Basic Derivative Properties


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1 Bsic Derivtive Properties Let s strt this section by remining ourselves tht the erivtive is the slope of function Wht is the slope of constnt function? c FACT 2 Let f () =c, where c is constnt Then f 0 () =0 Proof If f () =c, then Figure 2: The slope of horizontl line y = c is 0 t ech point, so the erivtive is 0 t ny point f 0 f ( + h) f () c c 0 () =lim = 0 Geometriclly this result is obvious The grph of the constnt function f () =c is horizontl line with height c The geometriclly the erivtive of f is slope of the grph of the function Becuse the grph is horizontl line, the slope of the function is 0 t ech point, ectly s we foun with the limit process Differentibility n Continuity If we think of continuous function s mening the the grph is unbroken there re no gps or holes, then ifferentible function is one tht it s smooth it hs no corners, let lone gps or holes Intuitively then, it woul seem tht ifferentible function must be continuous This is, inee, the cse THEOREM 20 [Differentible implies Continuous] If f is ifferentible t =, then f is continuous t = Proof Becuse f is ifferentible t =, we know two things First, by efinition of the erivtive, f 0 f () f () () =lim! eists (2) Secon, whenever 6=, then f () f () = or, whenever 6=, then f () = f () f () f () ( ), f () ( )+ f () (22)
2 mth 30 more on erivtives: y 3 2 Now tke the limit in eqution (22) using the efinition of the erivtive in eqution (2): f () f () lim f () =lim ( )+ f ()!! = f 0 () 0 + f () = f () Since lim! f () = f (), by efinition f is continuous t = Sometimes sying the sme thing in ifferent wy cn be more helpful Notice tht by the theorem bove we cn sy: If f is not continuous, then f cnnot be ifferentible, becuse if it were ifferentible Theorem 20 woul sy tht f h to be continuous In other wors, THEOREM 202 [Not Continuous implies Not Differentible] If f is not continuous t =, then f is not ifferentible t = A wor of cution Other wys to fil to be ifferentible Be creful Neither version of the theorem tells us wht hppens if we strt with continuous function In prticulr, if f is continuous t =, we cnnot sy whether f is ifferentible EXAMPLE 203 (Emple of continuous function tht is not ifferentible) Show tht f () = is continuous t = 0 but is not ifferentible there SOLUTION First we nee to show tht is continuous t 0 This mens showing tht lim = 0 This is obvious from the grph (see Figure 22), but let s check it Becuse is piecewise function, we nee to check the left n right limits s pproches 0 From the left From the right, lim = < 0 lim liner = 0 lim = > 0 lim liner = Since the left n right limits both re 0, it follows tht lim = 0 Since f (0) = 0 = 0, we see tht f () is continuous t = 0 To show tht f is not ifferentible t 0, we use the efinition of the erivtive of f t point, lim f () f (0) 0 Agin, becuse of the piecewise nture of the, we must check the onesie limits So From the left lim From the right, f () f (0) < 0 = 0 Figure 22: The function f () = is efine by (, if 0 =, if < 0 is continuous t = 0, but is not ifferentible there becuse there is corner The slope s pproches 0 from the left is n from the right is + f () f (0) 0 > 0 lim = f () f (0) Since the left n right limits re ifferent, it follows tht lim oes not 0 eist In other wors, the function is not ifferentible t = 0 Here re four emples of functions tht re continuous t = but which re not ifferentible Lecture0te Version: Mitchell206/09/304:42:54
3 3 () Figure 23: () The grph is continuous (unbroken) but hs corner, so there re two ifferent slopes t = epening on whether we pproch from the left or the right So f is not ifferentible t, tht is, f 0 () oes not eist (b) (b) The grph is continuous (unbroken) but hs verticl tngent t = (infinite slope) So f is not ifferentible t, tht is, f 0 () oes not eist (c) (c) The grph is continuous (unbroken) but hs cusp t = Sof is not ifferentible t, tht is, f 0 () oes not eist () () The grph is continuous (unbroken) but hs corner t = So there re two ifferent slopes t = epening on whether we pproch from the left or the right So f is not ifferentible t, tht is, f 0 () oes not eist
