3 D DIFFERENTIATION RULES

Transcription

1 3 D DIFFERENTIATION RULES 3.1 Derivatives ofpolnomials and Eponential Functions h 1. (a) e is the number such that lim e h- 1 = 1. h----+o (b).7 X X From the tables (to two decimal places),..7 h h - 1 1im h = 0.99 and lim h = h----+o h----+o Since 0.99 < 1 < 1.03,.7 < e <.8.. (a) 4 The function value at = 0 is 1 and the slope at = 0 is e 1 (b) f() = ex is an eponential function and g() = e is a power function. d~ (ex) = ex and d~ ( e ) = e -. e (c) f () = ex grows more rapidl than 9 () = when is large. 3. f () = is a constant function, so its derivative is 0, that is, f' () = O. 4. f () = V30 is a constant function, so its derivative is 0, that is, f' () = O. 5. f(t) = - ~t ==? f'(t) = 0 - ~ = -~ 7.f()= ==? f'()=3-4(1)+0= h() = ( - )( + 3) = - X - 6 ==? h'() = () = = X- /5 ==? ' = _~X(-/5)-1 5 = _~-7/5 5 = _ 57 / 5 157

2 158 D CHAPTER 3 DIFFERENTIATION RULES 14. R(t)=SC 3/5 =? RI(t)=S[_~t(-3/5)-1]=-3C 8 / A(8) == -5 1 == ::::} A'(8) == -1(-58-6 ) == or 60/ B() == C-6 ::::} B'() == C( -6-7) == -6c-7 18 == -h == X 1/3 ::::} ' == l- / 3 == _I_ 3 3X/3 0. f(t) == Vi - ~ == t 1 / _ t- 1 / ::::} f'(t) == It- 1 / _ (_It- 3 / ) == _1_ + _1_ [ t + 1] Vi Vi t Vi or t Vi 1. == a + b + c ::::} ' == a + b. == -v; ( - 1) == X 3/ - X 1/ ::::} ' == ~X1/ - ~X-1/ == ~X-1/(3-1) [factor out ~X-1/], 3-1 or == -v;. X / 4 1/ 3-1/ 3.== == + + ::::} -v; Y ' == ~X1/ + 4(1)-1/ + 3(_1)- 3/ == ~ -v; + ~ - _3_ [note that X 3/ == -v; -v; The last epression can be wntten as -;= + -;= - -;= == v v v -v; X /. X 1/ == -v;] - - -v; _ / ::::} 5. == 47r ::::} ' == 0 since 47r is a constant. 7. We first epand using the Binomial Theorem (see Reference Page 1). H() == ( + X- 1)3 == X(X- 1) + (X- 1)3 == ::::} H' () == ( -l- ) + (-3-4 ) == , v b - -3 v b c ::::} == ae - v - cv == ae v v 3

3 SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS ) 1 (1) 1 3 /3 30. V =.,f;;+ -v;; = (.,f;;) +.,f;; -v;; + -v;; = + X 1 / - / 3 + 1/ / = X + X 1 / 6 + X- =* ( 1 v' == 1 + (1. X - S / 6 ) - ~X-S/3 == 1 + l. X - S / 6 - ~X-S/3 or A 3{1;5 3W loa z == 10 + BeY == A-10 + BeY =? z' == -10A-11 + BeY == BeY 33. == V; == X 1/4 =? ' == l. X- 3 / 4 == _1_. At (1, 1), ' == -4 1 and an equation of the tangent line is 4 4VX3-1==i(-1) or ==i+~. 34. == =? ' == At (1, ), ' == 7 and an equation of the tangent line is - == 7( - 1) or == == 4 + e =? ' == e. At (0,), ' == and an equation of the tangent line is - == ( - 0) or == +. The slope of the normal line is - ~ (the negative reciprocal of ) and an equation of the normal line is -==-~(-O) or ==-~ == (1 + X) == =? ' == At (1,9), ' == 1 and an equation of the tangent line is - 9 == 1( - 1) or == 1-3. The slope of the normal line is --1 (the negative reciprocal of1) and an equation of the normal line is - 9 == - -1 ( - 1) or == ; 5 At (1, ), ' == 6-3 == 3, so an equation of the tangent line is - == 3( - 1) or == t---f--~+--t---f----t = -.,f;; =* ' = 1 - ~X-l/ = 1 - \_. v At (1, 0), ' == ~, so an equation of the tangent line is - 0 == ~ ( - 1) or == ~ - ~. Ot "7II"'=------I -1

