170 D CHAPTER 3 DI FF ERENTIATIONRULES (c) Graph of 1, q, g, h, and : The graph ofthe five functions as a piecewise-defined function : 50 g - 5 0 1-",------. ''---...:..,...~,..-+--j 150 1----'--'--c:,+'-' - -+---''_< -t---!150 L, - 100 This is the piecewise-defined function assignment on a TI-83 Plus calculator, where Y = 1, Y 6 = g, Y 5 = q, Y 7 = h, and Y 3 =. A compari son ofthe graphs in part I(c) and part (c): 0 1--71"'------f-~...- - -! 110 3. The Product and Quotient Rules 1. Product Rule: y = ( + 1)( 3 + 1) =? y' = ( + 1)(3 ) + ( 3 + 1)( ) = 3 4 + 3 + 4 + = 5 4 + 3 +. Multiplying first: y = ( + 1)( 3 + 1) = 5 + 3 + + 1 =? y' = 5 4 + 3 + (equivalent).. - 3 h X- 3X 3 /. Quotient Rule: F() = h = X 1 / X 1 / (1 - X 1 / ) _ ( - 3X 3 / ) (.lx- 1 / ) 1/ 9 1 1/ 3 1 1/ '( ) X - " - " + " " - 3 1-1/ F = = = = - - 3 (X1/) X X Simplifying first: F () = - ~h = h _ 3 = X 1 / - 3 =? F ' ( ) = ~ X - 1 / - 3 (equivalent). For this problem, simplifying first seems to be the better method. 3. By the Product Rule, f ( ) = ( 3 + )e X =? j'( ) = ( 3 + )(e )' + e X ( 3 + ) ' = ( 3 + )e X + ex(3 + ) = e X [( 3 + ) + (3 + )] = e X ( 3 + 3 + + ) ex 5. By the Quotient Rule, y = ex 6. By the Quotient Rule, y =- l + (+ 1)'
PR QR The notations ==> and ==> indicate the use ofthe Product and Quotient Rules, respectively. SECTION 3. THE PRODUCT AND QUOTIENT RULES 0 171 7 () == 3-1 QR, () == ( + 1)(3) - (3-1)() == 6 + 3-6 + == 5. 9 + 1 ==> 9 ( + 1) ( + 1) ( + 1) ) 8. f (t) == 4 +t ~ f' (t) == ( 4 + t () - (t) (t) == 8 + t - 4t == 8 - t t (4 + t ) (4 + t ) (4 + t ) 9. V() == ( 3 + 3)( 4 - ) @ V'() == ( 3 + 3)(4 3 - ) + ( 4 - )(6 ) == (8 6 + 8 3-6) + (6 6-1 3) == 14 6-4 3-6 3)(US 10. Y(u) == (u- + u- - u ) @ 3)(5u4 Y'(u) == (u- + u- - 4u) + (US - u )( -u- 3-3u- 4) == (5u - 4u- 1 + 5u - 4u- ) + (-u - 3u + 4u- 1 + 6u- ) == 3u + u + u- 11. F(y) = (y~ - :4)(y + 5y3) = (y- - 3y-4)(y + 5y3) @; p'(y) == (y- _ 3y-4)(1 + 15y) + (y + 5y3)(_y-3 + 1y-S) == (y- + 15-3y-4-45y-) + (-y- + 1y-4-10 + 60y-) == 5 + 14y- + 9y-4 Of 5 + 14/y + 9/y4 1. R(t) == (t + e t ) (3 - Vi) ~ R'(t) = (t + e t ) ( _~Cl/) + (3 ~ 0)(1+ e t ) t) t t = ( _~tl/ - ~cl/et) + (3 + 3e t - 0-0 e = 3 + 3e t - ~0-0 e - e / (0) 14 _ + 1 QR. Y - 3 + X - ==>, ( 3 + X - )(1) - ( + 1)(3 + 1) 3 + X - - 3 3-3 - -I - 3-3 - 3 y - - - ( 3 + X - ) - ( 3 + X - ) - ( 3 + X - ) 3 + 3 + 3 Of------- ( - 1)( + X + ) 15. == t + QR Y t4-3t + 1 ==>, (t 4-3t + 1)(t) - (t + )(4t 3-6t) t[(t 4-3t + 1) - (t + )(t - 3)] y == (t4-3t + 1) == (t 4-3t + 1) t(t 4-3t + 1 - t 4-4t + 3t + 6) t(_t 4-4t + 7) (t 4-3t + 1) (t 4-3t + 1)
