Introuction 4.3 Some Very Basic Differentiation Formulas If a ifferentiable function f is quite simple, ten it is possible to fin f by using te efinition of erivative irectly: f () 0 f( + ) f() However, tis process is quite teious. Also, as f gets more complicate, te limit gets increasingly more ifficult to evaluate. In tis section, some ifferentiation formulas are evelope to make life easier. First, some notation: NOTATION for te erivative prime notation; f as erivative f Tere are several notations use for te erivative. So far, te prime notation as been use: if f is ifferentiable at, ten te slope of te tangent line at te point (, f()) is te number f (). Te name of te erivative function is f ; f () is te function f, evaluate at. If y is a ifferentiable function of, ten its erivative can be enote, using prime notation, by y. For eample, if y = 2, ten y = 2. If it is esire to empasize tat y is being evaluate at a particular input c, one can write y (c). NOTATION for te erivative Leibniz notation; y as erivative y y ; evaluate at c is enote by eiter y (c) or y =c If y is a ifferentiable function of, ten its erivative can alternately be enote by y y. Tis is te Leibniz notation for te erivative. Rea as ee y, ee. For eample, if y = 2, ten y = 2. Again, if it is esire to empasize tat y y y is being evaluate at a particular input c, one can write (c) or =c. Tese latter two epressions can bot be rea as: ee y, ee, evaluate at c. In particular, te vertical bar is rea as evaluate at. Similarly, if f is a ifferentiable function of, its erivative in Leibniz notation is f (rea as ee f, ee ). If one wants to empasize tat tis erivative is being evaluate at a particular value, say c, ten one writes f f (c) or =c. One problem wit Leibniz notation is tat te name of te function an an output of te function are confuse. Wen one says: if y = 2, ten y = 2, te symbol y is really being use as bot te function name an its output. Strictly speaking, one soul write: if y = 2, ten y () = 2. However, tis is not common practice. an important use of Leibniz notation: te operator Te notation can be use to enote an instruction: acts on a ifferentiable function of to prouce its erivative. For eample, one can write: (3 1) = 3 an t (t2 ) = 2t an (2z + 1) = 2 z Tis notation is often use in stating ifferentation formulas. Also, it is convenient if you are aske to ifferentiate a function tat is not given a name. 204
copyrigt Dr. Carol JV Fiser Burns ttp://www.onematematicalcat.org 205 EXERCISE 1 practice wit notation Let f() = 2. Rewrite te following sentences about f, using prime notation. f 1. = 2 2. 3. 4. 5. 6. f (3) = 6 f =3 = 6 f t = 2t f t (3) = 6 f t t=3 = 6 Rewrite te following sentences using Leibnitz notation. 7. f () = 2 8. f (3) = 6 9. f (t) = 2t compiling some ifferentiation tools DIFFERENTIATION TOOL te erivative of a constant is 0 alternate notation We now begin to compile some tools tat will elp us ifferentiate functions more easily. Let f() = k, for k R. Ten f () = 0. Tis rule can be rewritten, using te operator, as follows: For every real number k : (k) = 0 PROOF Proof. Let f() = k, for k R. Ten, for every : Tus, f () = 0. f( + ) f() k k lim = 0 0 0 EXAMPLE Remember tat te symbol merely marks te en of te proof. If f() = π 2 5, ten f () = 0. If y = e 3, ten y ) = 0. = 0 ( 7 3 2 If f() = a + b, were a an b are constants, ten f () = 0. EXERCISE 2 Rewrite eac of tese eamples, using alternate notation.