4 160 0 CHAPTER 3 DIFFERENTIATIONRULES 39. f ( ) = ex - 5 =} 1' ( ) = ex - 5. Notice that l'() = 0 when f has a horizontal tangent, l' is positive when f is increasing, and l' is negative when f is decreasing. 40. f( ) = =} 1' ( ) = Notice that l'() = 0 when f has a horizontal tange nt and that l' is an even function while f is an odd func tion. 41. f ( ) = 3X l =} 1'() = 45 l Notice that l'() = 0 when f has a horizontal tangent, l' is positive when f is increas ing, and l' is negative when f is decreasing f ( ) = + l / = =} 1' ( ) = 1- - = 1-1/..\, / f f' / f Notice that l'() = 0 when f has a horizontal tangent, l' is positive r <, when f is increas ing, and l' is negative when f is decreasing / ~ / 43. (a) 50 (b) From the graph in part (a), it appears that l' is zero at Xl ~ - 1.5, X ~ 0.5, and X 3 ~ 3. The slopes are negative (so l' is negative) on (-00, Xl) and (X, X3). The slopes are positive (so l' is positive) on (X l, X) and ( X 3, (0 ). - 3 f o-...<;+7--<--i 5-10

5 (c) f( )= 4-3 ~ j'() = 4 3 ~ SECTION 3.1 DERIVATIVES OFPOLYNOMIALSAND EXPONENTIALFUNCTIONS D (a) 8 (b) From the graph in part (a), it appears that j' is zero at X l R; 0. and X R;.8. The slopes are positive (so j' is positive) on (-00, I) and ( X, (0). The slopes are negative (so j' is negative) on (X l, X ) (c) g( ) = ex - 3 ~ g'( ) = e" I I----f"'< f+----l f ( ) = ~ j'( ) = ~ j"( ) = G(r) = v:;. + ifr ~ G'(r) = ~r - I / + ~r - / 3 ~ G"(r) = _ ~ r - 3 / _ ~ r -5 /3 3 4 I f ( ) = X / ~ j'( ) = - X- / ~ j"() = ~ ~ - 5 / 4 \/.' Note that f' is negative when f is decreasing and positive when f is ~.,r-; r\ increasing. f" is alwas positive since j' is alwas increasing.... f Note that f' () = 0 when f has a horizontal tangent and that j"() = ""h'~" l 6 when j' has a horizontal tangen t (a) 8 = t 3-3t ~ v(t) = 8 ' (t) = 3e - 3 ~ a(t ) = v'(t) = 6t (b) a() = 6() = 1 m/s (c) v(t ) = 3t - 3 = 0 when t = 1, that is, t = 1 and a(l ) = 6 m/ s