17 D CHAPTER 3 DIFFERENTIATION RULES t t QR 16. y == == =? (t - 1) t - t + 1 I (t - t + 1) (1) - t (t - ) (t - 1) - t (t - 1) (t - 1) [(t - 1) - t] - t - 1 y == [(t - 1)] == (t - 1)4 == (t - 1)4 == (t - 1)3 1 QR (s + kes)(o) - (1)(1 + ke S) I 18. Y == --k- =? Y s + e S (s+ke s) 1/ 1 v.jv - 1 V 3 / - 1 We can change the form ofthe answer as follows: v - V- = v -.JV =.JV =.JV t QR, ( + t 1 / ) () - t (~rl/) 4 + t 1 / _ t 1 / 4 + t 1 / 1. f(t) = + y't =} f (t) = ( + y't ) = ( + y't ) = ( + y't) or /7. 1/. g(t) == ~ == _t t _ == t /3 _ t 1/6 =? gl(t) == ~t-1/3 _ l.t-5/6 t1/3 t1/3 t 1/3 3 6 3. f() == A B+ C ex X)( X)(l fl(x) == ( + e -ey - (1 - e + ex) +e ( + e X ) 4. f() == 1 - e ~ ) PR I ( + ex)[-(e X + ex 1)] - (1 + ex - e - e =? f () == ( + ex ) _e - e - e - e - 1 - ex + e + e _e - e - ex - 1 (+e ) (+e ) ), fl(x) == ( + e/)(l) - (l - e/ == + e/ - + e/ == e/ e 5. f () == --/- =? + c ( + ;) ( : c ) ( ; c) ( + e) 6: f () == a + b =? fl () == (e + d) (a) - (a + b)(e) == ae + ad - ae - be == ad - be _ e + d (e + d) (e + d) (e + d) 4e 4e 7. f() == =? fl(x) == + ex 4 3 == ( 4 + 4 3)e X [or 3e( + 4)J =? i"() == ( 4 + 4 3)e + ex(4 3 + 1 ) == ( 4 + 4 3 + 4 3 )e + 1 X == ( 4 + 8 3 + 1 )e X [or ex( + )( + 6)J 8. f() = X 5 / ex =}.f'() = 5 e / + ex. ~X3/ = (X 5 / + ~X3/) ex [or ~X3/ ex( + 5)] =}.f"() = (X 5 / + ~X3/ )e + ex (~X3/ + Jf 1/) = (X 5 / + 5X 3/ + JfX 1/ )e [or ~l/e (4 + 0 + 15)]
SECTION 3. THE PRODUCT AND QUOTIENT RULES D 173 () ::::} f'() = (1 + )() - = + 4 - = + ::::} 9. f() = 1 + (1 + X) (1 + X) (1 + X) )' j"() = (1 + X)(4 + ) - ( + )(1 + 4 + 4 = (1 +?( + 1) - ( + 1)(4 + 8) [(1 + )] (1+ X)4 (1 + )[(1 + X) - 4( + 1)] (1 + 4 + 4-4 - 4) (1 + ) 4 (1 + ) 3 (1 + ) 3 X)(l) X) X f'() = (3 + e - (e = 3 + ex - e" 30. f () = 3 + ex ::::} (3 + e) (3 + e) ::::} /I (3 + e X) [ex - (e X + e" 1)] - (3 + ex - e X )(9 + 6e + ex. ex)' f () = [(3+ ex )] X)(6e X + ex. ex + ex. ex) (3 + e X) (-e X) - (3 + ex - e (3 + e X )4 (3 + e X) (-e X) - (3 + ex - e X) (6e X + e X) (3 + e X) (-e X) - (3 + ex - e X) (e X) (3 + ex) (3+e )4 (3+e )4 (3 + ex)e X [(3 + e X)( -) - (3 + ex - e X)] ex(-3 - e - 6 - e + e X) (3+e )4 (3+e )3 ex(e X - e - 3-6) (3 + e X )3 ::::}, ( + 1)() - () (1) 31. y = --1 + y = ( + 1) (+1)' At (1, 1), y' = ~, and an equation of the tangent line is y - 1 = ~ ( - 1), or y = ~ + ~. ex 1 ::::} y' = 3. y e X - e X. = At (1, e), y' = 0, and an equation of the tangent line is y - e = O( - 1), or y = e. 33. y = e ::::} y' = (. e" + ex. 1) = e ( + 1). At (0,0), y' = e O(0 + 1) = 1 1 =, and an equation of the tangent line is y - 0 = ( - 0), or y =. The slope of the nonnalline is -~, so an equation of the normal line is y - 0 = -~( - 0), or y = -~. 