206 copyrigt Dr. Carol JV Fiser Burns ttp://www.onematematicalcat.org DIFFERENTIATION TOOL constants can be sli out of te ifferentiation process alternate notation Suppose tat f is ifferentiable at, an let k R. Recall tat te function kf is efine by te rule: (kf)() := k f() Ten: (kf) () = k f () In wors, te erivative of a constant times a ifferentiable function is te constant, times te erivative of te function. Tis rule can be rewritten, using a miture of te operator an prime notation, as: ( ) kf() = k f () PROOF Proof. Let f be ifferentiable at, an let k R. It is necessary to sow tat te function given by (kf)() = k f() is ifferentiable at. (kf)( + ) (kf)() lim 0 kf( + ) kf() 0 0 k f( + ) f() f( + ) f() 0 (efn of kf) (factor out k) = k lim (prop. of limits, f iff. at ) = k f () (f is iff at ) Tus, te function kf is ifferentiable at, an as erivative given by: (kf) () = k f () Wat mae tis proof work? Properties of limits! EXAMPLE DIFFERENTIATION TOOL ifferentiating sums an ifferences Observe wat mae tis proof work: since we knew, a priori, tat f was ifferentiable at (so tat lim 0 f(+) f() eists), we were able to use te property of limits to slie te constant out. Te properties of limits will play a crucial role in te proofs of all te ifferentiation formulas. If f() = 2, ten f () = 2 () = 2(1) = 2. If is ifferentiable at, an f() = 2(), ten f () = 2 (). If y = 1 2t = 1 2 1 y t, ten t = 1 2 t( 1 ) t. (Tis last eample can be complete after te statement of anoter ifferentiation tool, te Simple Power Rule.) Suppose tat bot f an g are ifferentiable at. Ten te functions f + g an f g are also ifferentiable at, an: (f + g) () = f () + g () (f g) () = f () g () In wors, te erivative of a sum is te sum of te erivatives, an te erivative of a ifference is te ifference of te erivatives.
copyrigt Dr. Carol JV Fiser Burns ttp://www.onematematicalcat.org 207 alternate notation Tis rule can be rewritten, using a miture of te operator an prime notation, as: ( ) f() + g() = f () + g () PROOF Proof. It is sown first tat, uner te state ypoteses, f +g is ifferentiable at. Recall tat te function f + g is efine by te rule (f + g)() := f() + g(). Since, by ypotesis, bot f an g are ifferentiable at, it is known tat f () an g () eist. Ten: (f + g) () (f + g)( + ) (f + g)() : 0 f( + ) + g( + ) f() g() 0 f( + ) f() + 0 f( + ) f() 0 = f () + g () g( + ) g() + lim 0 g( + ) g() (efn. of erivative) (efn of f + g) (regroup) (limit of a sum, ypoteses) To see tat f g is ifferentiable at, we can now cite earlier results. Note tat: (f g)() := f() g() = f() + ( g()) = f() + ( g)() So, te function f g can be written as a sum of two functions, wit names f an g. Ten: (f g) () = f () + ( g) () (Wy?) = f () + ( g ()) (Wy?) = f () g () EXERCISE 3 1. Prove te previous result yourself, witout looking at te book. You coul be aske to write own a precise proof on an in-class eam. 2. Uner wat ypoteses is te limit of a sum equal to te sum of te limits? Was tis result use in te previous proof? Were? Were te ypoteses met? 3. Re-prove te fact tat (f g) () = f () g () (uner suitable ypoteses), but tis time DON T cite earlier results. Just use te efinition of erivative.
208 copyrigt Dr. Carol JV Fiser Burns ttp://www.onematematicalcat.org Does te rule apply wen tere are more tan 2 terms? Altoug te previous result is state for only 2 terms, oes it tell us tat, say, (f + g + ) () = f () + g () + (), proviing tat f, g an are all ifferentiable at? Of course! Just pull out te ol treat it as a singleton trick: (f + g + ) () = ( (f + g) + ) () (group) = (f + g) () + () (use result once) = f () + g () + () (use result again) EXERCISE 4 Prove tat, uner suitable ypoteses: (f + g + + k) () = f () + g () + () + k () SIMPLE POWER RULE ifferentiating n For all positive integers n : n = n n 1 More generally, if n is a real number, an I is any interval on wic bot n an n n 1 are efine, ten n is ifferentiable on te interval I, an: n = n n 1 EXAMPLE EXAMPLE rewriting te function, to make it fit te Simple Power Rule Here are some very basic applications of te Simple Power Rule: If f() = 2, ten f () = 2 2 1 = 2. Here, te Simple Power Rule was applie wit n = 2. 3 = 3 3 1 = 3 2 If y = 1007, ten y = 10071006. Here, te Simple Power Rule was applie wit n = 1007. Te slope of te tangent line to te grap of f() = 7 at te point (2, 2 7 ) is f (2) = 7(2 6 ). Here are some more avance applications of te Simple Power Rule. Te Simple Power Rule is use wenever te function being ifferentiate looks like (or can be mae to look like) n. Te laws of eponents, an fractional eponent notation, are use etensively to rewrite functions, to get tem into a form were te Simple Power Rule can be applie. Te Algebra Review in tis section reviews te necessary tools. Problem: Differentiate f() = 1. Solution: Rewrite te function as f() = 1. Taking n = 1 in te Simple Power Rule, one obtains: f () = ( 1) 1 1 = 2 = 1 2 On wat interval(s) is tis formula vali? It is vali on any interval for wic BOTH 1 an 1 are efine. Bot epressions are efine on R {0}. Tus, 2 te formula is vali for all real numbers, ecept 0.