6 16 0 CHAPTER 3 DIFFERENTIATION RULES 50. (a) s = t 3-7t + 4t + 1 =} v(t ) = S'(t) = 6t - 14t + 4 =} a(t) = v' (t ) = 1t - 14 (b) a(1) = 1-14 = - m/s (c) P..:---: ::-" j 4-14 I I I I I I 51. The curve = has a horizontal tangent when ' = = 0 {=} 6( + X - ) = 0 {=} 6( + )( - 1) = 0 {=} = - or = 1. The point s on the curve are (-,1) and (1, - 6). 5. f ( ) = X + 3 has a horizontal tangent when j'() = = 0 {=} = - 6±~ = - 1± t J Y = =} m = ' = , but : 0 for all, so m : 5 for all Y = V;;= X / =} ' = ~ X l /. The slope of the line = is 3, so the slope ofan line parallel to it is also 3. Thus, ' = 3 =} ~ X l / = 3 =} V;;= =} = 4, which is the z -coordinate of the point on the curve at which the slope is 3. The v-coordinate is = 4 J4 = 8, so an equation of the tangent line is - 8 = 3( - 4) or = The slope ofthe line 1 - = 1 (or = 1-1) is 1, so the slope of both lines tangent to the curve is 1. 3 = 1 + =} ' = 3. Thus, 3 = 1 =} = 4 =} = ±, which are the z -coordinates at which the tangent lines have slope 1. The points on the curve are (,9) and (-, - 7), so the tangent line equations are - 9 = 1( - ) or = 1-15 and + 7 = 1( + ) or = The slope of = 1 + e - 3 is given b m = ' = e - 3. = I + ex- /'3_""=_ -i--_ -rr- _ -,... The slope of 3 - = 5 {=} = 3-5 is 3. This occurs at the point (In 3, 7-3 1n3) ~ (1.1, 3.7). - 3 I ft---+--/-+---o----j 4 = In 3-1 = ' The slope of = is given b m = ' = - 5. The slope of - 3 = 5 {=} Y = 3 X - 3 IS 3 ' so the desired normal line must have slope t,and hence, the tangent line to the parabo la must have slope - 3. This occurs if - 5 = - 3 =} = =} = 1. When = 1, = 1-5(1) + 4 = 0, and an equation of the normal line is - 0 = t ( - 1) or = t - t

7 SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS == I () == - ::::} I' () == 1 -. SOI' (1) == -1, and the slope of the normal line is the negative reciprocal of that ofthe tangent line, that is, -1/(-1) == 1. So the equation of the normal line at (1, 0) is - 0 == 1( - 1) {:? == -I. Substituting this into the equation of the parabola, we obtain -I == - {:? X == ±1. The solution == -1 is the one we require. Substituting == -1 into the equation of the parabola to find the -coordinate, we have == -. So the point of intersection is (-1, -), as shown in the sketch. 59. Let (a, a ) be a point on the parabola at which the tangent line passes through the point (0, -4). The tangent line has slope a and equation - (-4) == a( - 0) {:? == a - 4. Since (a, a ) also lies on the line, a == a(a) - 4, or a == 4. So a == ± and the points are (,4) and (-,4). 60. (a) If == +, then ' == + 1. If the point at which a tangent meets the parabola is (a, a + a), then the slope of the tangent is a + 1. But since it passes through (, -3), the slope must also be ~Y = a + a + 3. ~ a- Therefore, a + 1 = a + a + 3. Solving this equation for a we get a + a + 3 = a - 3a - a- B a - 4a - 5 == (a - 5)(a + 1) == 0 {:? a == 5 or -1. If a == -1, the point is (-1,0) and the slope is -1, so the equation is - 0 == (-l)( + 1) or == If a == 5, the point is (5,30) and the slope is 11, so the equation is - 30 == 11( - 5) or == ll - 5. (b) As in part (a), but using the point (, 7), we get the equation 1 a + a - 7 ::::} a - 3a - == a + a - 7 {:? a - 4a + 5 == O. a+ == a- The last equation has no real solution (discriminant == -16 < 0), so there is no line through the point (,7) that is tangent to the parabola. The diagram shows that the.. point (, 7) is "inside" the parabola, but tangent lines to the parabola do not pass through points inside the parabola j'() = lim f( + h) - f() = lim ~ -;; = lim - ( + h) = lim -h = lim -1 h---+o h h---+o h h---+o h( + h) h---+o h( + h) h---+o ( + h) n n-1 n- 6. (a) I() == ::::} I/() == n ::::} I//() == n (n - 1) ::::} ::::} I(n)() == n(n - l)(n - ) l n-n == n! 3 (b) I () == X -1 ::::} I I ( ) == (-1) - ::::} 1//() == (-1) (- ) - ::::}... ::::} I(n)() == (-1)(-)(-3)... (_n)-(n+1) == (_1)nn!-(n+1) or (-l)nn! n + 1

