3.4 The Chain Rule. . r: 1/2 dy dy du 1 1/2 ( ) 49. (a) - tanx ==--- ==}- sec x == ==. So sec x == --2-' d. d 1 + cot x

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1 184 D CHAPTER 3 DIFFERENTIATION RULES d d sinx COS X cos x - sin x (- sin x) cos X + sin'' x (a) - tanx ==--- ==}- sec x == ==. So sec x == ---' dx dx cos x cos? X cos? X cos X b) d d 1 (cos x) (0) - 1(- sin x ) sin x ( dx sec x == dx cos x ==}- sec x tan x == cos X. So sec x tan x == cos X d. d 1 + cot x (c) -d (sin z + cos x) == -d ==} x x cscx. esc x (- csc x) - (1 + cot x) (- esc x cot x) esc x [- csc X + (1 + cot x) cot x] cos x - SIn x == esc? X == esc? X. cot x-i So cos x - SIn x ==. cscx - csc X + cot X + cot x -1 + cot x cscx esc x 50. We get the following formulas for rand h in terms of B: sin ~ = ;0 =}- r = 10 sin ~ and cos ~ = 1~ ==}- h == locos " B A(O) Now A( B) == ~1fr and B(B) == ~ (r)h == rho So lim A(B) == lim ~1fr == 1.1f lim!... == 1.1f lim 10 sin(b/) B(B) rh h cos(b/) == ~1f lim tan(b/) == 0 R By the definition ofradian n1easure, S == rt), where r is the radius ofthe circle. By drawing the bisector ofthe angle B,we can. B d/ see t h at SIn - == - d. B S li S li rb == 11'ITl (B/ ) == 11'm B/ == 1 ==}- == r SIn -. 0 r im -d == im. (B/ ) (B/ ) ( /) r SIn sin sin B [This is just the reciprocal ofthe limit lim sin x == 1 combined with the fact that as B , f!.. x---+o x also.] 3.4 The Chain Rule. dy dy du 1. Letu == g(x) == 4x andy == f(u) == sm tz. Then dx == du dx == (cosu)(4) == 4cos4x.. r: 1/ dy dy du 1 1/ ( ). Letu==g(x) ==4+3xandY==fu () ==yu==u.then-==--==-u- 3 ==--== 3 3. dx du dx yu V4 + 3x.. 10 dy dy du 3. Letu == g(x) == 1- x andy == f(u) == u. Then dx == du dx == (10u 9)(-x) == -0x(1- x ) Let u = g(x) = sin x and y = f(u) = tanu. Then ~~ = ~~ ~~ = (scc'' u)( cos x) = sec (sin x). cos x, or equivalently, [sec(sin x )] cos X. dy dy du ()1/ 1 e vx 5. Let u == g(x) == VX and y == f(u) == e": Then - ==-- == (e U ) l.x- == evx.-- == --. dx du dx VX VX

2 . dy dy du 6. Let u == g(x) == e" and y == f(u) == SIn u. Then dx == du dx == (cos u)(ex) == e" cos ex. SECTION 3.4 THE CHAIN RULE F(x) = (x 4 + 3x - )5 '* F'(x) = 5(x 4 + 3x - )4. d~ (x 4 + 3x - ) = 5(x 4 + 3x - )4(4x + 6x) [or 10x(x 4 + 3x - )4(x + 3)J 8. F(x) = (4x - X )100 '* F'(x) = 100(4x - X )99. d~ (4x - x ) = 100(4x - X )99(4 - x) [or 00X 9 9 (x - )(x - 4)99J '( )_1( 3)-3/4 d ( 3)_ 1 ( _ +3x F X - 4" 1 + x + x. dx 1 + x + x - 4(1 + x + x 3)3/4. + 3x ) - 4(1 + x + X 3)3/4 + 3x 1 4 ) (t) == (t 4 + 1)3 == (t jet) = ij1 + tan t = (1 + tan t)1/3 '* l'(t) = ~ (1 + tan t)-/3 sec t = sec t 3 \/(1 + tan t) 14. y == a 3 + cos" x =>- y' == 3(cos X)(- sin x) [a 3 is just a constant] == -3 sin x cos x 16. y == 3cot(nB) =>- y' == 3[- csc (n B). n] == -3ncsc (nb) 17. g(x) == (1 + 4X)5(3 + x - X )8 => ) )8 g' (x) == (1 + 4x) 5. 8(3 + x - x 7 (1 - x) + (3 + x - x. 5(1 + 4x) 4. 4 == 4(1 + 4X)4(3 + x - x )7 [(1 + 4x)(1 - x) + 5(3 + x - x )J == 4(1 + 4X)4(3 + x - X )7 [( + 4x - 16x ) + (15 + 5x - 5x )J == 4(1 + 4X)4(3 + x - x )7(17 + 9x - 1x ) 18. h(t) == (t 4-1)3(t 3 + 1)4 =>h'(t) == (t 4 _1)3. 4(t 3 + 1)3(3t ) + (t 3 + 1)4. 3(t 4-1)(4t 3) == 1t (t 4-1)(t 3 + 1)3 [(t 4-1) + t(t 3 + I)J == 1t (t 4-1)(t 3 + 1)3(t 4 + t - 1)