copyrigt Dr. Carol JV Fiser Burns ttp://www.onematematicalcat.org 209 EXAMPLE Problem: Differentiate y =. Solution: Rewrite y, using fractional eponents, as y = 1/2. Taking n = 1 2 in te Simple Power Rule, one obtains: y = 1 2 1 2 1 = 1 2 1/2 = 1 2 1 = 1 1/2 2 On wat interval(s) is tis formula vali? Te epression is efine for 0. Te epression 1 2 is efine for > 0. BOTH epressions are efine on (0, ). Tus, te formula is vali for all positive real numbers. put te erivative in a form tat matces te original function form kn = nk n 1 It is always a goo iea to put te erivative in a form tat agrees, as closely as possible, wit te form of te original function. Since te original function in tis eample was given in raical form, y =, te erivative was also rewritten in raical form, y = 1 2. Using bot te Simple Power Rule an te fact tat constants can be sli out of te ifferentiation process yiels an etremely useful formula: kn = k n = k(n n 1 ) = kn n 1 Tus, for eample: If f() = 3 2, ten f () = 6. π11 = 11π 10 If y = 2, ten y = (1)( 2) 1 1 = 2 0 = 2. It is not necessary to write out all tese steps. You soul be able to recognize y = k as a line tat as slope k. Tus, y = k. Te slope of te tangent line to te grap of y = 3 5 at te point (1, 3) is y =1. Here, y = 154, so tat y =1 = 15(1) 4 = 15. EXERCISE 5 practice wit te Simple Power Rule For eac of te functions liste below, o te following: Write te function in te form f() = n. Differentiate, using te Simple Power Rule. On wat interval(s) is te formula obtaine for te erivative vali? Fin te equation of te tangent line to te grap of f wen = 1. 1. f() = 3 2. f() = 1 3. f() = 3 2
210 copyrigt Dr. Carol JV Fiser Burns ttp://www.onematematicalcat.org iea of proof of te Simple Power Rule Wen te eponent is a positive integer, te iea of te proof of te Simple Power Rule is very simple. Tis iea is illustrate by consiering a special case: Sow tat if f() = 3, ten f () = 3 2. f f( + ) f() () 0 ( + ) 3 3 0 0 ( 3 + one factor of {}}{ 3 2 + 0 (3 2 + 3 + 2 ) 0 (3 2 + 3 + 2 ) = 3 2 more tan one {}}{ 3 2 + 3 ) 3 (epan ( + ) 3 ) Pascal s Triangle A brief review of Pascal s Triangle, a tool for easily epaning (a + b) n for positive integers n, will enable you to repeat tis argument for iger values of n. Let a an b be any real numbers. Observe te following pattern: (a + b) 0 = 1 (a + b) 1 = (1)a + (1)b (a + b) 2 = (1)a 2 + 2ab + (1)b 2 (a + b) 3 = (a + b)(a + b) 2 = (a + b)(a 2 + 2ab + b 2 ) = a 3 + 2a 2 b + ab 2 + ba 2 + 2ab 2 + b 3 = (1)a 3 + 3a 2 b + 3ab 2 + (1)b 3 A triangle is forme. Eac new row is easily obtaine from te previous row by simple aition. It can be proven ( say, by inuction) tat tis pattern continues forever. fining ( + ) 7 For eample, suppose we want to epan ( + ) 7. Long multiplication woul be etremely teious. Instea, first write te appropriate types of terms in te epansion. Eac term as variable part i j, were te eponents a up to 7. Te first term as 7 an 0 ; te secon 6 an 1, te tir term as 5 an 2, an so on. So we get te term types: 7 6 5 2 4 3 3 4 2 5 6 7 Now, get te correct coefficients from Pascal s triangle (from te row beginning wit te numbers 1 7... ): Tus: (1) 7 + 7 6 + 21 5 2 + 35 4 3 + 35 3 4 + 21 2 5 + 7 6 + (1) 7 ( + ) 7 = 7 + 7 6 + 21 5 2 + 35 4 3 + 35 3 4 + 21 2 5 + 7 6 + 7
