The first mechanical calculator (construvps/ME505/IEM/02 02.pdf · The first mechanical calculator (constru. Objectives: After the completion of this section the student - should

Chapter II CALCULUS II.2 Differentiation

73

IIII..22 DDIIFFFFEERREENNTTIIAATTIIOONN Objectives: After the completion of this section the student

- should recall the definition and geometrical and physical sense of the derivative;

- should be able to to find the derivatives of functions using the differentiation rules;

- should recall the main theorems of differential calculus;

- should be able to use derivatives in applications.

Contents: 1. Definitions

- derivative at the point - the derivative function - higher order derivatives - geometrical sense - physical sense 2. Differentiation Rules 3. Theorems 4. Examples 5. Review Questions and Exercises 6. Differentiation Calculus with Maple

The first mechanical calculator (constructed by Blaise Pascale in 1642)

Chapter II CALCULUS II.2 Differentiation 74

II.2 DIFFERENTIATION The derivative is a local property of smooth functions. The derivative of the function ( )f x at the point 0x x= (denoted ( )0f x′ ) shows the instantaneous

rate of change of ( )f x at this point, which numerically is expressed by the

slope of the line tangent to the graph of ( )f x .

If instead of a fixed point 0x we use the variable x , then the derivative of

( )f x is a function ( )f x′ . It characterizes the function in the following way: The operation of finding a derivative is called differentiation. 1. DEFINITIONS: 1) The derivative at the point 0x x= : Let function ( )f x be defined near the point 0x x= . If the limit

( ) ( ) ( )0 00 h 0

f x h f xf x lim

h→

+ −′ = (2.1)

exists, the function ( )f x is said to be differentiable at the point 0x and

( )0f x′ is called the derivative of ( )f x at the point 0x .

2) If the function ( )f x is differentiable at all points of the set S , then it is said to be differentiable in S .

3) Let the function ( )f x be defined in the set S . Then if the limit

( ) ( ) ( )h 0

f x h f xf x lim

h→

+ −′ = (2.2)

exists for all x in S , then ( )f x′ is called the derivative of the function

( )f x with respect to the variable x , and the function ( )f x is said to be differentiable in S .

Note: ( )0f x′ is defined at the point 0x x= and it is a number.

( )f x′ is defined for the variable x S∈ and it is a function.

In the equation (2.2), numerator is called the increment of function ( )f x :

( ) ( )y = f x h f x∆ + − and denominator is called the increment of independent variable x :

x h∆ =

( )0f x = tanθ′

f 0′ > f 0′ <

θθ

tan 0θ > tan 0θ <

( )f x increases ( )f x decreases


75

Note: The calculation of the limit in the equation (2.2) yields the ratio of increments which become infinitely small. That is why the original name for differentiation is the calculus of infinitesimals.

Other notations for the derivative: dfdx

, ( )d f xdx

, xf , f , y′ , dydx

, ( )Df x .

Higher order derivatives: Let the function ( )y f x= be differentiable in the interval ( )a,b , then

The derivative ( )dy f xdx

′ = is called the 1st order derivative

If the 1st order derivative is a differentiable function it can be differentiated again, yielding

( ) ( )2

2

d d dy f x f xdx dx dx

′′ = = 2nd order derivative

In the same manner this process can be continued, producing

( ) ( ) ( )n 1 n

nn 1 n

d d dy f x f xdx dx dx

−

−

= =

nth order derivative

If the derivative of any order n of the function ( )y f x= exists then the function is called infinitely many times differentiable.

Geometrical sense: Let the function ( )y f x= be differentiable in the interval ( )a,b and let

( )0x a,b∈ . Then the point ( )( )0 0x , f x belongs to the graph of the function

( )y f x= . The ratio yx

∆∆

defines the slope of the secant of the graph which

with a decrease of x∆ approaches the slope of the tangent line to the graph at the point ( )( )0 0x , f x . Therefore,

( )0 x 0

yf x lim tanx∆

∆ α∆→

′ = =

the derivative of the function ( )y f x= at the point ( )0x a,b∈ is equal to the

slope of the line tangent to the graph of the function at the point ( )( )0 0x , f x . The derivative can be used for the derivation of the equation of the tangent line to the graph of a function (see example 4).

