158 Calculus and Structures

Transcription

1 58 Calculus and Structures

2 CHAPTER PROPERTIES OF DERIVATIVES AND DIFFERENTIATION BY THE EASY WAY. Calculus and Structures 59 Copyrigt

3 Capter PROPERTIES OF DERIVATIVES. INTRODUCTION In te last capter you were introduced to a comple procedure tat could be used to find te derivative of any function tat by a metod I refer to as te Derivative Macine. Altoug tis metod is completely general, eac new function presents you wit a different set of algebraic callenges in order to find te derivative. However, once te derivative of a function is determined, tese functions can be put togeter in different ways by adding or subtracting tem, multiplying tem by constants, or multiplying and dividing tem to create more comple functions. Computational metods can be derived to compute te derivatives of tese more comple functions if you ave knowledge of a few basic derivatives witout aving to use te derivative macine. It is tese simple computational metods wic as made calculus a useful subject. If we ad to use te metods of te last capter for eac new function, calculus would not ave its present utility. In tis capter we will begin to sow you ow easy it is to compute derivatives. Additional simplifications will be introduced in Capter.. THE POWER LAW From te Eamples and Problems of te previous Capter you discovered te following derivatives: ) Do you notice a pattern to tese derivatives? In fact it can be sown tat, r r r were r is any real number. For eample, ). ) 5 4 /, / 5, and 6 Calculus and Structures

4 Section.. NEW FUNCTIONS AND THEIR DERIVATIVES FROM OLD ONES Consider te functions f and g were f = and g =. We can define new functions f, 5g and f g) as follows: f) = f = and 5g) = 5g=5. f g) f g. Similarly, f 5g) f 5g 5. If = for all i.e., te constant function) ten we could also define, 5g - f + ) = 5g - f + = wic is a second degree polynomial or quadratic function. In general, given two functions f and g we can define multiplication of a function by a scalar number) kf and te sum and difference functions, f g) as follows, a) kf) = k f a) b) f g) f g. b).4 DERIVATIVES OF THE NEW FUNCTIONS Now we would like to sow ow to differentiate tese functions. We ave seen tat a function f wit a smoot grap can be written near point as, f f ) f ' ) o ) a) o ) for and lim were te tangent line to te grap of te function, y = f, is, y line f ) f ' ) wit te slope m of te line at being te derivative of f or f ) Consider a second function, g. In a similar manner, g g ) g' ) o ). b) Te derivatives of te slopes of te function in a and b are given as followed: Calculus and Structures 6

5 Capter PROPERTIES OF DERIVATIVES a) kf: Let kf ) kf ) ) kf )' ) o ) were m kf )' ) 4a) Using Eq. a and a, kf ) kf kf ) kf ' ) o ) 4b) Comparing Eq. 4a and 4b, Eample : kf )' kf ' ) 6) If f ten f '. Terefore f ) f and f )' f ' 6 Eample : If f ten f '. Terefore, f ) f and f )' f ' ) b) f g : f g) f g) ) f g)' ) o ) 8a) Using Eq. b, a, and b, f g) f g f ) g ) f ' ) g' )) o ) o ) 8b) From te fact tat o ) o ) o ) and comparing Eq. 8a wit 8b, f g)' ) f ' ) g' ) 9) Eample : Consider f and g were f ' and g' ten f g)' f ' g' Eample 4: f and g were f ' and g' ten 4 f d4 5 Eample 5: 5g)' 4 f ' 5g)' 4 5 ) 8 5 or ) 8 5 f, g, and were were f =, g =, and 6 Calculus and Structures

6 Section.4 d5 ) =. Te derivative of te quadratic function of Eq. is: Summary: Te work of computing derivatives can be greatly simplified by using te following tree rules: a) r r r b) kf )' kf ' ) c) f g)' f ' ) g' Additional rules will be introduced in Capter.5 ADDING A CONSTANT FUNCTION Consider a function f wit derivative f. If te constant function, were = c for all is added to f, te value of te derivative does not cange because = as we sowed in Section 9.4, i.e., saying tis in anoter way, f )' f ' ' f ' f ' d f c) df f '.6 ANTIDERIVATIES: df If f is te derivative of F, i.e., f ten we can say tat F is te antiderivative of f. Te notation of tis is : F f Te logic of tis notation will become evident in Captes 5 and 6. In te last section we sowed tat wen adding a constant to any function te derivative of te function is uncanged. Terefore, if F is an anti-derivative of f i.e., F = f) were F f ), ten anoter anti-derivative is f c for any value of c. Calculus and Structures 6

7 Capter PROPERTIES OF DERIVATIVES Here are some simple antiderivatives: / or + c / or or + c + c ) Do you notice a pattern among in tese antiderivatives: r r a) c r for r -. Also it is easy to sow tat corresponding to te two derivative rules are two rules for antiderivatives; b) kf k f, and f g ) f g c) Wit tese tree rules many anti-derivatives can easily be computed. Eample 6: Eample 7: c 5 5 ) 5 5 c dy Eample 8: Find te function y = f wit t e properties tat wen =. and y = dy Solution: y c ) Since y = wen =, = c c = - Terefore, y, 64 Calculus and Structures

8 Section.7.7 OTHER ANTIERIVATIVES Any time you know te derivative of a function you also know its antiderivative. For eample, in Capter 9 we discovered tat, de e terefore e e c d ln terefore c ln Note: Absolute value of is used because logs are not defined for negative numbers. Remark: Tis fills in te case of r not accounted for in property a of Section.6. d sin cos terefore cos sin c d cos sin terefore sin cos c Rules b and c of Section.4 continue to old as illustrated by te following eamples: Eample : Eample : Eample : e e sin 5cos c cos 5sin c 5 5 ) c 5ln.8 DERIVATIVES AND ANTIDERIVATIVES FROM THE BEAM PROBLEMS We ave stated witout proof in Capters 7 and 8 tat: a) Te value of te bending moment d M equals te slope of te sear force, i.e., V making use of te interpretation of derivative as te slope of a tangent line; and b) te rate of decrease of te sear force equals te force density, dv, making use of te interpretation of derivate as a rate of cange. Compute te derivatives of te following previously determined bending moments and sear stresses to ceck tese ypoteses. Calculus and Structures 65

9 Capter PROPERTIES OF DERIVATIVES Eample : Te cantilever problem, Problem of Section 7.8: M 5 5 V = 5 lb. for 5 i.e., a weigtless beam) Eample 4: Eample of Section 8.5: M for for 4 V for 4 8 for 4 for Eample 5: Eample of Section 8.: { { M 5 5 V 5 Eample 6 : Eample of Section 8.4: M 44 V 44 6 Problem : For Eamples 5 and 6, use antiderivatives to compute M given V, i.e., M V c and te fact tat M = wen =. For Eample find M given V and te fact tat M = - 5 wen = Hint: Follow te approac of Eample 8). 66 Calculus and Structures

10 Capter Problems Problems Calculus and Structures 67