34 Y = h =}- y' = ( + 1) (~) - h (1) = ( + 1) - () = 1 -. +1 (+1) h(+1) h(+1)' At (4,0.4), y' = ~~ = -0.03, and an equation of the tangent line is y - 0.4 = -0.03( - 4), or y ~ -0.03 + 0.5. The slope of the nonnalline is l~o, so an equation of the normal line is y - 0.4 = l~o ( - 4) } y = l~o X _ 4~O + ~ } 35. (a) y = f () = 1 1 + ::::} (b) 1.5, f () = (1 + ) (0) - 1( ) ( ) l+ tangent line at the point (-1, ~) is J'(-1) = ~ = -. ( ). So the slope of the l+ (-1,0.5) = ~ and its -4 1-------,~-t ----14 equation is y - ~ = ~ ( + 1) or y = ~ + 1. -0.5
174 D CHAPTER 3 DIFFERENTIATIONRULES 36. (a) y = f ( ) = 1 + '* (b) 0.75, (1 + ) 1 - ( ) l - f () = ( ) = ( ) ' So the slope of the l + l + tangent line at the point (3, 0.3) is j' (3) = ;o~ and its equation is y - 0.3 = - 0.08( - 3) or y = - 0.08 + 0.54. - 0.5 ex 37. (a) f ( ) = 3" - f-----f- - - ---i (b) 5 j' = 0 when f has a horizontal tangent line, j' is negative when f f is decreasing, and j' is positi ve when f is increasing. - 3 I---+,, ~_+-+-_+;;;_...=:<=..-+-_l 7-3, ( - 1)1 - ( ) _ - 1 38. (a) f( ) = - '* f ( ) = ( _ 1) = ( - 1) - 1 (b) 5 Notice that the slopes ofall tangents to f are negative and f - 3 e::--+----:- - -'=--l 3 j' ( ) < 0 always. - 5 39. (a) f ( ) = ( - l )e X '* j' () = ( - l )e X + e X (I ) = ex ( - 1 + 1) = e. (b) j" ( ) = (e X) + e X (I ) = ex( + 1) - -- tj'! ",,/ '. If --, - 3 -' - j' = 0 when f has a horizontal tangent and f It = 0 when j' has a horizontal tangent. j' is negative when f is decreasing and positive when f is increasing. f It is negative when j' is decreasing and positive when j' is increasing. I" is negative when f is concave down and positive when f is concave up., ( +1)(I ) - ( ) l - 1-40: (a) f ( ) = + 1 '* f () = ( + 1) ( + 1) 4 + + 1 '* f" ( ) = ( + 1)(-) - (1 - )( 4 3 + 4) = ( )( + 1)[- ( + 1) - (1 - )()] ( - 3) ( + 1)4 ( + 1)4 ( + 1)3 (b) We can see that our answers are plausible, since f has horizontal tangents where j' ( ) = 0, and j' has horizontal tangents where j"( ) = o.,,,, \,,, \ ' \ ' - 1.5