Physical sense: 1) Instantaneous velocity ( ) ( )d dsv t s t sdt dt

= = =

Acceleration ( ) ( ) ( )2

2

dv d d da t s t s tdt dt dt dt

= = =

Newton’s Law F mx′′= or F ma= 2) Empirical physical laws (modeling): they establish the dependence of some

quantity on the rate of its change:

dP kPdx

= k is a constant of proportionality∈

This equation can be used in the exponential population growth models, logistic model, modeling of extinction of radiation in the medium with absorption coefficient k , etc.

( )( )0 0 0

equation of the tangent liney f x x x y′= − +


2. DIFFERENTIATION RULES Let ( )u x and ( )v x be differentiable functions and let c ∈ , then

1) ( )cu cu′ ′= coefficient rule

2) ( )u v u v′ ′ ′+ = + sum rule

3) ( )uv u v uv′ ′ ′= + product rule

4) 2

u u v uvv v

′ ′ ′− =

quotient rule

5) Chain rule: Let ( )y f u x= be a composite function, then

dy df dudx du dx

=

Example: 2y sin x= ( ) 2u x x=

( ) ( )( ) 2d sinu duy cos u 2x 2x cos xdu dx

′ = = =

6) Let ( )1y f x−= be the inverse function of ( )y f x= , then

( )( )

11

1f xf f x

−−

′ = ′ derivative of inverse function

Method of calculation: Example: ( ) xf x e= , ( )1f x ln x− = • calculate ( )f x′ ( ) xf x e′ =

• write ( )1

f x′ x

1e

• replace x by ( )1f x− ln x

1 1xe

=

and simplify the expression • write the answer

( )( )

11

1f xf f x

−−

′ = ′ ( ) 1ln x

x′ =

7) Implicit differentiation: given ( )f x, y 0= find y′ Method of calculation Example: ( )f x, y xy sin y 1 0≡ + − = use differentiation rules:

• terms without y differentiate ( ) ( ) ( )xy sin y 1 0′ ′ ′+ − = as a function of x

• terms with y differentiate using ( )y xy cos y y 0′′+ + = chain rule

( )d dgg y ydx dy

′= ( )y x cos y y 0′+ + =

• solve for y′ ( )

yyx cos y

−′ =+


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8) Logarithmic differentiation: given ( ) ( )v xy u x= find y′ Application of logarithm and implicit differentiation yields:

v uy u v lnu vu′ ′ ′= +

9) Derivative of a function defined parametrically: Let dependence of y on x be given through the parameter t by

( )( )

x x ty y t

= =

a t b≤ ≤ Parametric function defines a curve C in the plane

The derivative of y with respect to x can be determined as (use chain rule):

( )

( )( )

x x tdy

y tdy dtdxdx x tdt

= ′

= = ′

provided that dy dx,dt dt

exist and ( )x t 0′ ≠

Equation of the tangent line to the parametric curve at the point ( ) ( )( )0 0 0 0x x t , y y t= = is defined as

( ) ( ) ( ) ( )0 0 0 0x x y t y y x t 0′ ′− − − =

10) Differential: Let ( )f x be differentiable function. Then

( )df f x dx′=

is called the differential of the function ( )f x . Table of Derivatives:

( )f x

( )f x′

( )f x

( )f x′

( )f x

( )f x′

c

0

sin x cos x

cos x sin x−

1sin x−

2

1

1 x−

2

1

1 x

−

−

1sin x−a 1ax −ax

xexe tan x 2sec x

1tan x−xa xa ln a

2

11 x+

cot x

2csc x−

1cot x−

2

11 x

−+

ln x1x csc x cot x−

csc x

1csc x−2

1

x 1 x

−

−

alog x1

x ln a sec x sec x tan x1sec x−

2

1

x 1 x−


3. THEOREMS: Theorem 2.1 a) If the function ( )f x is differentiable at 0x x= then it is

continuous at 0x x= .

b) If the function ( )f x is differentiable in the set S then it is continuous in S.