copyrigt Dr. Carol JV Fiser Burns ttp://www.onematematicalcat.org 211 EXERCISE 6 1. Use Pascal s triangle to epan ( + ) 9. 2. Use Pascal s triangle to epan ( ) 4. Hint: Write ( ) 4 = ( + ( )) 4, so te appropriate term types are: 4 3 ( ) 2 ( ) 2 ( ) 3 ( ) 4 3. Prove tat if f() = 4, ten f () = 4 3. Te complete proof of te Simple Power Rule woul take several pages, an we o not yet ave at our isposal all te necessary tools. However, a sketc of te proof is as follows: First prove te result wen is a positive integer. (An easier proof tan te one sketce above uses te prouct rule for ifferentiation.) Use te quotient rule for ifferentiation to eten te result to te negative integers. Use te formula for te erivative of an inverse function to eten te result to eponents of te form 1 n. Write p/q = ( 1/q ) p to eten te result to all rational eponents. Use te eponential function to make sense of irrational eponents: r = e r ln. (Here, we require to be positive.) Differentiate to complete te proof. DIFFERENTIATION TOOL ifferentiating e If f() = e, ten f () = e. Tus, te erivative of te eponential function is itself! Tis is a property of te eponential function tat is not sare by any oter function. Make sure you unerstan wat tis fact is saying: if you look at any point on te grap of te function e, ten te y-value of te point also tells you te slope of te tangent line to tat point!
212 copyrigt Dr. Carol JV Fiser Burns ttp://www.onematematicalcat.org iea of proof Let f() = e. Ten: If it can be sown tat lim 0 e 1 f( + ) f() e + e lim 0 0 e e e 0 lim 0 e ( e ) 1 e 0 ( e ) 1 = 1, ten we can complete te proof: = e lim 0 e 1 = e (1) = e A grap of g() := e 1 for values of close to 0 is sown, wic illustrates e te fact tat lim 1 0 = 1. DIFFERENTIATION TOOL ifferentiating ln If f() = ln, ten f () = 1. te result is believable Observe tat tis result is believable: wen is large, te slopes of tangent lines to te grap of ln are small; an wen is close to 0, te slopes are large an positive. EXAMPLE To ifferentiate functions involving e an ln, it is often necessary to first rewrite te function, using properties of eponents an logs. Tese properties are reviewe in te Algebra Review at te en of tis section. Problem: Differentiate f() = e 2+. Solution: First write f() = e 2+ = e 2 e. Ten, f () = e 2 e = e 2 e = e 2+. Anoter (easier) way to ifferentiate f will be possible after we stuy te Cain Rule for Differentiation.
copyrigt Dr. Carol JV Fiser Burns ttp://www.onematematicalcat.org 213 EXAMPLE Problem: Differentiate g() = ln 2. Solution: First write g() = ln 2 + ln. Ten: g () = 0 + 1 = 1 Anoter (easier) way to ifferentiate g will be possible after we stuy te Cain Rule for Differentiation. EXERCISE 7 Differentiate eac of te following functions. It will be necessary to first rewrite te functions, using properties of eponents an logaritms. 1. f() = e +5 ; interpret your result grapically. 2. f() = ln 7 3. Do you tink tat we ave te necessary tools yet to ifferentiate f() = e 2? Wy or wy not? 4. Do you tink tat we ave te necessary tools yet to ifferentiate g() = ln ( + 3)? Wy or wy not? A cart summarizing te tools evelope in tis section is given below: DIFFERENTIATION TOOLS prime notation if f() = k, ten f () = 0 (kf) () = k f () operator (k) = 0 ( kf() ) = k f () (f + g) () = f () + g () ( f() + g() ) = f () + g () (f g) () = f () g () ( f() g() ) = f () g () if f() = n, ten f () = n n 1 if f() = e, ten f () = e if f() = ln, ten f () = 1 n = n n 1 (e ) = e (ln ) = 1 ALGEBRA REVIEW raicals an fractional eponents, properties of logaritms raicals A raical is an epression of te form n, (*) for n = 2, 3, 4,.... Wen n = 2, (*) is written more simply as, an is rea as te square root of. Wen n = 3, 3 is rea as te cube root of. For n 4, n is rea as te n t root of.