8 164 0 CHAPTER 3 DIFFERENTIATION RULES 63. Let P() == a + b + c. Then PI(X) == a + band PIl(X) == a. pll() == =} a == =} a == 1. P' () == 3 =} (1)() + b == 3 =} 4 + b == 3 =} b == -1. P()==5 =} 1()+(-1)()+c==5 =} +c==5 =} c==3.sop()== == A + B + 0 =} ' == A + B =} " == A. We substitute these epressions into the equation " + ' - == to get (A) + (A + B) - (A + B + 0) == A + A + B - A - B - 0 == (-A) + (A - B) + (A + B - 0) == (1) + (O) + (0) The coefficients of on each side must be equal, so - A == 1 =} A == - ~. Similarl, A - B == 0 =} A == B == - ~ and A + B - 0 == 0 =} -1 - ~ - 0 == 0 =} 0 == - ~. 65. == f() == a 3 + b + c + d =} f'(x) == 3a + b + c. The point (-,6) is on f, so f( -) == 6 =} -8a + 4b - c + d == 6 (1). The point (,0) is on f, so f() == 0 =} 8a + 4b + c + d == 0 (). Since there are horizontal tangents at (-,6) and (,0), f' (±) == O. f' (-) == 0 =} 1a - 4b + c == 0 (3) and f' () == 0 =} 1a + 4b + c == 0 (4). Subtracting equation (3) from (4) gives 8b == 0 =} b == O. Adding (1) and () gives 8b + d == 6, so d == 3 since b == O. From (3) we have c == -1a, so () becomes 8a + 4(0) + (-1a) + 3 == 0 =} 3 == 16a =} a == fb. Now c == -1a == -1 (fb) == - ~ and the desired cubic function is == fb 3 - ~ == a + b + c =} '(X) == a + b. The parabola has slope 4 at == 1 and slope -8 at == -1, so '(l) == 4 =} a + b == 4 (1) and ' (-1) == -8 =} -a + b == -8 (). Adding (1) and () gives us b == -4 {:} b == -. From (1), a - == 4 {:} a == 3. Thus, the equation of the parabola is == c. Since it passes through the point (,15), we have 15 == 3() - () + c =} c == 7, so the equation is == f() == - if :::; 1 and f() == - + if > 1. Now we compute the right- and left-hand derivatives defined in Eercise.8.54:.. f~(l) == lim f(l + h) - f(l) == lim - (1 + h) - 1 == lim -h == lim -1 == -1 and h---+o- h h---+o- h h---+o- h h---+o- ' (1) - 1 f(l + h) - f(l) _ 1 (1 + h) - (1 + h) _ 1 h - 1 h - 0 f + - 1m - 1m - 1m - 1m -. h---+o+ h h---+o+ h h---+o+ h h---+o+ Thus, f' (1) does not eist since [': (1) :f- f+ (1), so f is not differentiable at 1. But f' () == -1 for < 1 f and f' () == - if > 1. o (1, 1) o

9 SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS D if < g() = : if-1~~1 { if > 1 lim g(-l+h)-g(-l) = lim [-1-(-1+h)]-1 h lim -h = lim (-) = - and h-+o- h h-+o- h h-+o- h-+o lim g(-l+h)-g(-l) = lim (-l+h?-l = lim -h+h = lim (-+h)=- h-+o+ h h-+o+ h h-+o+ h h-+o+ ' so 9 is differentiable at -1 and g' (-1) = -. lim g(l + h) - g(l) = lim (1 + h) - 1 = r h + h lim ( + h) = and h-+o- h h-+o- h h~- h h-+olim g(l + h~ - g(l) = lim (1 +~) - 1 = lim -hh = lim 1 = 1, so g' (1) does not eist. h-+o+ h-+o+ h-+o+ h-+o+ Thus, 9 is differentiable ecept when = 1, and if < -1 g' () == - X if -1 < < 1 { 1 if > 1 =g() -lox (a) Note that - 9 < 0 for < 9 {:} Il < 3 {:} -3 < < 3. So if < -3 if < -3 { { if Il > 3 f()= - if -3 < < 3 ~ f' () == - if -3 < < 3 _ 9 if :: 3 if > 3 = {~: if Il < 3 To show that l'(3) does not eist we investigate lim f ( 3 + h~ - f ( 3 ) b computing the left-and right-hand derivatives h-+o defined in Eercise f'-(3) = lim f(3+h)-f(3) = lim [-(3+h?+9]-O = lim (-6-h)=-6 and h-+o- h h-+o- h h-+o f+(3) = lim f(3 + h) - f(3) = lim [(3 + h? - 9] - 0 = lim 6h + h = lim (6 + h) = 6. h-+o+ h h-+o+ h h-+o+ h h-+o+ Since the left and right limits are different, lim f(3 + h~ - f(3) does not eist, that is, l'(3) h-+o does not eist. Similarl, i'(-3) does not eist. Therefore, f is not differentiable at 3 or at -3. (b) -3 0 / If'