3 186 D CHAPTER 3 DIFFERENTIATION RULES 19. y == (x - 5)4(8x - 5)-3 ~ y' == 4(x - 5)3()(8x - 5)-3 + (x - 5)4( -3)(8x - 5)-4(16x) == 8(x - 5)3(8x - 5)-3-48x(x - 5)4(8x - 5)-4 [This simplifies to 8(x - 5)3(8x - 5)-4( -4x + 30x - 5).] 0. y == (x + 1) (x + ) 1/3 ~ y' = x(x + ) 1/3 + (x + 1) G)(x + ) -/3 (x) = x(x + ) 1/3 [1 + 3(:: 1 )] x + 1)3 1. y == ( x - 1 ~ y' == 3 (x + 1). ~ (x + 1) == 3 (x + 1). (x - 1)(x) - (x + 1)(x) x - 1 dx x - 1 x - 1 (x - 1) == 3(x + 1). x[x -1- (x + 1)] == 3(x + 1). x(-) == -1x(x + 1) x - 1 (x - 1) x - 1 (x - 1) (x - 1)4. y == e- 5x cos 3x ~ y' == e- 5x (-3 sin 3x) + (cos 3x) (_5e- 5x) == _e- 5x (3 sin 3x + 5 cos 3x) 3. y == excos x ~ y' == excos x. ~ (x cos x) == excos x [x(- sin x) + (cos x). 1] == e" cos x(cos x - x sin x) dx x d ) 4. Using Formula 5 and the Chain Rule, y == ~ y' == x (In 10). dx (1 - x == -x(ln 10)10 1- X. 5. F(z) = Jz -1 = (z _1)1/ ~ z+1 z+1 p'(z ) ==!(z - 1)-1/. ~ (z - 1) ==! (z + 1)1/. (z + 1)(1) - (z - 1)(1) z + 1 dz z + 1 z - 1 (z + 1) l(z+i)1/ z+l-z+1 1 (z + 1)1/ 1 " (z - 1)1/' (z + 1) " (z - 1)1/. (z + 1) (z - 1)1/(Z + 1)3/ (y _ 1)4 6. G(y) == (y + y)5 ~ G'(y) (y + y)5. 4(y - 1)3.1- (y - 1)4. 5(y + y)4(y + ) [(y + y)5j (y + y)4(y - 1)3 [(y + y) - 5(y - 1)(y + I)J (y + y)10 (y - 1)3 [(y + 4y) + (-5y + 5)J (y - 1)3(_3y + 4y + 5) (y + y)6 (y+y)6

4 SECTION 3.4 THE CHAIN RULE D 187 r 7. y= ~ =} yr + 1 ~~_r vr + 1 Another solution: Write y as a product and make use of the Product Rule. y = r(r + 1)-1/ =} The step that students usually have trouble with is factoring out (r + 1) -3/. But this is no different than factoring out x 5 from x + x ; that is, we are just factoring out a factor with the smallest exponent that appears on it. In this case, - ~ is smaller than - ~. 9. y = sin(tan x) =} y' = cos(tan x). d~ (tan x) = cos(tan x). sec (x). d~ (x) = cos(tan x) sec (x) 30. G( ) = (L)5 =} G'(y) = S(L)4.(y + 1)(y) - y(1) = 5. y8. y(y + - y) = 5 y 9(y + ) y y+1 y+1 (y+1) (y+1)4 (y+1) (y+1)6 31. Using Formula 5 and the Chain Rule, y = sill1rx =}. d. y' = SIll 1rX (In ). - dx (sin 1rx) = SIll 1rX (In ). cos 1rX. 1r = SIll 1rX ( 1r In ) cos 1rX 3. y = tarr' (38) = (tan 38) =} y' = (tan 38). d~ (tan 38) = tan 38. sec = 6 tan 38sec y = sec x + tan x = (secx) + (tanx) =} y' = (sec x) (sec x tan x) + (tan x) (sec x) = sec x tan x + sec x tan x = 4 sec x tan x.1,.1 1( 1) y = x SIn - =} y = SIn - + x COS = SIn COS X X X x X X X X 1 e ) 35. y = cos ( -- l+e x