214 copyrigt Dr. Carol JV Fiser Burns ttp://www.onematematicalcat.org meaning of n Te purpose of raicals is to uno eponents. Tat is, raicals provie a sort of inverse to te raise to a power operation. Unfortunately, te raise to a power functions f() = n are only 1 1 if n is o. Wen n is even, special consierations nee to be mae. ODD roots First consier f() = 3. Here, f is 1 1, an its inverse is te cube root function, f 1 () = 3. Tat is: For all real numbers an y : y = 3 y 3 = Reprasing, y is te cube root of if an only if y, wen cube, equals. Tus, 3 8 = 2, since 2 is te unique number wic, wen cube, equals 8. Also, 3 8 = 2, since 2 is te unique number wic, wen cube, equals 8. Inee, for all real numbers, an for n = 3, 5, 7, 9,..., n n =, since is te unique real number wic, wen raise to an o n t power, equals n.
copyrigt Dr. Carol JV Fiser Burns ttp://www.onematematicalcat.org 215 EVEN roots Wen n is even, f() = n is NOT 1 1. Consier, for eample, f() = 2. Here, (as for all even powers), R(f) = [0, ). Given z R(f), tere are TWO inputs wic, wen square, give z. Matematicians ave agree to coose te NONNEGATIVE number wic works. Precisely, we ave: For all 0 an for all real numbers y : y = y 0 an y 2 = Tat is, y is te square root of if an only if y is nonnegative, an y, wen square, equals. Tus, 4 = 2, since 2 is nonnegative, an 2 2 = 4. Te epression 4 is not efine, because tere is NO real number, wic wen square, equals 4. Wat is 2? Tere are TWO real numbers wic, wen square, give 2 : an. We nee to coose wicever is nonnegative. Te absolute value comes to te rescue: For all real numbers : 2 = Inee, for all nonnegative real numbers, an for all n = 2, 4, 6, 8,..., we ave: n n = EXERCISE 8 practice wit raicals 1. Consier tis matematical sentence: For all real numbers an y : y = 3 y 3 = (*) Tis sentence compares two component sentences. Wat are tey? Wat is (*) telling us tat tey ave in common? Wat is (*) telling us (if anyting) wen y = 2 an = 8? How about wen y = 2 an = 8? 2. Consier tis matematical sentence: For all 0 an for all real numbers y : y = y 0 an y 2 = (**) Wat two component sentences are being compare? Wat o tey ave in common? Wat is (**) telling us (if anyting) wen y = 2 an = 4? How about wen y = 2 an = 4? Evaluate te following roots. Be sure to write complete matematical sentences. State any necessary restrictions on an y. 3. 5 32 4. 4 ( 2) 4 5. 6. 6 6 9 9
216 copyrigt Dr. Carol JV Fiser Burns ttp://www.onematematicalcat.org fractional eponent notation Wen working wit raicals in calculus, it is usually more convenient to use fractional eponent notation rater tan raical notation. Wenever n is efine, it can be alternately written as 1 n. Tus: 1 5 = 5 2 3 = 1/3 for all real numbers 4 = 1/4 for all nonnegative real numbers Ten, using properties of eponents (wic are summarize below for your convenience), one can make sense of arbitrary rational eponents: p q = ( p ) 1 q = q p or p q = ( 1 q ) p = ( q ) p, provie tat bot q p an ( q ) p are efine. Use wicever representation is easiest for a given problem. PROPERTIES OF EXPONENTS Assume tat a, b, n an m are restricte to values for wic eac epression is efine. a m a n = a m+n (same base, multiplie, a eponents) a m a n = a m n (same base, ivie, subtract eponents) (a m ) n = a mn (power to a power, multiply eponents) (ab) m = a m b m ( a b )m = am b m a n = 1 a n a 0 = 1 for a 0 (prouct to a power, eac factor gets raise to te power) (quotient to a power, bot numerator an enominator get raise to te power) (efinition of negative eponents) (efinition of zero eponent) EXERCISE 9 Convince yourself tat eac of tese eponent laws makes sense. Just look at special cases, were convenient. For eample, for positive integers m an n : a m a n = m factors of a n factors of a {}}{{}}{ (a... a) (a... a) = m+n factors of a {}}{ a m+n