::::} fl(x) == (1 + )() - (1) == + - == + ::::} 41. f() == 1 + (1 + X) (1 + X) + + 1 SECTION 3. THE PRODUCT AND QUOTIENT RULES D 175 1"() = ( + + 1)( + ) - ( + )( + ) = ( + )( + + 1 - - ) ( + + 1) [(X + 1)] 1 so 1"( 1 ) = (1 + 1)3 ~ 8 - "4' ( + 1)(1) (+l)4 (+l)3' X I () e". 1 - X. ex e" (1 - ) 1-4 g() ==-::::} g == == ==--::::}. ex (e) ( e) ex /I ex. (-1) - (1 - ) e" e" [-1 - (1 - )] - 9 () == (e ) == (e ) == ~ ::::} 11/ ex. 1 - ( - )e X ex [1 - ( - )] 3-9 () == (e ) == (e ) == ~ ::::} g(4) X _ e X (-I)-(3-)eX _ e X[-I-(3-)] _ -4 ( ) - (e) - ( e) - ex The pattern suggests that 9 (n) () == ( - n) (-1)n. (We could use mathematical induction to prove this formula.) ex 43. We are given that f(5) == 1, I' (5) == 6, g(5) == -3, and gl(5) ==. (a) (fg)'(5) == f(5)gl(5) + g(5)fl(5) == (1)() + (-3)(6) == - 18 == -16 (b) (L)' (5) = g(5)1'(5) - f(5)g'(5) = (-3)(6) - (1)() = _ 0 9 [g(5)] (-3) 9 (5) == f(5)gl(5) - g(5)fl(5) == (1)() - (-3)(6) == 0 ()( fl)1 c f [f (5)] (1) 44. We are given that f() == -3, g() == 4, fl () == -, and gl() == 7. (a) h() == 5f() - 4g() ::::} hl() == 5fl(X) - 4g l(), so hi() == 5fl () - 4g 1 () == 5(-) - 4(7) == -10-8 == -38. (b) h() == f()g() ::::} hl() == f(x)gl(x) + g(x)fl(x), so h l() == f()gl() + g()fl() == (-3)(7) + (4)(-) == -1-8 == -9. (c) h() = f() =}- h'() = g()1'() - f()g'(), so 9 ( ) [g ( )] hi() == g()fl () - f()gl () == 4(-) - (-3)(7) == -8 + 1 == 13. [g()] 4 16 16 d) h( ) == g() hl( ) == [1 + f()] gl(x) - g(x)fl(x) ( X 1 + f ()::::} [1 + f () ] ' so hl() == [1 + f()] gl() - g() fl() == [1 + (-3)](7) - 4(-) == -14 + 8 == -6 == _~ [1 + f(x)] [1 + (-3)] (-) 4 45. f() == exg() ::::} fl(x) == exgl() + g()e X == ex[gl() + g()]. fl(o) == eo[gl(o) + g(o)] == 1(5 + ) == 7
176 D CHAPTER 3 DIFFERENTIATION RULES 46. ~ [h(x)] == h'() - h() 1 =? ~ [h(x)] == h'() - h() == (-3) - (4) == -10 == -.5 d d X = 4 4 47. (a) From the graphs of f and g, we obtain the following values: f(l) == since the point (1,) is on the graph of f; 9 (1) == 1 since the point (1, 1) is on the graph of g; f' (1) == since the slope of the line segment between (0, 0) and (, 4) is 4-0 = ; g' (1) = -1 since the slope of the line segment between (-,4) and (,0) is 0 (4 ) = -1. - 0 - - Now u() == f()g(), so u'(l) == f(l)g'(l) + g(l) f'(l) == (-1) + 1 == O. (b) v() = f()/g() so v' (5) = g(5)1' (5) - f(5)g' (5) = (-~) - 3. i = -~ = _~, [g (5)] 4 3 48. (a) P() == F() G(), so P'() == F() G'() + G() F'() == 3 ~ + 0 == ~. (b) Q( ) == F( )jg() Q'(7) == G(7) F'(7) - F(7) G'(7) == 1. ±- 5 (-~) ==! 10 == 43, so [G(7)] 1 4 + 3 1 49. (a) y == g() =? y' == g'() + g(). 