0.1 Differentiation Rules

Transcription

1 0.1 Differentiation Rules From our previous work we ve seen tat it can be quite a task to calculate te erivative of an arbitrary function. Just working wit a secon-orer polynomial tings get pretty complicate - imagine computing te erivative of a fift-orer polynomial. It is te same now as it was wit limits. Rater tan trying to calculate every possible limit iniviually, we sougt rules upon wic we coul combine known limits, in orer to save ourselves a great eal of work. We ave alreay notice some suc rules (an use tem) up to tis point, but ere we will formally acknowlege an evelop tese rules a bit furter. Te first ting tat we notice is tat te erivative of a line is just te slope of te line. Te immeiately corollary is tat te erivative of a constant is 0 (because a constant is just a orizontal line). Tere are two more basic ifferentiation results wic we will simply state at tis point - we will not gain a wole lot by trying to prove tem or justify tem as true. Tese are results for te erivatives of power an exponential functions. Teorem (Basic Derivatives). We ave te following results for erivatives x c = 0, c R x xp = p x p 1, p R, p 0 3. x ex = e x We ave alreay seen te erivative of a power function for p = 1 an p = 2, but te above result generalizes to all oter p as well. Te last of te above erivatives ses some ligt on te elusive number e. Te exponential function e x is suc tat its erivative is itself - te instantaneous rate of cange of e x at all points is equal to its current value. Tis is quite a remarkable property for a function to ave, an interestingly enoug te exponential function unerlies many pysical processes, suc as growt an ecay. At te moment we on t ave te tools to sow wy tis is true, but we ll use tis useful fact, aing one more function to our arsenal of ifferentiable functions. Example 1. Fin te erivative of x 2. Solution We apply te power rule to fin x x2 = 2x 2 1 = 2x. Example 2. Fin te erivative of x. Solution First we rewrite x = x 1/2 an ten apply te power rule. x x1/2 = 1 2 x1/2 1 = 1 2 x 1/2 = 1 2 x Since te square root function is not real value for x < 0, te function is not efine for x < 0, an its erivative is not efine for x 0, as ivision by 0 is forbien. Example 3. Fin te erivative of x 1. Solution Since x 1 = 1 x is not efine for x = 0, neiter is its erivative. Neverteless, we can apply te power rule to fin te value of te erivative at all oter points, to fin x x 1 = x 2. 1

2 Example 4. Fin te erivative of x 3/4. Solution Tis function, an tus its erivative, is not efine for x 0. We apply te power rule to fin te erivative for oter points, x x 3/4 = 3 4 x 7/4. Having looke at erivatives of tese basic builing blocks, we like to generalize our results to combinations of tese functions. We very easily gain results for linear combinations of suc functions. Teorem (Linearity of Differentiation). Suppose f an g are ifferentiable functions, wit erivatives f an g, respectively. It follows tat x (α f(x)) = αf (x). x (f + g)(x) = f (x) + g (x). Proof αf(x + ) αf(x) f(x + ) f(x) (α f(x)) = α lim = α f (x). x (f + g)(x + ) (f + g)(x) f(x + ) f(x) + g(x + ) g(x) (f + g)(x) x 0 f(x + ) f(x) g(x + ) g(x) + lim = f (x) + g (x). Te above properties are calle linearity, insofar tat ifferentiation respects linear combinations of functions. If we ifferentiate a linear combination of functions, te result is te same linear combination of teir erivatives. Te above properties migt also be referre to as te constant prouct an sum rules for ifferentiation. Since all polynomial functions are simply linear combination of power functions, tese rules give us te means to ifferentiate any polynomial, witout going troug te aruous task of starting from te efinition. Example 5. For a general object, surface area an volume are relate by S = cv 2/3. were c is a constant of proportionality, wic epens on te given object. Fin te erivative of surface area wit respect to volume, an escribe ow te proportionality between surface area an volume canges wit canging volume. Solution We must be careful in tat we cannot immeiate apply te power rule in tis situation, as we ave a power function multiplie by a constant. However, by linearity of ifferentiation, we can factor out te te constant, an ten apply te prouct rule. We fin S V = c V V 2/3 = 2 3 cv 1/3. Te rate at wic te surface area canges is proportionate to 1 3 V. Tus, as volume increases, te surface area increases more an more slowly; as an object gets large, te ratio of surface area to volume ecreases. 2

