Proof of the Power Rule for Positive Integer Powers


 Kenneth Melton
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1 Te Power Rule A function of te form f (x) = x r, were r is any real number, is a power function. From our previous work we know tat x x 2 x x x x 3 3 x x In te first two cases, te power r is a positive integer. In te last two, r is a rational number. In eac case, it appears tat x x r Tis is te Power Rule. We will first prove it for te case wen te power is a positive integer. Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 2/26 Proof of te Power Rule for Positive Integer Powers Consier two special case first. For r = 0, x 1 = Since, if f (x) = 1, ten. An for r = 1, Since, if f (x) = x, ten x x = f (x + ) f (x) = Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 3/26
2 Proof of te Power Rule for Positive Integer Powers Continue Now let f (x) = x n were n is an integer wit n > 1. Ten, using te binomial expansion, we ave f (x + ) f (x) Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 4/26 Proof of te Power Rule for Positive Integer Powers Continue Hence f (x) So tat, for any integer n wit n 0 we ave x x n = Even toug we ave only prove te Power Rule ere for n an integer wit n 0, we will use it for any power r. As we go along we will use oter rules to complete te proof of te Power Rule. Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 5/26
3 Some General Differentiation Rules Te Power Rule tells us ow to ifferentiate a specific class of functions. It gives us a specific formula for all functions in tat class. Oter ifferentiation rules tell us ow to ifferentiate combinations of any functions wose erivatives we know. Let c be a constant, ten some of tese rules are x x x c = [cf (x)] = [f (x) ± g(x)] = Combining tese rules wit te Power Rule we can ifferentiate any polynomial function. Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 6/26 Example 36 Using te Power Rule Fin te inicate erivative. (a) For y = 4x 5 3x 4 + 2x 3 4x 2 + 5x + 6 fin y x. (b) For f (x) = 4 x x 3 fin f (x). (c) For q(x) = 4x 3 fin q (1). Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 7/26
4 Solution: Example 36(a) Using te Power Rule togeter wit te Constant Multiple an Sum an Difference Rules gives: y x It is not necessary to follow troug all of te steps sown above. In fact, after te first couple of times you ifferentiate a polynomial you can go irectly to final result. Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 8/26 Solution: Example 36(b) & (c) Part (b): Rewrite te function as f (x) Ten te erivative is f (x) Part (c): Te Power Rule applies for any real number power. Tus q (x) Ten evaluating te erivative at gives q (1) Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 9/26
5 Example 37 Derivative Function of a Polynomial Let f (x) = x ( x 2 3 ). (a) Fin f (x) an f (x). (b) (c) () Determine te values of x for wic f (x) = 0. Give te corresponing points on te grap of f. Explain in terms of te grap of te f wat te points foun in part (b) represent. Explain te relationsip between te grap of f an te grap of te secon erivative. Solution (a): Rewrite te function as f (x) = x 3 3x. Ten applying te rules as in last example gives f (x) = Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 10/26 Solution: Example 37(b) & (c) Solution (b): If f (x) = 0, ten. Tis gives. Te points of te grap of f were f (x) = 0 are Solution (c): At a point were f (x) = 0, te slope of. Tis is a point were te grap of te function. Te grap of f looks like tis. At tese points, te, like tis. Tese points correspon to points were te grap of te erivative. Like tis. Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 11/26
6 Solution: Example 37(c) continue Furter note tat were te grap of f (x) is above te xaxis, f (x) > 0, te grap of f (x) as positive slope, so te grap of. An were te grap of f (x) is below te xaxis, f (x) < 0, te grap of f (x) as negative slope, so te grap of f (x) is. Tis is inicate using te symbols sown on te iagram, were an Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 12/26 Solution: Example 37() Were te secon erivative of te function is positive, tat is, were its grap is, te grap of te function is. Were te secon erivative of te function is negative, tat is, were its grap is, te grap of te function is. Te graps look like tis. In tis iagram we sow were te grap is an In orer to be able to pursue iger erivatives in more ept, an to see more interesting applications of te erivative, we first nee to evelop some more rules for ifferentiation. Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 13/26
