MATH2231-Differentiation (2)

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1 -Differentiation () The Beginnings of Calculus The prime occasion from which arose my iscovery of the metho of the Characteristic Triangle, an other things of the same sort, happene at a time when I ha stuie geometry for not more than si months...at that time I was quite ignorant of Cartesian algebra an also of the metho of inivisibles; inee I i not know the correct efinition of the center of gravity. For, when by chance I spoke of it to Huygens, I let him know that I thought that a straight line rawn through the center of gravity always cut a figure into two equal parts...huygens laughe when he hear this, an tol me that nothing was farther from the truth. So I, ecite by this stimulus, began to apply myself to the stu of the more intricate geometry. (From a 1680 letter from Gottfrie Leibniz to Ehrenfrie Walter von Tschirnhaus ( )) Builing on the work of many mathematicians over the centuries who consiere the problems of etermining the areas of regions boune by curves an of fining the maimum or minimum values of certain functions, two geniuses of the last half of the seventeenth century, Isaac Newton ( ) an Gottfrie Leibniz ( ), create the machinery of the calculus, the founation of moern mathematical analysis an the source of application to an increasing number of other isciplines. The maimum- minimum problem an the area problem, along with the relate problems of fining tangents an etermining volumes, ha been attacke an solve for various special cases over the years. No one ha evelope any algorithm which woul enable these problems to be solve easily in new situations. It was not until the avent of analytic geometry in the first half of the seventeenth century, the possibility suenly opene up of constructing all sorts of new curves an solis. After all, any algebraic equation etermine by a curve, an a new soli coul be forme, for eample, by rotating a curve aroun any line in its plane. Derivative an Tangent Line Problem I. Major Calculus Problems During Seventeenth Century A. Tangent line problems B. Velocity an acceleration problems C. Minimum an maimum problems D. Area problems 1

2 II. Tangent Line A. Definition 1. A line is tangent to a curve at a point P if it touches, but oes not cross, the curve at point P (works for general curves).. A line is tangent to a curve if the line touches or intersects the curve at eactly one point (works for circle). B. Fining the tangent line at a point P 1. Fining the slope of the tangent line at point P.. Approimate the slope (in 1.) using a secant line through the point of tangency an a secon point on the curve. m sec f ( c ) f ( c) ( c ) c f ( c ) f ( c) y C. Definition of Tangent Line with Slope m 1. If f is efine on an open interval containing c, an if the limit y f lim lim ( c ) f ( c ) m 0 0 eists, then the line passing through ( c, f ( c )) With slope m is the tangent line to the graph of f at the point ( c, f ( c)).. The slope of the tangent line to the graph of f at the point ( c, f ( c )) is also calle the slope of the graph of f at c. D. Eamples 1. Fin the slope of the graph of a linear function Given: f ( ) 5 at the point (, 9 ) f Fin: lim ( ) f ( ) 0

3 . Tangent lines to the graph of a nonlinear function Fin the slopes of the tangent lines to the graph of f ( ) at the point ( 0, -5) & (, -1). Given: f ( ) 5 f Fin: lim ( ) f ( ) 0 E. Vertical tangent line If f is continuous at c an f lim ( c ) f ( c ) f Or lim ( c ) f ( c ) 0 0 The vertical line, c, passing through (c, f ( c)) is a vertical tangent line to the graph of f. If the omain of f is the close interval a, b, you can eten the efinition of a vertical tangent line to inclue the enpoints by consiering continuity an limits from the right (for a ) an from the left ( for b). III. Derivative of a Function A. Definition for erivative f f f '( ) lim ( ) ( ) 0 Provie that the limit eists. * f '( ) Is rea as " f prime of " 3

4 *The process of fining the erivative of a function is calle ifferentiation. B. Differentiable: A function is ifferentiable at if its erivative eists at an ifferentiable on an open interval ( a, b) if it is ifferentiable at every point in the interval. C. Other notations for erivative: f y '( ),, ', [ f ( )], D y [ ] is rea as "the erivative of y with respect to " =lim y lim f ( ) f ( ) f '( ) 0 0 D. Eamples 1. Given: f ( ) 3 Fin the erivative of f ( ) by the limit process f f f '( ) lim ( ) ( ) 0 4