5 188 D CHAPTER 3 DIFFERENTIATION RULES 36. f(t) = Vt ~4 = C ~4Y/ =?, 1( t ) -1/ d ( t ) 1(t + 4) 1/ (t + 4)(1) - t (t) f (t) =" t + 4. dt t + 4 =" -t- (t + 4) (t + 4)1/ t t 4 - t t 1/ (t + 4) t1/(t + 4)3/ (sin8) 37. y = cot = [cot(sin8)] ~ y' = [cot (sin 0)]. ~ [cot(sin 0)] = cot (sin 0). [~csc(sin 0). cos 0] = - cos 0 cot (sin 0) csc (sin 0) 39. f( t) = tan(e t ) + e t a n t 40. y = sin(sin(sinx)) =? y' = cos(sin(sinx)) d~ (sin(sinx)) = cos (sin(sin x)) cos(sinx) cos x 43. g(x) = (ra r x + n)p ~ 45. y = cos vlsin(tan 1rx) = cos(sin(tan 1rX))1/ ~ y' = - sin(sin(tall7rx) )1/. d~ (sin(tall7rx))1/ = - sin(sin (tan 7rX)) 1/. ~ (sin(tall'7l"x)) -1/. d~ (sin(tall7rx)) - sin visin(tan1rx) d - sin visin(tan1rx) =. cos(tan1rx). -d tan 1rX =. cos(tan 1rx). sec (1rx) <t: visin(tan1rx) x visin(tan1rx) -1r cos( tan 1rx)sec (1rx) sin J sin(tan 1rx) JSin(tan 1rx) 46. y = [x + (x + sin' X)3J4 ~ y' = 4 [x + (x + sirr' X)3J3. [1+ 3(x + sirr' X). (1+ sinx cosx)j

6 h(x) == vx ==> h'(x) == -(X + 1)- 1/(X) = ~ '* x + 1 ".JX x [~(X + 1)-1/(X)] (x + 1) -1/ [(X + 1) _ X] 1 h (X) = (VX + 1 ) = (X + 1)1 (x + 1)3/ SECTION 3.4 THE CHAIN RULE D 189 c x Cx Cx 48. y == xe ==> y' == x. e. e + e. 1 == e Cx (ex + 1) ==> a x a x a x 49. y == e sin (3x ==> y' == e. (3 cos (3x + sin (3x. ae == e a x ((3cos (3x + a sin (3x) ==> a x y" == e ( - (3 sin (3x + a(3 cos (3x) + ((3 cos (3x + a sin (3x). ae a x == e ( - (3 sin (3x + a(3 cos (3x + a(3 cos (3x + a sin (3x) == e a x (a sin (3x - (3 sin (3x + a(3 cos (3x) a x == e [(a - (3)sin (3x + a(3 cos (3xJ a x 51. y == (1 + X)10 ==> y' == 10(1 + X)9 == 0(1 + X)9. At (0, 1), y' == 0(1 + 0)9 == 0, and an equation of the tangent line is y - 1 == 0(x - 0), or y == 0x y == sin x + sirr' x ==> y' == cos x + sin x cos x. At (0,0), y' == 1, and an equation of the tangent line is y - 0 == l(x - 0), or y == x. 53. y == sin(sin x) ==> y' == cos(sin x). cos x. At (1r, 0), y' == cos(sin1r). cos 1r == cos(o). (-1) == 1(-1) == -1, and an equation of the tangent line is y - 0 == -l(x - 1r), or y == -x + tt. e-x e-x 54. y == x ==> y' == x (_e- X ) + e- x (x) == xe- x - x. At (1, ~), y' == e-1 - e-1 == ~. So an equation of the tangent line is y - ~ == ~ (x - 1) or y == ~ x. 55. (a) y == 1 +e- x (b) eo (1) 1. f h At (0, 1), y' = (1 + eo) (1 + 1) == == ' So an equation 0 t e tangent line is y - 1 == ~ (x - 0) or y == ~ x + 1. x 56. (a) For X > 0, Ixi = x, andy = f(x) = ~ '* (b) - x 4 'x _ ~(1)-x(~)(-x)-1/(-x). (_X )1/ f ( ) - (V _ x ) ( - x )1/ ( - x ) + x -1.5 t ~=----.,.' ( X )3/ So at (1, 1), the slope of the tangent line is f' (1) == and its equation is y - 1 == (x - 1) or y == x (0,1) 3 F=-~---+--I l