The inverse statement is not true: a continuous function is not necessarily differentiable, for example, a function with a cusp.

Theorem 2.2 (Roll’s Theorem)

Let the function ( )y f x= be differentiable in the open

interval ( )a,b and continuous in the closed interval [ ]a,b ,

and let ( ) ( )f a f b= .

Then there exists a point ( )a,bξ ∈ such that

( )f 0ξ′ =

Theorem 2.3 (Lagrange’s Theorem or the Mean Value Theorem) Let the function ( )y f x= be differentiable in the open

interval ( )a,b and continuous in the closed interval [ ]a,b .

Then there exists a point ( )a,bξ ∈ such that

( ) ( ) ( )f b f af

b aξ

−′ =

−

Theorem 2.4 (The Extreme-Value Theorem for a Continuous Function)

Let the function ( )y f x= be continuous in the closed interval

[ ]a,b . Then the function ( )y f x= is bounded on [ ]a,b , and

it attains its extrema on [ ]a,b , i.e.

there exist the points [ ]1 a,bξ ∈ and [ ]2 a,bξ ∈ such that ( )

[ ]( )1 x a ,b

f max f x Mξ∈

= ≡

( )

[ ]( )2 x a ,b

f min f x mξ∈

= ≡

0

derivative does notexist at x x=

( )f 0ξ′ =

( ) ( ) ( )f b f af

b aξ

−′ =

−


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Differentiation Procedure: How the derivatives of the functions can be found: 1) Application of the definition (2.2):

- write increments ( ) ( )y = f x h f x∆ + − and x h∆ =

- write and simplify the ratio ( ) ( )f x h f xyx h

∆∆

+ −=

- evaluate the limit to find ( ) ( ) ( )h 0

f x h f xf x lim

h→

+ −′ =

This method is needed to establish the basic results and differentiation rules and as exercise for better understanding of the operation of differentiation; it is not usually used in engineering practice.

2) By application of the tables of derivatives, of the differentiation rules and of

the properties of the derivative. 3) Using commercial mathematical software (Maple, Matlab, Mathcad,

Mathematica etc) or scientific calculators. 4. EXAMPLES: 1) Using the definition (2.2), find the derivative of a constant function

( )f x c= Solution: Write: ( )f x c=

( )f x h c+ = (still the same constant) ( ) ( )y = f x h f x = c - c = 0∆ + −

( )y 0 0 for h 0x h

∆∆

= = ≠ (identically equal to zero)

Therefore, ( ) ( )h 0

0f x c = lim 0h→

′′ = = .

We established the following rule of differentiation: c 0′ = for any c ∈

2) Using the definition (2.2), find the derivative of a linear function

( )f x ax b= + Solution: Write: ( )f x ax b= +

( ) ( )f x h a x h b ax ah b+ = + + = + + y =ax ah b ax b ah∆ + + − − =

y ah ax h

∆∆

= ≡

Therefore, ( ) ( )h 0

f x ax+b = lim a a→

′′ = = .

We could expect this result because we know that in the linear function y ax b= + coefficient a represents the slope of the line defined by the function.


3) Using the definition (2.2), find the derivative of a quadratic function ( ) 2f x x= at 0x 2= Solution: Write: ( ) 2

0f x 2 4= =

( ) ( )2 20f x h 2 h 4 4h h+ = + = + +

2 2y =4 4h h 4 4h h∆ + + − = +

2y 4h h 4 h

x h∆∆

+= = +

Therefore, ( ) ( ) ( )0

20 h 0

x 2

f x x = lim 4 h 4→

=

′′ = + = .

4) Find the equation of the line tangent to the graph of the function ( ) 2f x x=

at the point ( )1,1 . Solution: Recall the point-slope form of the equation of a line ( )0 0y m x x y= − +

where m is the slope of the line and ( )0 0x , y are the coordinates of the point lying on the line.