1 == g'() + g(), g() 1 - g'() g() - g'() (b) y == g() =? Y == [g ( )] == [g ( )] (c) y == g(), g'()-g() l g'()-g() =? y == (X) == 50. (a) y == f() =? y' == f'() + f()() (b) y == f(), f'() - f()() f'() - f() =? Y == (X) == 3, f()() - f'() =? Y == [f ( )] (d) y == 1 +f() =? yx y' yx [1'() + f()] - [1 + f()] (yx) \; y X 3 / f' () + X 1 / f() - ~X-l/ - ~Xl/ f() X 1 / f() + f'() - 1 X X1 / X 3 / 1 1 51. If y == f () == _-, then l' () = ( +t)() P( ) = ( 1 1)' When = a, the equation of the tangent line is +1 +1 + y - _a_ = ( 1) ( - a). This line passes through (1,) when - _a_ = ( 1 ) (1 - a) ~ a+1 a+1 a+1 a+1 (a + 1) - a(a + 1) == 1 - a {:? a + 4a + - a - a-i + a == 0 {:? a + 4a + 1 == O. -4 ± y!4-4(1)(1) -4 ± y1 The quadratic formula gives the roots of this equation as a = (1) = = - ± V3,
SECTION 3. THE PRODUCT AND QUOTIENT RULES D 177 so there are two such tangent lines. Since 1(- ± V3) = - ± V3 - ± V3. -1=1= V3 -±V3+1-1±V3-1=FV3 ±V3=FV3-3 -1±V3 1=FV3 1-3 - 6 the lines touch the curve at A (- + V3, 1- v'3) ~ (-0.7, -0.37) andb(- - V3, 1+ v'3) ~ (-3.73,1.37). -6 == -I, ( + 1) (1) - ( - 1) (1). 5. y + 1 =? Y == ( + 1) == ( + 1)' If the tangent Intersects 6 the curve when == a, then its slope is /(a + 1). But if the tangent is parallel to - y =, that is, Y = ~ - 1, then its slope is ~. Thus, (a: 1) = ~ =* (a + 1) == 4 =? a + 1 == ± =? a == 1 or -3. When a == 1, y == 0 and the equation of the tangent is y - 0 == ~( - 1) or y == ~ - ~. When a == -3, y == and the equation of the tangent is y - == ~ ( + 3) or y == ~ + ~. -6 53. If P(t) denotes the population at time t and A(t) the average annual income, then T(t) == P(t)A(t) is the total personal income. The rate at which T(t) is rising is given by T' (t) == P(t)A'(t) + A(t)P'(t) =? T'(1999) == P(1999)A'(1999) + A(1999)P'(1999) == (961,400)($1400/yr) + ($30,593) (900/yr) == $1,345,960,000/yr + $81,455,600/yr == $1,67,415,600/yr So the total personal income was rising by about $1.67 billion per year in 1999. The term P(t)A'(t) ~ $1.346 billion represents the portion of the rate of change of total income due to the eisting population's increasing income. The term A(t)P'(t) ~ $81 million represents the portion of the rate of change of total income due to increasing population. 