5 . Given: f ( ) Fin the erivative of f ( ) by the limit process. Fin the slope of the graph of f at the points (1, 1 ) an ( 4, ). Discuss the behavior of f at ( 0, 0 ). f f f '( ) lim ( ) ( ) 0 3. Given: f ( ) Fin the erivative of f ( ) by the limit process. 5

6 IV. Differentiability an Continuity f f c A. Alternative form of erivative: f '( c) lim ( ) ( ) c c The eistence of the limit in this alternative form requires that one-sie f limits lim ( ) f ( c ) f an lim ( ) f ( c ) eists an are equal. c c c c f Is ifferentiable on the close interval [a, b] if it is ifferentiable on (a, b) an if the erivative from the right at a an the erivative from the left at b both eist. B. If a function is not continuous at c, it is also not ifferentiable at c. Eample: f ( ) the greatest integer function is not continuous at 0 therefore, it is not ifferentiable at 0. f Verification: lim ( ) f ( 0 ) lim f lim ( ) f ( 0 ) lim C. Differentiability implies continuity If f is ifferentiable at 0 erivative from the left 0 0 erivative from the right c, then f is continuous at c. f( ) f( c) f f c lim[ f( ) f( c)] lim[( c)( )] [lim ( c)][lim ( ) ( ) ] ( 0)[ f'( c)] 0 c c c c c c As c, f ( ) f ( c) 0 lim f ( ) f ( c ) c f Is continuous at c 6

7 D. Continuity oes not necessarily imply ifferentiability Eample: f ( ) 5 is continuous at 5. f lim ( ) f ( 5 ) lim f lim ( ) f ( 5 ) lim erivative from the left erivative from the right 5 are not equal. Therefore, f is not ifferentiable at 5 an the graph of f oes not have a tangent line at the point ( 5, 0 ). 1 3 Eample: f ( ) is continuous at 0 lim ( ) ( ) 3 f f lim lim The tangent line is vertical at 0 f.is not ifferentiable at 0. 1 *A function is not ifferentiable at a point at which its graph has a sharp turn or a vertical tangent. 7

8 Basic Differentiation Rules an Rates of Change I. The Constant Rule: The erivative of a constant function is 0. If c is a real number. [ c] 0 E. 1. y 10. f ( ) 6 f '( ) 3. t( j) 5 t'( j) 4.y k, k is a constant y' II. The Power Rule: If n is a rational number, then the function f ( ) n is ifferentiable an n [ ] n n1 For f to be ifferentiable at number such that n1 is efine on an interval containing 0. Proof: n n [ ] lim ( ) 0 n 0, n must be a E. 1. [ ]. f ( ) 4 f '( ) 3. g( ) 4 g'( ) 4. y 1 3 8

9 5. Fin the slope of the graph of f ( ) 3 a. If b. If 6. Fin an equation of the tangent line to the graph of f ( ) 3 when III. Constant Multiple Rule: If f is a ifferentiable function an c is a real number, then cf is also ifferentiable an [ cf ( )] cf '( ) Proof: cf cf cf [ ( )] lim ( ) ( ) 0 E. 1. y 5 t. f ( t) y y 3 5. y y y ( 3) 5 8. y ( ) 3 f '( t) y' y' y' 9

10 IV. The Sum an Difference Rule: The erivative of the sum/ifference of two ifferentiable functions is ifferentiable an is the sum/ifference of their [ erivative. f ( ) g ( )] f '( ) g '( ) [ f ( ) g ( )] f '( ) g '( ) Proof: f g f g f g [ ( ) ( )] lim [ ( ) ( )] [ ( ) ( ) ] 0 E. 1. f ( ) 3 4 f '( ). g ( ) g'( ) V. Derivatives of Sine an Cosine Functions: [sin ] cos Recall: lim sin 0 [cos ] sin 1 1 lim cos 0 0 Proof: sin( ) sin( ) [sin ] lim 0 10

11 E. 1. y 4sin. y sin 4 y' y' 3. y cos y' VI. Rates of Change Derivative can be use to etermine the rate of change of one variable with respect to another. In escribing the motion of an object moving in a straight line, moving to the right or upwar is consiere to be in the positive (+) irection an moving to the left or ownwar is consiere to be in the negative (-) irection. Function s that gives the position (relative to the origin) of an object as a function of time t is a position function. Over a perio of time, t, the object changes its position by s s( t t ) s( t ) is tan ce A. Rate time average velocity s t E. Given a ball roppe from a height of 100 feet, its height s at time t is escribe the position function s 16t 100 (such that s is in feet, t is in secons) Fin the average velocity of the following time intervals. a. [1,] b. [1. 1.8] c. [,.5] 11