7 190 D CHAPTER 3 DIFFERENTIATIONRU LES 57. (a) j (x ) = x) - x = x( _ X )1/ '* 1'(x) = x 1( - X ) - 1/( _x) + ( _ X )1/. 1 = ( _ X )- 1/ [_x + ( _ x)] = - x ) _ x (b) < fL - -\ j f l' = 0 when j has a horizontal tangent line, l' is negative when j is decreasing, and l' is positive when j is increasing (a) From the graph of j, we see that there are 5 horizontal tangents, so there must be 5 zeros on the graph of 1'. From the symmetry of the grap h of j, we must have the graph of l' as high at x = 0 as it is low at x = 7f. The interva ls of increase and decrease as well as the signs of l' are indica ted in o'----+.i...t "-+-../ tr the figure. tt in ftn!tt n, i + 1'- i + i-li+ i o 71" f' x " (b)j(x) = sin (x + sin x ) '* l'(x) = cos(x + sin x ) d~ (x + sin x ) = cos(x + sin x )(1+ cos x ) 59. For the tangent line to be horizontal, l'(x ) = O. j(x ) = sin x + sirr' x '* t'(x) = cos x + sin x cos x = 0 -R cos x (l + sin x ) = 0 -R cosx = 0 or sin x = - 1, so x = ~ + mr or 3; + mf, where n is any integer. Now j (~) = 3 and j e") = - 1, so the points on the curve with a horizontal tangent are ( ~ + mf, 3) and e" + mf, - 1),. where n is any integer. 60. j (x ) = sin x - sin x '* 1' (x ) = cos x - cos x = 4 cos X - cos x -, and d cos'' x - cos x - = 0 -R (cos x - 1)(4 cosx + ) = 0 -R cos x = lor cos x = - ~. So x = mf or (n + 1)7r ±?}, n any integer. 61. F (x ) = j (g(x )) '* F '(x) = 1'(g(x)) g'(x), so F'(5) = 1'(g(5)) ' g'(5) = 1' (-) 6 = 4 6 = 4 6. h(x) = J4 + 3j(x) '* h'( x) = ~(4 + 3j(X))-1 /. 31'( x ), so h'( l) = ~(4 + 3j(1))- 1/. 31' (1) = ~ ( )-1 / 3 4 = k = ~ 63. (a) h(x ) = j (g(x )) '* h' (x ) = 1' (g(x )) g'( x), so h' (l ) = 1' (g(l )). g' (l ) = 1' () 6 = 5 6 = 30. (b) H (x ) = g(j (x )) '* H' (x ) = g'(j(x )). 1' (x ), so H'(l) = g'(1(l)). 1'(1) = g'(3) 4 = 9 4 = 36.