54. (a) f(0) == 10,000 means that when the price of the fabric is $0/yard, 10,000 yards will be sold. f' (0) == -350 means that as the price of the fabric increases past $0/yard, the amount of fabric which will be sold is decreasing at a rate of 350 yards per (dollar per yard). ~ (b) R(p) == pf(p) =? R'(p) == pf'(p) + f(p)'l =? R'(0) == 0f'(0) + f(0) 1 == 0(=-350) + 10,000 == 3000. This means that as the price of the fabric increases past $0/yard, the total revenue is increasing at $3000/($/yard). Note that the Product Rule indicates that we will lose $7000/($/yard) due to selling less fabric, but this loss is more than made up for by the additional revenue due to the increase in price. We will sometimes use the form f'9 + f g' rather than the form f g' + 9f' forthe Product Rule. 55. (a) (fgh)' == [(fg)h]' == (fg)'h + (fg)h' == (f'9 + fg')h + (fg)h' == f' gh + fg'h + fgh' (b) Putting 1 = 9 = h in part (a), we have d~ [f()]3 = (f1f)' = l'11+ 11'1+ ffl' = 3111' = 3[f(Wt'().
178 D CHAPTER 3 DIFFERENTIATION RULES d (e 3) d (c) - = - (e )3 = 3(e X) ex = 3e e = 3e 3 d d 56. (a) We use the Product Rule repeatedly: F = f 9 ::::} F' = f' 9 + f g' ::::} F" = 'J"9 + f' g') + (f' g' + f gil) = f" 9 + f'g' + f gil. (b) F'" = f'" g + f"g' + (f"g' + f'g") + f'g" + fg'" = f'" g + 3f"g' + 3f'g" + fg'" F(4) = f (4 )9 + t"g' + 3 'J"g' + f" gil) + 3 'J"gil + I' g'") + f' g'" + f g(4) ::::} = f (4 )9 + 4f'"g' + 6f"gil + 4f'g'" + fg (4 ) (c) By analogy with the Binomial Theorem, we make the guess: F(n) = r:9 + nj<n-l)g' + (~) j(n-)g" +... + (~) j(n-k)g(k) +... + n]'g(n-l) + j g(n), n ) _ n! _ n(n-1)(n-) (n-k+1) where (k - k! (n - k)! - k!. 57. For f () = e", f' () = ex + e" () = e" ( + ). Similarly, we have f" () = ex( + 4 + ) f'"(x) = e X( + 6 + 6) r:() = ex( + 8 + 1) f (5 ) () = e" ( + lo + 0) It appears that the coefficient of in the quadratic term increases by with each differentiation. The pattern for the constant terms seems to be 0 = 1 0, =. 1,6 = 3., 1 = 4. 3, 0 = 5 4. So a reasonable guess is that Proof' Let Sn be the statement that fcn)() = e" [ + n + n(n - 1)]. 1. Sl is true because l' () = ex( + ).. Assume that Sk is true; that is, f Ck) () = ex[ + k + k(k - 1)]. Then j(k+1)() = d~ X [j(k)(x)] = e ( + k) + [ + k + k(k - l)]e X This shows that Sk+1 is true. = e X[ + (k + ) + (k + k)] = e X[ + (k + l) + (k + l)k] 3. Therefore, by mathematical induction, Sn is true for all n; that is, f Cn) () = ex[ + n + n (n - 1)] for every positive integer n. d d ~ (_1_) _ g(). d (1) - 1. d [g()] 9 (). 0-1. g' () 0- g'() g' () 58. (a) d g() - [g(x)] [Quotient Rule] [g()] [g(x)] - [g(x)] (b) y = _1_ ::::} y' = _ 1 + ke s s-s- ke» (s+ke s) n-1 n n-1-n -n-1 (c) d~ (- n ) = d~ C1n) = - ~:?~ [by the Reciprocal Rule] = --- = -n = -n X n