The slope of the tangent line to the graph of the function ( ) 2f x x= at the point ( ) ( )0 0x , y 1,1= is equal to the value of the

derivative of the function at the point 0x x 1= = . From the table of derivatives:

( ) ( )2f x x 2x′′ = =

( )f 1 2 1 2 = m′ = ⋅ = Therefore, the equation of the tangent line is

( ) ( )0 0y m x x y = 2 x-1 1 2x 1= − + + = − 5) Find the derivative of the function ( ) 2f x x sin x=

Solution: ( )f x′ 2x sin x ′ =

[ ]2 2x sin x x sin x′ ′ = + product rule 22x sin x x cos x= + table 6) Find the derivative of the function ( ) ( )2f x sin 2x 1= +

Solution: ( )f x′ ( )2sin 2x 1 ′ = +

( )2 2cos 2x 1 2x 1 ′ = + + chain rule ( )24x cos 2x 1= + table


81

7) Find the second derivative of the function

( ) 2xf x e=

Solution: ( )f x′ 2xe ′ =

2x 2e x ′ = chain rule

2x2xe= table

( )f x′′ 2x2xe ′ =

[ ] [ ]2 2x x2x e 2x e ′′ = + product rule

[ ]2 2x x2 e 2x 2xe = + previous result

( ) 2x2 1 2x e= + 8) Find the derivative of the function (logarithmic differentiation) ( ) cos xf x x= Solution: ( ) cos xln f x ln x = take logarithm

( )( ) ( )cos xln f x ln x ′′ = implicit differentiation

( ) ( ) ( )1 f x cos x ln x

f x′′ =

( ) ( )cos xf x sin x ln x f xx

′ = − +

( ) cos xcos xf x sin x ln x xx

′ = − +

9) Related Rates:

The volume of a sphere is defined by 34V r3

π= , where r is a radius.

What is the rate of change of volume if the radius is changing with the rate cm5 s

? Given dr 5dt

= , find dVdt

assuming that both are functions of time.

Differentiate the equation for volume using the chain rule:

2 2dV 4 dr dr3r 4 rdt 3 dt dt

π π= =

Substitute dr 5dt

=

2dV 20 rdt

π=

So, the rate of change of the volume of a sphere will be proportional to the square of its radius.


5. REVIEW QUESTIONS: 1) What is the derivative of the function ( )f x ?

2) What is the derivative of the function ( )f x at the point 0x x= ? 3) What is the geometrical sense of the derivative? 4) What is the role of the limit in the definition of the derivative? 5) What is differentiation? What is a differentiable function at a point, in the

set? 6) Give the geometrical examples of a continuous function not differentiable at

a point. 7) How do derivatives characterize the properties of the functions?

8) How can you interpret the notation dydx

?

9) What rules of differentiation do you recall?

10) Prove the coefficient rule: ( )cu cu′ ′= 11) What is the second derivative? 12) How many derivatives can function have? 13) Derive the equation of the line tangent to the graph of the function ( )f x .

14) Is the function y x= differentiable? 15) How many times can the following functions be differentiated for all real numbers: a) 2y x= b) ( )y sin x= c) xy e= d) 2y 1 x= + 16) Derive the formula for logarithmic differentiation.

EXERCISES: 1) Using the definitions find the derivative of the following functions: a) y x= c) y cos x=

b) 1yx

= d) ( )2y x 2= +

2) Using the definitions find the derivatives at the indicated point for the

following functions:

a) y x= at 0x 1= c) y sin 2x= at 0x4π

=

b) 2

1yx

= at 0x 2= d) 1yx

= at 0x 4=

3) Find the derivatives of the following functions: a) 3 2y 2x 4x 3x 5= + + + g) 2y 1 sin x= +

b) y x ln x= h) 2

sin(ln x )yx 1

=+

c) 2y x cos x= i) ( )2y ln x 5= +

d) 3

1y ln xx

= k) ( )2

1sin xy e=

e) 2

sin xy3x 2

=+

l) 2xy x=

f) ( )y cot 2x= m) ( )xy sin 2=


83

4) Find the equation of the line tangent to the graph of the function ( )f x at the given point and sketch the graph:

