160 Chapter 3: Differentiation

Transcription

1 3. Differentiation Rules Differentiation Rules Tis section introuces a few rules tat allow us to ifferentiate a great variety of functions. By proving tese rules ere, we can ifferentiate functions witout aving to apply te efinition of te erivative eac time. Powers, Multiples, Sums, an Differences Te first rule of ifferentiation is tat te erivative of every constant function is zero. RULE 1 Derivative of a Constant Function If ƒ as te constant value ƒsx = c, ten ƒ x = sc = 0. x EXAMPLE 1 y If ƒ as te constant value ƒsx = 8, ten c (x, c) (x, c) y c f x = s8 = 0. x Similarly, 0 x x x x a- p b = 0 an x a3b = 0. FIGURE 3.8 Te rule s>xsc = 0 is anoter way to say tat te values of constant functions never cange an tat te slope of a orizontal line is zero at every point. Proof of Rule 1 We apply te efinition of erivative to ƒsx = c, te function wose outputs ave te constant value c (Figure 3.8). At every value of x, we fin tat ƒsx + - ƒsx ƒ sx :0 c - c :0 :0 0 = 0.

2 160 Capter 3: Differentiation Te secon rule tells ow to ifferentiate x n if n is a positive integer. RULE Power Rule for Positive Integers If n is a positive integer, ten x xn = nx n - 1. To apply te Power Rule, we subtract 1 from te original exponent (n) an multiply te result by n. EXAMPLE Interpreting Rule ƒ ƒ x x 1 x x 3 3x x 4 4x 3 Á Á HISTORICAL BIOGRAPHY Ricar Courant ( ) First Proof of Rule Te formula z n - x n = sz - xsz n z n - x + Á + zx n - + x n - 1 can be verifie by multiplying out te rigt-an sie. Ten from te alternative form for te efinition of te erivative, ƒsz - ƒsx ƒ sx z:x z - x = nx n - 1 z n - x n z:x z - x z:x sz n z n - x + Á + zx n - + x n - 1 Secon Proof of Rule If ƒsx = x n, ten ƒsx + = sx + n. Since n is a positive integer, we can expan sx + n by te Binomial Teorem to get ƒsx + - ƒsx ƒ sx :0 :0 :0 nxn :0 cnx n = nx n - 1 cx n + nx n nsn - 1 nsn - 1 sx + n - x n :0 nsn - 1 x n - + Á + nx n n x n - + Á + nx n n - x n x n - + Á + nx n - + n - 1 Te tir rule says tat wen a ifferentiable function is multiplie by a constant, its erivative is multiplie by te same constant.

3 3. Differentiation Rules 161 RULE 3 Constant Multiple Rule If u is a ifferentiable function of x, an c is a constant, ten u scu = c x x. y y 3x In particular, if n is a positive integer, ten x scxn = cnx n - 1. Slope 3(x) 6x 3 (1, 3) Slope 6(1) y x Slope x Slope (1) (1, 1) FIGURE 3.9 Te graps of y = x an y = 3x. Tripling te y-coorinates triples te slope (Example 3). Denoting Functions by u an Y Te functions we are working wit wen we nee a ifferentiation formula are likely to be enote by letters like ƒ an g. Wen we apply te formula, we o not want to fin it using tese same letters in some oter way. To guar against tis problem, we enote te functions in ifferentiation rules by letters like u an y tat are not likely to be alreay in use. x EXAMPLE 3 (a) Te erivative formula says tat if we rescale te grap of y = x by multiplying eac y-coorinate by 3, ten we multiply te slope at eac point by 3 (Figure 3.9). (b) A useful special case Te erivative of te negative of a ifferentiable function u is te negative of te function s erivative. Rule 3 wit c = -1 gives Proof of Rule 3 x s3x = 3 # x = 6x x s -u = x s -1 # u = -1 # su =-u x x. cusx + - cusx cu x :0 usx + - usx = c lim Limit property :0 = c u u is ifferentiable. x Te next rule says tat te erivative of te sum of two ifferentiable functions is te sum of teir erivatives. RULE 4 Derivative Sum Rule If u an y are ifferentiable functions of x, ten teir sum u + y is ifferentiable at every point were u an y are bot ifferentiable. At suc points, u su + y = x x + y x. Derivative efinition wit ƒsx = cusx