4 mth 30 more on erivtives: y Minute Review: Noticing Pttern In the lst few ys we hve clculte severl erivtives Using the clcultions from our previous clss, the lb ticket, n lb fill in the erivtives of ech of these power functions f () p = /2 f 0 () If you spot pttern, if f () = 4 wht shoul f 0 () be? If f () = 4, wht shoul f 0 () be? Bonus: If f () = p, wht shoul f 0 () be? Other Nottion for Derivtives The evelopment of clculus is often ttribute to two people, Isc Newton n Gottfrie Leibniz, who inepenently worke its fountions Although they both were instrumentl in its cretion, they thought of the funmentl concepts in very ifferent wys It is interesting to note tht Leibniz ws very conscious of the importnce of goo nottion n put lot of thought into the symbols he use Newton, on the other hn, wrote more for himself thn nyone else Consequently, he tene to use whtever nottion he thought of on tht y This turne out to be importnt in lter evelopments Leibniz s nottion ws better suite to generlizing clculus to multiple vribles n in ition it highlighte the opertor spect of the erivtive n integrl As result, much of the nottion tht is use in Clculus toy is ue to Leibniz Nonetheless, there re severl wys re commonly use to inicte the erivtive of function This prgrph is tken from The History of Clculus" t mthuheu/~tomfore/clchistory html NOTATION Assume tht y = f () is ifferentible function Then the erivtive function of f my be enote by ny of the following: f 0 () =y 0 = y 0 () = y = f ( f ()) = = D ( f ()) The nottion y Dy is ment to remin us of the slope formul D tht motivtes the efinition of the erivtive The nottion D ( f ()) is especilly useful becuse we cn use it to isply both the originl function n its erivtive quite compctly For emple, inste of writing the sentence If f () = 2, then f 0 () =2, (23) we cn write D ( 2 )=2 or (2 )=2 (24) This rises n importnt point with which beginning clculus stuents sometimes hve ifficulty: using proper nottion It is ll too esy to incorrectly write f () = 2 = 2 (25) This confuses the originl function with its erivtive Obviously the two functions 2 n 2 re not the sme, but tht s wht the equl sign mens in eqution (25) So be sure to use the correct nottion s in (23) or (24) bove Lecture0te Version: Mitchell206/09/304:42:54
5 5 NOTATION To reference the erivtive of function t specific point when =, use the ny of the following nottion: f 0 () =y 0 () = y = = f = For emple, if f () = 2, we hve seen tht f 0 () =2 So to inicte the slope or instntneous rte of chnge in f t = 5, we might write f 0 (5) =0 or y =5 = 0 Let s use our new nottion in etermining nother erivtive EXAMPLE 24 Determine the erivtive of f () = p Note: The omin of f () is (0, ) SOLUTION Using the efinition of the erivtive, pple p p p = +h lim h!0 p p p +h +h p h h!0 p p p +h +h p h p p + + h p p + + h ( + h) p + h p ( p + p + h) h p + h p ( p + p + h) p p p p h!0 + h ( + + h) Cont = p p ( p + p ) (2 p ) 2 3/2 2 3/2 In other wors, D /2 = 2 3/2 This fits the erlier pttern seen in the tble on pge 4 It ppers tht D ( r ) = r r It ppers tht D ( r ) = r r The Derivtive of f () = n With bit of creful lgebr n the use of some bsic limit properties, we cn now etermine generl formul for the erivtive of f () = n, t lest when n is positive integer (From the pttern tht we hve observe, we epect the nswer to be f 0 () =n n ) Before we begin, we nee to mke n observtion bout fctoring This is best illustrte with the following emple EXAMPLE 25 Multiply out n simplify the prouct ( )( ) SOLUTION We fin ( )( )= = 4 4 (26) Notice by iviing by we cn epress this prouct s quotient = 4 4
6 mth 30 more on erivtives: y 3 6 YOU TRY IT 2 Notice the cncelltion of ll but the first n lst terms in eqution (26) bove Simplify the prouct ( )( ) For ny positive integer n >, Determine the following prouct ( )( n + n n 3 n 2 + n ) Finlly rewrite ech epression s equivlent nswer to you try it n n n, respectively As quotients, = 5 5 n n + n n 3 n 2 + n = n n Now we re rey to prove the generl result THEOREM 26 (Power Rule for Derivtives) Let n be positive integer The erivtive of f () = n is f 0 () =n n Proof Let n be positive integer n let be ny rel number We will etermine the formul for f 0 () By efinition, f 0 f () () =m tn! f () You Try It 2 n! + n n 3 n 2 + n = n n Poly =! n n z n times } { n + n + n + + n + n Tht is, f 0 () =n n, so f 0 () =n n Tht is, for ny positive integer n, [n ] = n n Wow! This mens we cn clculte lot s of erivtives without hving to evlute limit epression EXAMPLE 27 Determine the erivtives of f () = 9, g(t) =t 99, n f (s) =s 203 SOLUTION Using Theorem 26, we know tht [n ] = n n so D ( 9 )=9 8, D t (t 99 )=99t 98, n D s (s 203 )=203s 202 Notice the use of proper nottion, incluing the correct vrible nmes Though we hve not proven it yet, there is more generl version of the power rule tht pplies to ny nonzero eponent tht hs the sme form s the Power Rule for Derivtives THEOREM 28 (Generl Power Rule for Derivtives) Let r be ny nonzero rel number Then [r ] = r r EXAMPLE 29 Determine the erivtives of f () = 2/3, g(t) =t 9/2, f (s) =s 203, n g(w) = w 5/4 Lecture0te Version: Mitchell206/09/304:42:54