3 Example 6. Fin te erivative of 2e x + x 4. Solution Here we will use linearity of ifferentiation, in orer to see x (2ex + x 4 ) = 2 x ex + x x4 = 2 e x + 4x 3. Example 7. Fin te erivative of ax 3 + bx 2 + cx +, were a, b, c, R. Solution First we apply te sum rule, ten te constant prouct rule, an finally te power rule to fin tat x (ax3 + bx 2 + cx + ) = x ax3 + x bx2 + x cx + x = ax3 x + bx2 x + cx x + = 3ax2 + 2bx + c In some situations, we may wis to fin te erivative of polynomial like (x + 2) 4. Multiplying troug all of tese terms is a very long task. Fortunately, we can use te binomial teorem to greatly simplify te task. Teorem (Te Binomial Teorem). Te expansion of (x + y) n is a polynomial of egree n, an (x + y) n = x n + nx n 1 y + n! (n 2)!2 xn 2 y 2 n! (n k)!k! xn k y k nxy n 1 + y n. Te above equation is extremely aunting, so we will write te expansion for te first few values of n. (x + y) 2 = x 2 + 2xy + y 2 (x + y) 3 = x 3 + 3x 2 y + 3xy 2 + y 3 (x + y) 4 = x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + y 4 Te first result soul be very familiar, but te oters are unlikely so. Example 8. Fin te erivative of (2x + 1) 3. Solution One strategy woul to be expan te polynomial by an, but we can use te binomial teorem to fin tat (2x + 1) 3 = (2x) 3 + 3(2x) 2 + 3(2x) + 1 = 8x x 2 + 6x + 1. Now we apply our results from earlier to calculate te erivative. x (2x + 1)3 = x (8x3 + 12x 2 + 6x + 1) = 24x x + 6. In aition to fining te erivatives of linear combination of known functions, we can also fin te erivative of proucts of known functions. One woul probably guess immeiately tat te result is simply te prouct of te iniviual erivatives. However, we can verify immeiately tat tis is not true. Tink about multiplying te functions f(x) = x an g(x) = x togeter. Eac of tese iniviual erivatives are 1, yet te erivative of te prouct is 2x. Tus, te process of fining te erivative of a prouct of functions is not as simple as just fining te prouct of te erivatives. In orer to unravel tis mystery we ll nee to look at tings a bit ifferently. Let s suppose tat we ave two functions f an g. One way of interpreting te prouct f(x) g(x) is as an area of a square, were te value of eac of tese functions represents te lengt of one of te sies. In orer to fin te erivative of tis function, we re going to nee to look at (f g)(x + ) (f g)(x) f(x + ) g(x + ) f(x) g(x) lim. 0 3

4 If we tink about moifying tese functions by a little bit, to consier f(x + ) an g(x + ) ten we ave a new square, wit tese new respective lengts. For te sake of illustration, let us suppose tat f(x + ) an g(x + ) are larger tan f(x) an g(x) (tis is not require, but we woul nee to raw a number of pictures for eac ifferent situation oterwise). Wen we o so, we can view te original square as resiing insie tis new, larger square we ave create (see figure 1). Figure 1: Illustration of te areas involve in te erivative of a prouct of functions. Our next task is to look at tis new, larger area, in terms of te original area. Using te square f(x)g(x) as our guie, we break up tis larger square into four pieces. Wen we write te total area in terms of tese four squares, we fin tat or f(x + )g(x + ) = f(x)g(x) + ( g(x + ) g(x) ) f(x) + ( f(x + ) f(x) ) g(x) + ( f(x + ) f(x) )( g(x + ) g(x) ), f(x + )g(x + ) f(x)g(x) = f(x)g(x) + ( g(x + ) g(x) ) f(x) + ( f(x + ) f(x) ) g(x) + ( f(x + ) f(x) )( g(x + ) g(x) ). If we look at te left-an sie of te above equation, tis is exactly te form tat sows up in te efinition of te erivative of te prouct. We simply nee to ivie by an look in te limit as 0. We fin ten tat f(x + )g(x + ) f(x)g(x) (f g)(x) x 0 (g(x + ) g(x))f(x) (f(x + ) f(x))g(x) + lim (f(x + ) f(x))(g(x + ) g(x)) + lim = g (x)f(x) + f (f(x + ) f(x))(g(x + ) g(x)) (x)g(x) + lim. Now we still nee to evaluate tis final limit, corresponing to te little corner of te big square. Intuitively, as we let 0, bot sies of tis square are srinking on te orer of, so tis little square srinks muc faster tan te oters, an becomes negligible. Alternatively, we can tink 4