4 16 Capter 3: Differentiation EXAMPLE 4 Proof of Rule 4 Derivative of a Sum y = x 4 + 1x y x = x sx4 + x s1x = 4x We apply te efinition of erivative to ƒsx = usx + ysx: [usx + + ysx + ] - [usx + ysx] [usx + ysx] x :0 usx + - usx c :0 usx + - usx :0 Combining te Sum Rule wit te Constant Multiple Rule gives te Difference Rule, wic says tat te erivative of a ifference of ifferentiable functions is te ifference of teir erivatives. x su - y = u [u + s -1y] = x x Te Sum Rule also extens to sums of more tan two functions, as long as tere are only finitely many functions in te sum. If are ifferentiable at x, ten so is u 1 + u + Á u 1, u, Á, u n + u n, an x su 1 + u + Á + u n = u 1 x + u x + Á + u n x. + ysx + - ysx ysx + - ysx + lim :0 = u x + y x. y + s -1 x = u x - y x EXAMPLE 5 Derivative of a Polynomial y = x x - 5x + 1 y x = x x3 + x a4 3 x b - x s5x + x s1 = 3x # x = 3x x - 5 Notice tat we can ifferentiate any polynomial term by term, te way we ifferentiate te polynomial in Example 5. All polynomials are ifferentiable everywere. Proof of te Sum Rule for Sums of More Tan Two Functions We prove te statement x su 1 + u + Á + u n = u 1 x + u x + Á + u n x by matematical inuction (see Appenix 1). Te statement is true for n =, as was just prove. Tis is Step 1 of te inuction proof.

5 3. Differentiation Rules 163 Step is to sow tat if te statement is true for any positive integer n = k, were k Ú n 0 =, ten it is also true for n = k + 1. So suppose tat Ten x su 1 + u + Á + u k = u 1 x + u x + Á + u k x. x (u 1 + u + Á + u k + u k + 1 ) (++++)++++* ()* Call te function Call tis efine by tis sum u. function y. (1) = x su 1 + u + Á + u k + u k + 1 x Rule 4 for su + y x y y x 4 x Eq. (1) Wit tese steps verifie, te matematical inuction principle now guarantees te Sum Rule for every integer n Ú. EXAMPLE 6 = u 1 x + u x + Á + u k x + u k + 1 x. Fining Horizontal Tangents Does te curve y = x 4 - x + ave any orizontal tangents? If so, were? ( 1, 1) 1 1 (0, ) (1, 1) 0 1 FIGURE 3.10 Te curve y = x 4 - x + an its orizontal tangents (Example 6). x Solution Te orizontal tangents, if any, occur were te slope y>x is zero. We ave, Now solve te equation y x = x sx4 - x + = 4x 3-4x. y x = 0 for x: 4x 3-4x = 0 4xsx - 1 = 0 x = 0, 1, -1. Te curve y = x 4 - x + as orizontal tangents at x = 0, 1, an -1. Te corresponing points on te curve are (0, ), (1, 1) an s -1, 1. See Figure Proucts an Quotients Wile te erivative of te sum of two functions is te sum of teir erivatives, te erivative of te prouct of two functions is not te prouct of teir erivatives. For instance, x sx # x = x sx = x, wile x sx # x sx = 1 # 1 = 1. Te erivative of a prouct of two functions is te sum of two proucts, as we now explain. RULE 5 Derivative Prouct Rule If u an y are ifferentiable at x, ten so is teir prouct uy, an y suy = u x x + y u x.