a) y x 1= − at x 2= c) ( )y ln x 1= − at x 0=

b) xy e= at x 1= d) y sin x= at x3π

=

5) Which function grows faster at the given point (sketch the graph):

a) y x= or y sin x= at x 0= c) y ln x= or y x 1= − at x 1= b) 2y x= or y tan x= at x 0= d) 3y x 1= + or xy e= at x 0= 6) Find the derivative of the function defined implicitly: a) 2 2x y 4xy 1+ − = b) 2y x sin xy= + 7) Find the derivative of the function defined parametrically:

a) ( )( )

x t a sin ty t bcos t

ωω

= =

b) ( )

( ) 2t

x t ty t te

= =

8) Find the second derivative of the following functions: a) y x ln x= c) ( )2y x sin 3x ln x= + 9) Find the equation of the line tangent to the parametric curve

( )

( )x t sin t

y t 2cos t = =

0 t 2π≤ ≤

at the point

( )( )

1 1

1 1

x x ty y t

= =

, 1t 3π

=

and sketch the graph.

10) The surface area of a sphere is 2V 4 rπ= , where r is a radius. What is the

rate of change of surface area if the radius is changing with the rate m2 s

?

11) Find the velocity and acceleration of a point if its equation of motion is

( )s t t cos 2t= +

12) Prove the following corollaries from Lagrange’s Theorem:

a) If ( )f x 0′ = for all ( )x a,b∈ , then ( )f x c= for all ( )x a,b∈ , c ∈

(is a constant function).

b) If ( ) ( )f x g x′ ′= for all ( )x a,b∈ , then ( ) ( )f x g x c= + , c ∈ for all

( )x a,b∈ (functions differ by some constant).


6. DIFFERENTIATION WITH MAPLE:

Application of the Definition: > limit(((x+h)^2-x^2)/h,h=0);

2 x

Command diff: > diff(cos(x),x);

− ( )sin x

Operator D: > f:=sqrt(x);

:= f x

> D(f); 12

( )D xx

Derivative at the point: > fp(x):=diff(f,x);

:= ( )fp x 12 x

> fp(2):=subs(x=2,fp(x));;

:= ( )fp 2 24

Higher order derivatives: > diff(f,x$3);

3

8 x( )/5 2

> diff(x*exp(2*x),x$4); + 32 e

( )2 x16 x e

( )2 x

Equation of Tangent Line (Example 4): > g:=x^2;x[0]:=1;y[0]:=1;

:= g x2

:= x0 1

:= y0 1

> m:=subs(x=x[0],diff(g,x)); := m 2

> t(x):=m*(x-x[0])+y[0]; := ( )t x − 2 x 1

> plot({g,t(x)},x=-1..2);

>


85

Differentiation of Implicit Function: > F(x,y):=sin(x*y(x))=x;

:= ( )F ,x y = ( )sin x ( )y x x

> FP:=diff(F(x,y),x);

:= FP = ( )cos x ( )y x

+ ( )y x x

d

dx ( )y x 1

> isolate(FP,diff(y(x),x));

= ddx ( )y x

− 1

( )cos x ( )y x ( )y x

x

Mechanical calculator designed by Gottfried Leibnitz The quotient rule in the first edition of L’Hospital’s book


L'Hospital on Leibniz: "I must here in justice own, (as Mr. Leibnitz (sic) himself has done in Journal des Scavans for August, 1694) that the learned Sir Isaac Newton likewise discovered something like the Calculus Differentialis, as appears by his excellent Principia, published first in the year 1687 which almost wholly depends upon the use of the said Calculus. But the method of Mr. Leibnitz's is much more easy and expeditious, on account of the notation he uses, not to mention the wonderful assistance it affords on many occasions."


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Documents

The first mechanical calculator (construvps/ME505/IEM/02 02.pdf · The first mechanical calculator (constru. Objectives: After the completion of this section the student - should