12 B. Instantaneous Velocity at t c can be approimate by calculating the average velocity over a small interval, [1, 1+t], as t 0 s t t s t 1. Velocity of an object at time t is: v( t ) lim ( ) ( ) s'( t ) t0 t. Spee of an object is the absolute value of its velocity. 3. Position of a Free-Falling Object (neglect air resistance) uner 1 influence of gravity: s( t ) gt v 0t s 0 v 0 : Initial velocity of the object s 0 : Initial height of the object g: Acceleration ue to gravity (-3ft per secon per secon or -9.8 m per secon per secon) E. At t 0, a iver jumps from the iving boar at 3 ft. above the water. The position of the iver is given by: s( t) 16t 16t 3 A. When oes the iver hit the water? B. What is the iver s velocity upon impact? (s in ft., t in secretary) 1

13 The Prouct an Quotient Rules an Higher-Orer Derivatives I. The Prouct Rule: The erivative of two ifferentiable functions a an b is ifferentiable. The erivative of ab is the first function times the erivative of the secon, ae to the secon function times the erivative of the first. Proof: a b a b [( a( ) b( )] lim ( ) ( ) ( ) ( ) 0 E. 1. Fin the erivative of: f ( ) ( 5 4 )( 6 ). Fin the erivative of: g( ) cos 3. Fin the erivative of: y 4cos sin II. The Quotient Rule: The erivative of the quotient a of two ifferentiable b functions a an b is ifferentiable at all values of for which b( ) 0. The erivative a is given by the enominator times the erivative of the numerator minus the b numerator times the erivative of the enominator, all ivie by the square of the Denominator. a( ) b a a b [ s t b b( ) ] ( ) '( ) ( ) '( ).. ( ) [ b( )] 0 13

14 5 E. 1. Fin the erivative of: f ( ) 1. Fin the erivative of: y 6 3. Fin the erivative of: ( ) 4. Fin the erivative of: y 6 14

15 S.Nunamaker III. Constant Multiple Rule vs. Quotient Rule: E. 1. Fin the erivative of: y 3 6. Fin the erivative of: y Fin the erivative of: y 3 ( ) 7 4. Fin the erivative of: y

16 IV. The Power Rule: n [ ] [ ] n k 1 n1 s.t. n is a negative integer an n E. 1. Fin the erivative of: y 3 4 k. Fin the erivative of: y V. The Derivatives of Trigonometric Functions A. [sin ] cos B. [cos ] sin C. [tan ] sec D. [sec ] sec tan E. [cot ] csc F. [csc ] csc cot E. 1. Fin the erivative of: y tan. Fin the erivative of: y sec 16

17 3. Fin the erivative of: y 1 cos sin 4. Fin the erivative of: y csc cot VI. Higher Orer Derivatives Position Function: s( t) Velocity Function: s'( t) = v( t) Acceleration Function: a( t) v'( t) s''( t) First Derivatives: y f ', '( ),, [ f ( )], D y [ ] Secon Derivatives: y f y '', ''( ),, [ f ( )], D [ y] Nth Derivatives: y f n y n ( n) ( n) n, ( ),, [ f ( )], D y n n [ ] E. Fin the ratio of earth s gravitational force to moon s, given position function of a falling object on the moon is s( t) 0. 81t, acceleration ue to gravity on earth is -9.8 meters per secon per secon. 17

18 E. Determine the velocity an acceleration function of each of the following: 1. s t 5t 3. s 4t 3 3. s 4 t t 4. s 64t 16t 1 5. s gt v t s 0 0 ( g, v, s 0 0 are constants) 18

19 The Chain Rule I. The Chain Rule If y f ( u) is a ifferentiable function of u an u g( ) is a ifferentiable function of, u u or equivalently, [ f ( g ( ))] f '( g ( )) g '( ). E. Fin for y ( 1 ) u Using chain rule, u y u 3 1,, ( 1)( ) = 4 4 u Using power rule, y ( 1) II. The General Power Rule If y [ u( )] n s.t. u Is a ifferentiable function of an n is a rational number. n n[ u( )] 1 u E. For y ( 3 ) ( 3 )( 6 ) For y 4 ( 1 ) ( 1) ( ) 4 For y 8 ( 5) ( )( 5) ( ) 3( 5) ( 5) 3 19

