becauseit is differentiable at a, and thus f(a+h) f(a) as h goes to 0.



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The chain rule

Ex: Let k(x) =x2 + 1. Find f(x)

Ans:

Denote f(x) =x2 + 1 andg(x) =x. Then

k(x) =g(f(x)) = (g f)(x).

Applying the Chain Rule,

k(x) =g(f(x)) f(x).

g(x) = x = g(x) = 1

2x = g(f(x)) = 1

2f(x) = 1

2(x2+1)

f(x) =x2 + 1 = f(x) = 2x= k(x) = 2x

2(x2 + 1)=

x

x2 + 1



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The power rule combined with the chain rule

For any R, we have

d

dxu = u1

du

dx

Ex: Differentiate y= (x3

1)100.

Ans: We have

dy

dx =

d

dx[(x3 1)100] = 100(x3 1)1001(x3 1)

= 100(x

3

1)99

(3x

2

)= 300(x3 1)99x2


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Exercises

3.4:

112, 4750, 68, 75


I li i diff i i


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Implicit differentiation

If a function is given as an expression of the variable y=f(x), then

we can use definition and differentiation rules to compute y.

E.g., given y=x+ 1. Then y = 1

2x+1

However, if the function y is given implicitly as a relation between xand y, then we need to use the method of implicit differentiation

E.g., given x3 +y3 = 6xy. We need to find y?

E.g., Differentiating both sides, noting that y is a function ofx,

(x3 +y3)x= (6xy)x 3x2 + 3y2y = 6y+ 6xy

(y2 2x)y= 2y x2

y=2y x2

y2 2x


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Implicit differentiation

Ex: Find y ifx4 +y4 = 16

Ans: Differentiating both sides to obtain

(x4 +y4)x= (16)x 4x3 + 4y3y = 0= y =

x3

y3

Differentiating both sides of the blue equation

y = x3

y3

x

= (x3)xy

3 x3(y3)xy6

= 3x2y3 3x3y2y

y6

= 3x2y3

3x3y2

x3

y3y6 = 3x2

x4 +y4

y7

= 48x2

y7


D i ti f i t i t i f ti


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Derivative of inverse trigonometric functions

(sin1 x) = 11x2 (cos

1 x) = 11x2

(tan1 x) = 11+x2

(cot1 x) = 11+x2

(sec1 x) = 1xx21 (csc

1 x)= 1xx21


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Exercises

3.5

14, 2122, 3336, 40


Rates of changes in Natural and Social Sciences


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Rates of changes in Natural and Social Sciences

see Section 3.7 Page 221


Linear approximation


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x

y

y = f(x)

Pf(a)

a

t

Qf(x)

x

f (a)(x a) + f(a))

y = f (a)(xa) + f(a)

The approximation f(x) f(a)(x a) +f(a) is called the linearapproximation of f at a.

The function L(x) =f(a)(x a) +f(a)is the linearization of f




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Ex: Write the linearization of f(x) =x+ 3 at x= 1, then use it to

approximate the values

3

.

95 and

4

.

05Ans:

The linearization at x= 1 is

L(x) =f(1)(x

1) +f(1) = 1

21 + 3(x

1) +

1 + 3 = x

4+

7

4

The corresponding linear approximation is

x+ 3

x

4

+7

4

(when x is near 1)

In particular,

3.95 =

0.95 + 3 0.954 + 74 = 1.9875

and

4.05 =

1.05 + 3 1.054 + 74 = 2.0125


Exercises


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Exercises

3.10:

110


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