copyrigt Dr. Carol JV Fiser Burns ttp://www.onematematicalcat.org 217 EXAMPLE working wit fractional eponents Problem: Rewrite using fractional eponent notation. State any necessary restrictions on an y. Were possible, write in two ifferent ways. 1. y = 5 3 2. f() = 3 5 Solutions: 1. y = 5 3 = ( 3 ) 1/5 = 3 1 5 = 3/5 2. Observe tat D(f) = { > 0}. For suc : Alternately, if esire: 3 f() = 5 = 1/2 5/3 = 3 6 + 10 6 = 1/2 ( 5 ) 1/3 = 1 2 + 5 3 = 13 6 6 6 = 13 6 6 6 = 7/6 = ( 7 ) 1/6 = 6 7 7/6 = ( 1/6 ) 7 = ( 6 ) 7 All te steps were sown in te above isplay. You will probably be able to o a number of tese steps in your ea. properties of ln a precise view of functions Net, some properties of logaritms are reviewe. Wenever f is a function, ten every input as a unique corresponing output. In oter wors, wenever two inputs are te same (an peraps just ave ifferent names), ten tey must ave te same output. Precisely, wenever f is a function wit omain elements a an b : a = b = f(a) = f(b) (1) Tus, wenever te sentence a = b is true, so is te sentence f(a) = f(b). a precise view of a 1 1 function If f is, in aition, a 1 1 function, ten every output as a unique corresponing input. In oter wors, wenever two outputs are te same, ten tey must ave come from te same input. Precisely, wenever f is a 1 1 function wit omain elements a an b : f(a) = f(b) = a = b (2) Tus, wenever te sentence f(a) = f(b) is true, so is te sentence a = b. Putting (1) an (2) togeter, wenever f is a 1 1 function wit omain elements a an b : a = b f(a) = f(b) Tus, if two inputs are te same, so are te corresponing outputs (te function conition); an wenever two outputs are te same, so are te corresponing inputs (te 1 1 conition).
218 copyrigt Dr. Carol JV Fiser Burns ttp://www.onematematicalcat.org EXAMPLE Te function f() = e is 1 1 an as omain R. Tus, for all real numbers an y : = y e = e y Te function g() = ln is 1 1 an as as its omain te set of positive real numbers. Tus, for all positive real numbers an y : = y ln = ln y e an ln are inverse functions In aition, recall tat e an ln are inverse functions. Tus, a point (, y) lies on te grap of f() = e eactly wen te point (y, ) lies on te grap of g() = ln. Tat is, for all y > 0 an for all real numbers : y = e = ln y We are now in a position to verify some important properties of logaritms, wic are summarize below for convenience: PROPERTIES OF LOGARITHMS Assume tat a an b are restricte to values for wic eac epression is efine ln(ab) = ln a + ln b ln a b = ln a ln b ln a b = b ln a sample proof Te first equation says tat te log of a prouct is te sum of te logs. Here is its proof. Te remaining proofs are left as eercises. Let a > 0 an b > 0, so tat all tree epressions ln(ab), ln a, an ln b are efine. Ten: y = ln a + ln b e y = e ln a+ln b (e is a 1 1 function) e y = e ln a e ln b (properties of eponents) e y = ab (e an ln uno eac oter) ln e y = ln ab (ln is a 1 1 function) y = ln ab Tus, te sentences y = ln a + ln b an y = ln ab always ave te same trut values. Tat is, ln ab = ln a + ln b. EXERCISE 10 1. In wors, wat oes ln a b = ln a ln b say? Prove it. Be sure to justify every step of your proof. 2. Prove tat: ln a b = b ln a Be sure to write complete matematical sentences, an justify every step of your proof.
copyrigt Dr. Carol JV Fiser Burns ttp://www.onematematicalcat.org 219 QUICK QUIZ sample questions 1. Differentiate f() =. Write te erivative using bot prime notation, an Leibniz notation. 2. TRUE or FALSE: ( π 2 7+ 3 ) = 0 3. TRUE or FALSE: Te slope of te tangent line to te grap of y = 3 at te point (2, 8) is 12. Sow any work leaing to your answer. 4. Epan (a b) 4, using Pascal s Triangle. 5. Let g() = e + ln. Fin g (). KEYWORDS for tis section Prime notation for te erivative, Leibniz notation for te erivative, te operator, te erivative of a constant, constants can be sli out of te ifferentiation process, ifferentiating sums an ifferences, te Simple Power Rule for ifferentiation, Pascal s Triangle, ifferentiating e an ln, raicals an fractional eponent notation, properties of ln. END-OF-SECTION EXERCISES Differentiate te following functions. Feel free to use any tools evelope in tis section. 1. f() = (2 + 1) 3 2. g() = +1 7 { 3 2 2 + 1 1 3. () = 4 2 < 1 Wat is D()? Wat is D( )? { 3 2 2 + 1 1 4. () = 3 1 < 1 Wat is D()? Wat is D( )?