7 Te Derivative of te Exponential Function A function of te form f (x) = a x, were te base a is any positive real number, is an exponential function. Let f (x) = 2 x. Fin f (x). Using Formula 3 we ave f (x) To evaluate, or at least estimate, tis limit, make a table similar to tat in Example 7. From te table you will fin tat lim So tat YUK!! x 2x Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 14/26 Te Derivative of te Exponential Function Continue In te same way, you will fin tat lim So tat Still YUK!! x 3x Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 15/26
8 Te Derivative of te Exponential Function Continue From te previous results we guess tat tere is a number between, call it e, for wic e 1 So tat lim 0 x ex Te exponential function wit te special number e as base, calle te natural exponential function, is te only function wose erivative is te function itself. Tis property arises in many applications of te exponential function. Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 16/26 A Limit Formula for e Recall tat in Example 29 we estimate tat lim 0 ln(1 + ) Using te properties of logaritms we can write tis as = 1 Since ln x is a continuous function, we can intercange te limit an ln in te result above to write = 1 Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 17/26
9 A Limit Formula for e continue But ln x = log e x, so tat ( ln lim (1 + )1/ 0. Tus, we must ave ) Try evaluating te expression for smaller an smaller values of, an you will get values closer an closer to. Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 18/26 Te Derivative of e rx Let f (x) = e rx were r is any real number. Te erivative of f (x) is f (x) = Cange variables in te limit above to. Ten te limit is lim 0 e r 1 = Hence x erx Tis is te Exponential Function Rule. Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 19/26
10 Example 38 A Learning Curve In a learning experiment subjects are given a list of wors wit efinitions. Te subjects are teste every ay to etermine te total number of wors tey ave learne up to tat point. At te en of te stuy it is foun tat, on average, te number of wors, N, learne after t is given by N = 1000 (1 e t/20) (a) (b) Fin an expression for te rate of cange wit respect to time of te number of wors learne by te subjects as a function of te time t in ays. Sketc te grap of N as a function te time t. From te grap etermine te maximum number of wors tat te subjects coul learn. Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 20/26 Continuing Example 38 (c) () Fin te time wen a subject as learne alf of te maximum number of wors. Determine te rate of cange of te number of wors learne at tis time, an explain wat tis value tells you. Sow tat N satisfies te ifferential equation N t = 1 (1000 N) 20 an explain te meaning of tis equation. Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 21/26
11 Solution: Example 38(a) Te rate of cange wit respect to time of te number of wors learne is te erivative of te function giving N in terms of t. Applying Constant Multiple Rule an te Sum an Difference Rule gives N t Now applying te Exponential Function Rule wit r = 1/20 gives N t Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 22/26 Solution: Example 38(b) First note tat N is. Tis t means tat te grap of N is. Furter, recalling te limits at infinity for te exponential function, we ave lim t 50e t/20 an ( lim e t/20) t Hence, te for te grap of N, an te grap. So far we ave tis for te grap. Furter, te erivative gets smaller an smaller for larger t, so te grap of. Te grap looks like tis. Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 23/26
12 Solution: Example 38(b) Comment Te secon erivative is 2 N t 2 wic is always so tat te grap is. Tis is anoter confirmation tat te grap is always below te orizontal asymptote. Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 24/26 Solution: Example 38(c) Since te maximum number of wors tat can be learne is 1000, we want to fin t for wic 1000 (1 e t/20) Solving for e t/20 gives logaritm of tis equation to give To solve for t take te natural t 20 = So it takes about, on average, for te subjects to learn alf of te wors. At tis time te rate of cange is N t = 25 Te units of tis rate of cange are an it means tat from te. Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 25/26
13 Solution: Example 38() In part (a) we foun tat N t = 50e t/20 Now N = 1000 (1 e t/20) Combining tis wit te expression above for N t gives N t = Tis equation says tat te rate at wic a subject learns wors is te rate of learning.. Tis means tat Clint Lee Mat 112 Lecture 10: Differentiation Power & Exponential Functions 26/26
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