6 164 Capter 3: Differentiation Picturing te Prouct Rule If u(x) an y(x) are positive an increase wen x increases, an if 7 0, y(x ) y y(x) 0 u(x) y u(x)y(x) u y y(x) u u(x) u u(x ) ten te total sae area in te picture is usx + ysx + - usxysx = usx + y + ysx + u - u y. Diviing bot sies of tis equation by gives As : 0 +, leaving usx + ysx + - usxysx = usx + y - u y. + ysx + u u # y : 0 # y x = 0, y suy = u x x + y u x. Te erivative of te prouct uy is u times te erivative of y plus y times te erivative of u. In prime notation, suy =uy +yu. In function notation, EXAMPLE 7 Fin te erivative of Using te Prouct Rule Solution We apply te Prouct Rule wit u = 1>x an y = x + s1>x: x c1 x ax + 1 x b = 1 x ax - 1 x b + ax + 1 x ba- 1 x b Proof of Rule 5 [ƒsxgsx] = ƒsxg sx + gsxƒ sx. x = - 1 x x 3 = 1 - x 3. y = 1 x ax + 1 x b. usx + ysx + - usxysx suy x :0 To cange tis fraction into an equivalent one tat contains ifference quotients for te erivatives of u an y, we subtract an a usx + ysx in te numerator: usx + ysx + - usx + ysx + usx + ysx - usxysx suy x :0 ysx + - ysx cusx + :0 usx + # ysx + - ysx lim :0 :0 + ysx As approaces zero, usx + approaces u(x) because u, being ifferentiable at x, is continuous at x. Te two fractions approac te values of y>x at x an u>x at x. In sort, y suy = u x x + y u x. y suy = u x x + y u x, an x a1 x b =-1 x by Example 3, Section.7. usx + - usx + ysx # usx + - usx lim. :0 In te following example, we ave only numerical values wit wic to work. EXAMPLE 8 Derivative from Numerical Values Let y = uy be te prouct of te functions u an y. Fin y s if us = 3, u s = -4, ys = 1, an y s =. Solution From te Prouct Rule, in te form y =suy =uy +yu,

7 3. Differentiation Rules 165 we ave y s = usy s + ysu s = s3s + s1s -4 = 6-4 =. EXAMPLE 9 Differentiating a Prouct in Two Ways Fin te erivative of y = sx + 1sx Solution (a) From te Prouct Rule wit u = x + 1 an y = x 3 + 3, we fin x CAx + 1BAx 3 + 3BD = sx + 1s3x + sx 3 + 3sx = 3x 4 + 3x + x 4 + 6x = 5x 4 + 3x + 6x. (b) Tis particular prouct can be ifferentiate as well (peraps better) by multiplying out te original expression for y an ifferentiating te resulting polynomial: y = sx + 1sx = x 5 + x 3 + 3x + 3 y x = 5x4 + 3x + 6x. Tis is in agreement wit our first calculation. Just as te erivative of te prouct of two ifferentiable functions is not te prouct of teir erivatives, te erivative of te quotient of two functions is not te quotient of teir erivatives. Wat appens instea is te Quotient Rule. RULE 6 Derivative Quotient Rule If u an y are ifferentiable at x an if ysx Z 0, ten te quotient u>y is ifferentiable at x, an u y x au y b = x - u y x y. In function notation, x c ƒsx gsxƒ sx - ƒsxg sx = gsx g. sx EXAMPLE 10 Fin te erivative of Using te Quotient Rule y = t - 1 t + 1.

8 166 Capter 3: Differentiation Solution We apply te Quotient Rule wit Proof of Rule 6 y t u = t - 1 an y = t + 1: = st + 1 # t - st - 1 # t st + 1 = t 3 + t - t 3 + t st + 1 = 4t st + 1. usx + x au y b ysx + - usx ysx :0 :0 ysxusx + - usxysx + ysx + ysx To cange te last fraction into an equivalent one tat contains te ifference quotients for te erivatives of u an y, we subtract an a y(x)u(x) in te numerator. We ten get x au y b ysxusx + - ysxusx + ysxusx - usxysx + :0 ysx + ysx usx + - usx ysx :0 - usx ysx + ysx t au ysu>t - usy>t y b = y ysx + - ysx Taking te limit in te numerator an enominator now gives te Quotient Rule.. Negative Integer Powers of x Te Power Rule for negative integers is te same as te rule for positive integers. RULE 7 Power Rule for Negative Integers If n is a negative integer an x Z 0, ten x sxn = nx n - 1. EXAMPLE 11 (a) Agrees wit Example 3, Section.7 x a1 x b = x sx -1 = s -1x - =- 1 x (b) x a 4 x 3 b = 4 x sx -3 = 4s -3x -4 =- 1 x 4