20 Try: Fin the first erivative of the following functions: 1. y ( 6 5) 4. y 1 3. y csc 4 4. f ( ) ( 9 ) 3 5. y f ( t ) ( 1 t 3 ) 7. y ( 3 1) 4 8. g( ) 3 9. y y y ( ) 1. y ( 1 3 ) 3 0

21 III. Trigonometric Functions an the Chain Rule [sin u] (cos u) u' [cos u] (sin u) u' [tan u] (sec u) u' [csc u] (csc ucot u) u' [sec u] (sec u tan u) u' [cot u] (csc u) u' E. If y sin 4 y' cos 4 [ 4] (cos 4)( 4) 4cos 4 Try: Fin the erivative of each of the following functions: 1. y cos. g( ) sec 3. y cos( 1) 4. y tan 3 5. f ( ) cos 4 6. y cos 7. y (cos 3) 8. y cos( ) 9. f ( ) tan g( ) (csc )(cot ) 1

22 Implicit an Eplicit Functions Eplicit form: variable y is eplicitly written as a function of. (E.y Implicit form: function y I. Differentiating with respect to 1 is implicitly efine by y 1. 1 ) E. [ ] y y [ ] y [ ] 1 y y y y [ ] [ ] [ ] ( 3 ) y ( ) 3 y y II. Implicit Differentiation 3 that contains both an y can be prouce by implicit ifferentiation. Guielines for implicit ifferentiation: 1. Differentiate both sies of the equation with respect to.. Collect all terms involving on the left sie of the equation an move all other terms to the right sie of the equation. 3. Factor out of the left sie of the equation. 4. Solve for by iviing both sies of the equation by the left-han factor that oes not contain.

23 3 E. Given: y y 5y y y y [ 3 5 ] [ 4] y y y [ 3 ] [ ] [ 5 ] [ ] [ 4] 3y y ( 3y y 5) 4. ( 3y y 5) 3 E. Given y y 3y 38, fin 1.. y y y [ 3 3 ] [ 38] y [ 3 ] [ y] [ 3y ] 0 3y ( y4) ( 3 y y ( 3)) 0 3y 4y 6y 3y 0 3y 6y 3y 4y ( 3y 6y) 3y 4y 3y 4y ( 3y 6y) 3

24 Try: Fin by implicit ifferentiation an evaluate the erivative at a point: Equation Point 1. y 4 (-4, -1) 3. y 0 ( 1, -1 ) 3. ( y) y (-1, 1 ) y y ( 1, 1 ) 4

25 III. Fining the Slope of a Graph Implicitly E. Determine the slope of the tangent line to the graph of 1 the point (, ). y [ 4 ] [ 4] ( 4)( y) 0 8y 0 8y 8y 4y ), 1 4( ) 1 4y 4 at Try: Fin the slope of the graph of 3( y ) 100y at ( 3, 1 ) 5

26 IV. Fining the Secon Derivative Implicitly E. Given : y 9, fin an y 1. y [ ] [ 9] y 0 y y y. y y ( 1) ( )( ) y ( ) y y [ ] y y y y y y y y 3 y y y ( ) 9 (Since y 9 ) 3 3 y y Try: Fin y in terms of an y. 1. y 5. y y 3 6

27 V. Fining a Tangent Line to a Graph Given: ( y ) y Fin: the tangent line to the graph at (, ) Solution: y y 4 ( y 4 ) ( y ) 3 4 ( y ) y ( ) y ( y ) y y 4 3 ( y y) y 4 3 y 3 4 ( y y) ( y ) = y( 1) ( y ) y( 1 ), ( ) 4 4 ( 1 ) 4 ( 11 ) (, ), m 3 3 y m b 3 b b ( 1 3) ( (, ), y 3 Is the equation of the tangent (, ) 7

Section 2.1 The Derivative and the Tangent Line Problem

Section 2.1 The Derivative and the Tangent Line Problem Chapter 2 Differentiation Course Number Section 2.1 The Derivative an the Tangent Line Problem Objective: In this lesson you learne how to fin the erivative of a function using the limit efinition an unerstan