5 of multiplying tis term by a convenient coice of 1, /, an since bot f (x) an g (x) exist, it follows tat (f(x + ) f(x))(g(x + ) g(x)) lim 0 f(x + ) f(x) = f (x)g (x) lim = 0. 0 Tus, tis final term vanises. In summary, we ave te following result. g(x + ) g(x) Teorem (Prouct Rule for Derivatives). Suppose tat f an g are functions ifferentiable at x. It follows tat x (f g)(x) = f(x)g (x) + f (x)g(x). Example 9. Fin te erivative of p(x) = (x 1)(x + 1). Solution Let f(x) = x 1, an g(x) = x + 1. Since tese are bot linear functions wit slope 1, it follows tat f (x) = 1 = g (x). Now we can apply te prouct rule p (x) = f(x)g (x) + g(x)f (x) = (x 1) 1 + (x + 1) 1 = 2x. In tis situation, we can also multiply te two polynomials out, an fin tat p(x) = x 2 1 Applying te sum an prouct rules as we i in te last section we fin wic is consistent wit te prouct rule. p (x) = 2x + 0 = 2x Example 10. Suppose te volume of a plant is escribe by V (t) = t, were t is measure in ays, an V is measure in cm 3. Let te ensity of te plant ecrease accoring to ρ(t) = t, were ρ is measure in grams per cm 3. Describe te cange of plant s mass over time. Solution First recall tat mass is relate to volume an ensity by M(t) = ρ(t)v (t) Now we can apply te prouct rule to fin te rate of cange of te mass M (t) = ρ(t)v (t)+ρ (t)v (t) = ( t)12+(100+12t)( 0.05) = t 5 0.6t = t Looking at te rate of cange of mass as a function of time, we can see tat initially te mass is increasing, but te rate it is increasing at ecreases wit time. Eventually, wen t > 23 6, te erivative becomes negative, so te mass begins to ecrease. In aition to aving a means of fining te erivative of a prouct of two functions, we can also fin te erivative of a quotient of known functions. Tis process is less-intuitive, so we will not try an justify it at tis point. Wen we work wit te cain rule in te next section, we will see tat te quotient rule follows simply from te prouct an cain rules. 5

6 Teorem (Quotient Rule for Derivatives). Suppose u an v are bot ifferentiable at x, an v(x) 0. It follows tat u(x) x v(x) = v(x)u (x) u(x)v (x) [v(x)] 2. Tis rule can be remembere as low ig minus ig low over low square. Hig is te numerator of te quotient, were low is te enominator. Finally, represents te erivative of. Example 11. Fin te erivative of x2 +1 x 1. Solution Let u(x) = x 2 + 1, an v(x) = x 1. It follows tat u (x) = 2x an v (x) = 1. Now we can use te quotient rule u(x) x v(x) = v(x)u (x) u(x)v (x) v(x) 2 = (x 1) 2x (x2 + 1) 1 (x 1) 2 = 2x2 2x x 2 1 (x 1) 2 = x2 2x 1 (x 1) 2. Example 12. Te set of Hill functions is a family of functions use to escribe stimulus response in te stuy of biology. Hill functions take on te general form (x) = xn 1 + x n were n Z + (is a positive integer 1, 2,...). Calculate te erivative of te Hill functions for n = 1 an n = 2. Solution We will begin wit n = 1. Let u(x) = x an v(x) = 1 + x. It follows tat u (x) = v (x) = 1. Using te quotient rule we fin (1 + x) 1 x 1 = x (x + 1) 2 = 1 (x + 1) 2 Tis erivative is always positive for x > 0, an (0) = 1. Now let n = 2, so u(x) = x 2 an v(x) = 1 + x 2. It follows tat u (x) = v (x) = 2x. From te quotient rule we fin x = (1 + x2 ) 2x x 2 2x (x 2 + 1) 2 = 2x (x 2 + 1) 2. Tis erivative is also always positive for x > 0, but as te value (0) = 0. Using te above teorems an te erivatives for a few basic functions, we can now compute te erivatives of a large number of functions. In aition to using limits to evaluate erivatives, we can go troug te opposite process. If we can recognize a limit simply as te erivative of a known function, ten we can save a lot of work in evaluating te limit. 2(x + ) 2 2x 2 Example 13. Evaluate lim. Solution One way to work troug tis problem is to just blinly begin evaluating te limit. However, if we recognize te form of tis limit is simply te erivative of 2x 2, ten we can use te power rule an save ourselves all of te work. We fin tat 2(x + ) 2 2x 2 lim = x 2x2 = 4x. 6