8 SECTION 3.4 THE CHAIN RULE D (a) F(x) == f(f(x)) ~ F'(x) == f'(f(x)) f'(x), so F'() == f'(f()) f'() == f'(l) 5 == 4 5 == 0. (b) G(x) == g(g(x)) ~ G'(x) == g'(g(x)). g'(x), so G'(3) == g'(g(3)). g'(3) == g'() 9 == 7 9 == (a) u(x) == f(g(x)) ~ u'(x) == f'(g(x))g'(x). So u'(l) == f'(g(l))g'(l) == f'(3)g'(1). To find f'(3), note that f is linear from (,4) to (6,3), so its slope is 3-4 == - ~. To find g' (1), note that 9 is linear from (0,6) to (,0), so its slope 6-4 is ~ =~ = -3. Thus, 1'(3)g'(1) = (-i) (-3) = ~. (b) v(x) == g(f(x)) ~ v'(x) == g'(f(x))f'(x). So v'(l) == g'(f(l))f'(l) == g'()f'(1), which does not exist since g' () does not exist. (c) w (x) == 9(g(x)) ~ w' (x) == g'(g(x) )g'(x). So w' (1) == g'(g(1) )g' (1) == g' (3) g' (1). To find g' (3), note that 9 is linear from (,0) to (5,), so its slope is ~ =~ = ~. Thus, g'(3)g'(1) = G) (-3) = (a) h(x) = f(f(x)) ~ h'(x) == f'(f(x))f'(x). So h'() == f'(f())f'() == f'(1)f'() ~ (-1)(-1) == 1. (b) g(x) = f(x ) =} g'(x) = J'(x ). d~ (x ) = 1'(x )(x). So g'() = 1'( )(.) = 41'(4) ;::::: 4() = (a) F(x) = f(ex) =} F'(x) = l'(ex) d~ (ex) = l'(ex)e X (b) G(x) == ef(x) ~ G' (x) == ef(x) ~ f(x) == ef(x) f' (x) dx 68. (a) F(x) = f(x") =} F'(x) = i'(x") d~ (x") = l'(x")o!x"-l (b) G(x) == [f (x)]lx ~ G' (x) == a [f (x)]lx-1 f' (x) 69. r(x) == f(g(h(x))) ~ r'(x) == f'(g(h(x))). g'(h(x)). h'(x), so r'(l) == f'(g(h(l))). g'(h(l)). h'(l) == f'(g()). g'(). 4 == f'(3) 5 4 == == f(x) == xg(x ) ~ f'(x) == xg'(x ) x + g(x ). 1 == x g'(x ) + g(x ) ~ f"(x) == x g"(x ) x + g'(x ) 4x + g'(x ) x == 4x 3g"(X ) + 4xg'(x ) + xg'(x ) == 6xg'(x ) + 4x 3g"(X ) 71. F(x) == f(3f(4f(x))) ~ F'(x) = 1'(3f(4f(x))). d~ (3f(4f(x))) = 1'(3f(4f(x))). 31'(4f(x)). d~ (4f(x)) == f'(3f(4f(x))). 3f'(4f(x)). 4f'(x), so F'(O) == f'(3f(4f(0))) 3f'(4f(0)) 4f'(0) == f'(3f(4. 0)) 3f'(4 0) 4 == f'(3 0) 3 4 == 3 4 == F(x) == f(xf(xf(x))) ~ F'(x) = l'(xf(xf(x))). d~ (xf(xf(x))) = l'(xf(xf(x))). [x. l'(xf(x)). d~ (xf(x)) + f(xf(x)). 1J == f'(xf(xf(x))). [xf'(xf(x)). (xf'(x) + f(x) 1) + f(xf(x))], so F'(l) == f'(f(f(l))). [f'(f(l)). (f'(l) + f(l)) + f(f(l))] == f'(f()). [f'(). (4 + ) + f()] == I' (3). [ ] == == 198.

9 19 0 CHAPTER 3 DIFFERENTIATION RULES 73. y == Ae- x + Bxe?" =? y/ == A( _e- X ) + B[x(-e- X ) + e- x. 1] == -Ae- x + Be- x - Bxe- x == (B - A)e- X - Bxe- x =? y" == (B - A)(_e- X ) - B[x(_e- X ) + e- x 1] == (A - B)e- X - Be- x + Bxe- x == (A - B)e- X + Bxe- x, SO y" + y' + y == (A - B)e- X + Bxe- x + [(B - A)e- X - Bxe- X ] + Ae- x + Bxe- X == [(A - B) + (B - A) + A]e- X + [B - B + B]xe- X == O. (r + 6) (r - 1) == 0 =? r == 1 or The use of D, D,..., D" is just a derivative notation (see text page 157). In general, D f(x) == f'(x), D f(x) == 4f"(x),..., D" f(x) == n f(n)(x). Since f(x) == cos x and 50 == 4(1) +, we have f(50)(x) == f() (x) == - cos x, so D 50 cos x == _ 50 cos x. 76. f(x) == xe- x, f' (x) == e- x - xe- x == (1 - x )e- X, f" (x) == _e- x + (1 - x) (_e- X ) == (x - )e- x. Similarly, f'ii(x) == (3 - x)e- X, f(4)(x) == (x - 4)e- X,, f(looo)(x) == (x )e- x. 77. s(t) == 10+ ~sin(101rt) =? the velocity aftert seconds isv(t) == s/(t) == ~cos(101rt)(101r) == 57f cos(101rt) cru/s. 78. (a) s == A cos(wt + 6) =? velocity == s' == -wa sin(wt + 6). n1r - 6. (b) If A -# 0 and w -# 0, then s' == 0 {:} sin(wt + 6) == 0 {:} wt + 6 == nat {:} t ==--,ti an Integer. w 79. (a) B(t) == sin 1rt =? db == (0.35 cos 1rt) (1r) == 0.71r cos 1rt == 71r cos 1rt 5.4 dt db 71r 1r (b) At t == 1, -d == -4 cos -4 ~ t L(t) == 1+.8sin(J~(t-80)) =? L'(t) ==.8cos(J;5(t-80))(J;5)' On March 1, t == 80, and L'(80) ~ hours per day. On May 1, t == 141, and L'(141) ~ , which is approximately one-half of L' (80). 81. s(t) == e-1. 5t sin 1rt =? v( t) == s' (t) == [e-1.5t (cos 1rt) (1r) + (sin 1rt)e-1. 5t (-1.5)] == e -1.5t (1r cos 1rt sin 1rt) 15 Graph of position o f ~-_= I Graph of velocity o I--\ ,f---+--~-----:;..,...=::- I -1-7 kt 8. (a) lim p(t) == lim 1 kt 1 == 1, since k > 0 =? -kt =? e- -+ O. t.s-sco t--+oo 1 + a.c: 1 + a 0 kt dp - (1 + ae -kt) --- dt (1 + ae-kt ) (b) p(t) == (1 + ae-kt)-l - _ - -kt)-( - k ae - _ kae