10 166 D CHAPTER 3 DIFFERENTIATION RULES 70. If 1, then h() = I 11 + I + 1 = -I + + = + 1. If - < < 1, then h() = -( 1) + + = 3. If ~ -, then h() = -( 1) - ( + ) = - 1. Therefore, h()= { if < - if-<<1 if 1 ::::} h'() = { - ~ if < - if - < < 1 if > 1 To see that h' (1) = lim h() - h(l) does not eist, -d -I = h()-h(l) 3-3 h'() observe that lim = lim -- = 0 but c -+l- -I -+l- 3-1 Y = h() - a 1 h() - h(l) li - S' '1 1 1l Itl = 1m -- =. 1m1 ar, l+ -I -+l+ -I - a 1 h' (- ) does not eist. 71. Substituting = 1 and = 1 into = a + b gives us a + b = 1 (1). The slope of the tangent line = 3 - is 3 and the slope of the tangent to the parabola at (, ) is ' = a + b. At = 1, ' = 3 ::::} 3 = a + b (). Subtracting (1) from () gives us = a and it follows that b = -1. The parabola has equation = = 4 + a 3 + b + e + d ::::} (O) = d. Since the tangent line = + 1 is equal to 1 at = 0, we must have d = 1. ' = a + b + e ::::} ' (0) = e. Since the slope of the tangent line = + 1 at = 0 is, we must have e =. Now (l) = 1 + a + b+ e + d = a + b+ 4 and the tangent line = - 3 at = 1 has -coordinate -1, so a + b + 4 = -1 or a + b = -5 (1). Also, '(I) = 4 + 3a + b + e = 3a + b + 6 and the slope of the tangent line = - 3 at = 1 is -3, so 3a + b + 6 = -3 or 3a + b = -9 (). Adding - times (1) to () gives us a = 1 and hence, b = -6. The curve has equation = = f() = a ::::} f' () = a. So the slope of the tangent to the parabola at = is m = a() = 4a. The slope of the given line, + = b {:} = - + b, is seen to be -, so we must have 4a = - {:} a = -~. So when =, the point in question has -coordinate -~. = -. Now we simpl require that the given line, whose equation is.. + = b, pass through the point (, -): () + (-) = b {:} b =. So we must have a = ~~ and b =. 74. The slope of the curve = e h is ' = e/ and the slope of the tangent line = ~ + 6 is ~. These must be equal at the v point of tangenc ( a, c ~ ), so ~ = ~ '* c = 3~. The -coordinates must be equal at = a, so e = 3V4 = 6.