9 3. Differentiation Rules 167 Proof of Rule 7 Te proof uses te Quotient Rule. If n is a negative integer, ten n = -m, were m is a positive integer. Hence, x n = x -m = 1>x m, an x sxn = x a 1 x m b y y x x (1, 3) 1 y x 4 FIGURE 3.11 Te tangent to te curve y = x + s>x at (1, 3) in Example 1. Te curve as a tir-quarant portion not sown ere. We see ow to grap functions like tis one in Capter 4. 3 x EXAMPLE 1 Tangent to a Curve Fin an equation for te tangent to te curve at te point (1, 3) (Figure 3.11). Solution Te slope at x = 1 is = 0 - mxm - 1 x m = -mx -m - 1 = nx n - 1. Te slope of te curve is y x ` x = 1 y = x + x = c1 - x x = 1 Te line troug (1, 3) wit slope m = -1 is = x m # x A1B - 1 # x Axm B sx m y - 3 = s -1sx - 1 y = -x y = -x + 4. Quotient Rule wit u = 1 an y = x m Since m 7 0, x sxm = mx m - 1 Since -m = n y x = x sx + x a1 x b = 1 + a- 1 x b = 1 - x. = 1 - = -1. Point-slope equation Te coice of wic rules to use in solving a ifferentiation problem can make a ifference in ow muc work you ave to o. Here is an example. EXAMPLE 13 Coosing Wic Rule to Use Rater tan using te Quotient Rule to fin te erivative of expan te numerator an ivie by x 4 : y = sx - 1sx - x x 4 y = sx - 1sx - x x 4, = x3-3x + x x 4 = x -1-3x - + x -3.

10 168 Capter 3: Differentiation Ten use te Sum an Power Rules: y x = -x - - 3s -x -3 + s -3x -4 =- 1 x + 6 x 3-6 x 4. Secon- an Higer-Orer Derivatives If y = ƒsx is a ifferentiable function, ten its erivative ƒ sx is also a function. If ƒ is also ifferentiable, ten we can ifferentiate ƒ to get a new function of x enote by ƒ. So ƒ =sƒ. Te function ƒ is calle te secon erivative of ƒ because it is te erivative of te first erivative. Notationally, ƒ sx = y x = x ay x b = y x = y =D sƒsx = D x ƒsx. D Te symbol means te operation of ifferentiation is performe twice. If y = x 6, ten y =6x 5 an we ave y = y x = x A6x5 B = 30x 4. How to Rea te Symbols for Derivatives y y prime y y ouble prime y square y x square x y y triple prime y sn y super n n y to te n of y by x to te n x n D n D to te n Tus D Ax 6 B = 30x 4. If y is ifferentiable, its erivative, y =y >x = 3 y>x 3 is te tir erivative of y wit respect to x. Te names continue as you imagine, wit y sn = x y sn - 1 = n y x n = D n y enoting te nt erivative of y wit respect to x for any positive integer n. We can interpret te secon erivative as te rate of cange of te slope of te tangent to te grap of y = ƒsx at eac point. You will see in te next capter tat te secon erivative reveals weter te grap bens upwar or ownwar from te tangent line as we move off te point of tangency. In te next section, we interpret bot te secon an tir erivatives in terms of motion along a straigt line. EXAMPLE 14 Fining Higer Derivatives Te first four erivatives of y = x 3-3x + are First erivative: Secon erivative: Tir erivative: Fourt erivative: y =3x - 6x y =6x - 6 y =6 y s4 = 0. Te function as erivatives of all orers, te fift an later erivatives all being zero.