7 7 SOLUTION Using Theorem 28, we know tht [r ] = r r So () D ( 2/3 )= 2 3 /3 (b) D t (t 9/2 )= 9 2 t7/2 (c) D s (s 203 )= 203s 204 () D w w 5/4 = D w w 5/4 = 5 4 w 9/4 Be especilly creful with negtive eponents to subtrct from the eponent Be sure to use proper nottion, incluing the correct vrible nmes YOU TRY IT 22 Determine the erivtives of f () = 5/9, g(t) = 9 t, n h(w) =w 4/7 nswer to you try it 22 f 0 () = 5 9 4/9, g 0 (t) = 9t 20, n h 0 (w) = 4 7 w /7 Algebric Properties of Derivtives As we i with limits, we will now evelop some bsic properties of erivtives of sums n constnt multiples of ifferentible functions Precisely becuse erivtives re limits, we cn employ the corresponing limit properties to help us evelop these rules THEOREM 20 (Constnt Multiple n Sum/Difference Properties) Assume tht f n g re ifferentible functions n tht c is constnt Then () [cf()] = cf0 (); (2) [ f () ± g()] = f 0 () ± g 0 () In other wors, the erivtive of sum is the sum of the erivtives Theorem 20 is bout orer of opertions For emple, prt (2) sys tht we tke the sum first n then clculte the erivtive or we cn tke the iniviul erivtives first n then them The result will be the sme Most of the time it is esier to etermine the iniviul erivtives first n then But occsionlly, the other orer is simpler How woul you escribe the orer of opertions in prt ()? Theorem 20 is very powerful tool Before we give its proof, let s emine how to use this result EXAMPLE 2 Determine the erivtive of h() =9 7 6 /2 + 5, where > 0 SOLUTION Using our erlier work, we know tht 7, /2 n 5 re ll ifferentible functions So we my pply Theorem 20 First we split h() into three pieces, D (9 7 6 /2 + 5) Theorem 20, prt 2 = D (9 7 ) D (6 /2 )+D (5) then fctor out the constnts Theorem 20, prt = 9D ( 7 ) 6D ( /2 )+D (5) n now use the Generl Power Rule n the erivtive of constnt function Theorem 28, Emple = 9(7 6 ) 6 2 /2 + 0 = /2 This is substntilly simpler thn ctully hving to use limit process to etermine the erivtive To be ble to use Theorem 20 we nee to emonstrte why it is true
8 mth 30 more on erivtives: y 3 8 Proof To prove prt of Theorem 20, ssume tht f is ifferentible function n tht c is constnt Then by efinition of the erivtive, cf( + h) cf() D (cf()) Using the Constnt Multiple Property for Limits f ( + h) f () = c lim An finish by using the the fct tht f is ifferentible = cf 0 () To prove prt 2 of Theorem 20, tht f n g re ifferentible functions Then by efinition of the erivtive, [ f ( + h)+g( + h)] [ f ()+g()] D ( f ()+g()) which we cn rerrnge s [ f ( + h) f ()] + [g( + h) g()] n then use the Sum Property of Limits f ( + h) f () g( + h) g() + lim Finish by using the the fct tht both f re ifferentible = f 0 ()+g 0 () Here re few more emples EXAMPLE 22 Determine the erivtives of the following functions, where they eist () f () = (b) g() = p 2 (c) s(t) =(t + )(t 3 + 2) () f (s) = 8s2 + 7s + 2 s SOLUTION In ech of these we mke use of the Generl Power Theorem, the Sum/Difference Theorem, n the Constnt Multiple Theorem for erivtives () For f () = , D ( ) Thm 20(2) = D (3 2 ) D (2 3 )+D (8) Thm 20() = 3D ( 2 ) 2D ( 3 )+D (8) Thm 26, Const = = (b) For g() = p 5 D /3 Thm 20(2) 2 2, rewrite in eponent form Thm 20() = Thm 26, Const = = D (0 /3 ) D (2) 5 = D 0 0 D ( 5 )+0D ( /3 ) D (2) / /3 3 Lecture0te Version: Mitchell206/09/304:42:54
9 9 In the net two we combine some steps (c) For s(t) =(t + )(t 3 + 2), first multiply out D t (t + )(t 3 + 2) = D t t 4 + t 3 + 2t + 2) Thm 20(2) = D t t 4 + D t (t 3 )+D t (2t)+D t (2) Thm 20(), Thm 26, Const = 4t 3 + 3t = 4t 3 + 3t () For the finl one, first simplify by iviing through by s 8s 2 + 7s + 2 D s = D s 8s s ) s Thm 20(2) = D t t 4 + D ( 8s)+D s (7)+D s (2s ) Thm 20(), Thm 26, Const = s 2 = 8 2s 2 Theorem 20 will be even more useful once we hve etermine (generl) erivtive formuls for functions other thn just powers of or constnts Such itionl functions inclue eponentil n log functions, the trig functions, n the inverse trig functions In the net section we look t erivtives of the eponentil function
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