11 SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS f is clearl differentiable for < and for >. For <, f' () ==, so [': () == 4. For >, f' () == m, so f~() == m. For f to be differentiable at ==, we need 4 == f~() == f~() == m. So f() == 4 + b. We must also have continuit at ==, so 4 == f() == lim f() == lim (4 + b) == 8 + b. Hence, b == (a) == c ~ == ~. Let P == (a, :::"). The slope of the tangent line at == a is ' (a) == - c Its equation is a a c c() c c... c S.. he z-i - - == - - a or == ----;;- + -, so Its -Intercept IS -. ettmg == 0 gives == a, so t e z-intcrccpt IS a. a a a.: a a The midpoint of the line segment joining (0, ~) and (a,0) is (a, ~) = P. (b) We know the - and -intercepts of the tangent line from part (a), so the area of the triangle bounded b the aes and the tangent is ~ (base) (height) == ~ == ~ (a) (c/a) == c, a constant. l 77. Solution 1: Let f () = X Then, b the definition of a derivative, l'(1) = lim f () - f (1) = lim X I I But this is just the limit we want to find, and we know (from the Power Rule) that f' () == , so f' (1) == 1000(1)999 == So lim - == I Solution : Note that ( ) == ( - 1)( X X X 1). So X li ( - 1)( + X X + 1) Ii ( ) lim ---- == irn 1 == un I X == == 1000, as above. \..I 1000 ones 78. In order for the two tangents to intersect on the -ais, the points of tangenc must be at equal distances from the -ais, since the parabola == is smmetric about the -ais. Sa the points of tangenc are (a, a ) and (-a, a ), for some a > O. Then since the derivative of == is d/ d ==, the left-hand tangent has slope -- a and equation - a == -a( + a), or == -a - a, and similarl the right-hand tangent line has equation - a == a( - a), or == a - a. So the two lines intersect at (0, _a ). Now if the lines are perpendicular, then the product of their slopes is -1, so (-a)(a) == -1 ~ a == ~ ~ a==~. So the lines intersect at (O,-i). 79. == ~ ' ==, so the slope of a tangent line at the point (a, a ) is ' == a and the slope ofa normal line is -1/(a),... a - c a - c 1 for a -I O. The slope of the normal hne through the points (a, a ) and (0, c) IS --, so-- == -- ~ a-o a a a - C == - ~ ~ a == c - ~. The last equation has two solutions if c > ~, one solution if c == ~, and no solution if c < ~. Since the -ais is normal to == regardless of the value of c (this is the case for a == 0), we have three normal lines if c > ~ and one normal line if c :s; ~.

12 168 0 CHAPTER 3 DIFFERENTIATION RULES 80. From the sketch, it appears that there ma be a line that is tangent to both curves. The slope of the line through the points P(a, a ) and b - b + - a Q(b, b - b + ) is b _ a. The slope of the tangent line at P is a [' = ] and at Q is' b - [' = - ]. All three slopes are equal, so a = b - {:} a = b - 1. Also, b _ = b - - : + - b b + - (b - 1) a =? b - = b - b + - b + b - 1 =? -a =? b - = b _ (b - 1) b = 3 =} b = ~ and a = ~ - 1 = ~. Thus, an equation of the tangent line at Pis - (~) = (~) ( - ~) or = - %. APPLIED PROJECT Building a Better Roller Coaster 1. (a) f() = a + b + c =? f'() = a + b. The origin is at P: f(o) = 0 =? c=o The slope of the ascent is 0.8: f'(o) = 0.8 =? b = 0.8 The slope of the drop is -1.6: f' (100) = -1.6 =? 00a + b = -1.6 (b) b = 0.8, so 00a + b = -1.6 =? 00a = -1.6 =? 00a = -.4 =?.4 a = - 00 = Thus, f() = (c) Since L 1 passes through the origin with slope 0.8, it has equation = The horizontal distance between P and Q is 100, so the -coordinate at Q is f(100) = -0.01(100) + 0.8(100) = -40. Since L passes through the point (100, -40) and has slope -1.6, it has equation + 40 = -1.6( - 100) or = (d) The difference in elevation between P(O, 0) and Q(100, -40) is 0 - (-40) = 40 feet ,/F---~~_ Q(100,-40) f (a) Interval Function First Derivative Second Derivative (-00,0) L 1() = 0.8 L~() = 0.8 L~() = 0 [0,10) g() = k 3 + l + m + n g'() = 3k + l + m g"{x) = 6k + l [10,90] q() = a + b + c q'() = a + b q" () = a (90,100] h( ) = p 3 + q + r + s h'() = 3p + q + r h"() = 6p + q (100,00) L () = L;() = -1.6 L~() = 0 There are 4 values of (0, 10, 90, and 100) for which we must make sure the function values are equal, the first derivative values are equal, and the second derivative values are equal. The third column in the following table contains the value of each side of the condition - these are found after solving the sstem in part (b).