10 SECTION 3.4 THE CHAIN RULE D 193 (c) p=o.8 From the graph ofp(t) = (1 + 10e- o. 5t )- 1, it seems thatp(t) = 0.8 (indicating that 80% of the population has heard the rumor) when t ~ 7.4 hours. o dv dv ds dv dv 83. By the Chain Rule, a(t) = dt = ds dt = ds v (t) = v (t) d.s' The derivative dv / dt is the rate of change ofthe velocity with respect to time (in other words, the acceleration) whereas the derivative dv / ds is the rate of change ofthe velocity with respect to the displacement. 84. (a) The derivative dv/ dr represents the rate of change ofthe volume with respect to the radius and the derivative dv/ dt represents the rate of change ofthe volume with respect to time dv dv dr dr (b) SInce V = "3 7TT, dt = dr dt = 47fr di: 85. (a) Using a calculator or CAS, we obtain the model Q = ab t with a ~ and b ~ (b) Use Q' (t) = ab t In b (from Formula 5) with the values of a and b from part (a) to get Q' (0.04) ~ /--lao The result of Example in Section.1 was -670 /--lao 86. (a) P = ab t with a = X 10-0 and b = , 3,000 (P in thousands) where P is measured in thousands ofpeople. The fit appears to be very good. ( b) F r 1800 = = m = = o m, ' So p'(1800) ~ (m1 + m)/ = thousand people/year. For 1850 m = 3,19-17,063 = 61.9 m = 31,443-3,19 = _ 1840 ' So p'(1850) ~ (m1 + m)/ = 719 thousand people/year. (c) Using p'(t) = ab t In b (from Formula 5) with the values of a and b from part (a), we get p'(1800) ~ and p'(1850) ~ These estimates are somewhat less than the ones in part (b). (d) P(1870) ~ 41, The difference of3.4 million people is most likely due to the Civil War ( )... I () 45 (t - ) 8. h. 1". h. h 1 hemati fi 87. (a) Denve grves 9 t = (t + 1)10 wit out simp ifying. WIt eit er Map e or Mat ematica, we rst get g' (t) = 9 t ~ );9-18 ( (t - ~:o, and the simplification command results in the expression given by Derive. t + 1 t + 1 (b) Derive gives y' = (x 3 - X + 1)3(x + 1)4(17x 3 + 6x - 9x + 3) without simplifying. With either Maple or Mathematica, we first get y' = 10(x + 1)4(x 3 - X + 1)4 + 4(x + 1)5(x 3 - X + 1)3(3x - 1). Ifwe use Mathematica's Factor or Simplify, or Maple's factor, we get the above expression, but Maple's simplify gives the polynomial expansion instead. For locating horizontal tangents, the factored form is the most helpful.

11 194 0 CHAPTER 3 DIFFERENTIATION RULES 4 )1/ (3x4-1) X: -x (a) f (x ) = ( X -x 1 Derive gives 1' (x ) = ( 4 )(x 4 + X 1.) whereas either Ma ple or Mathem atica x 4 + X + 1 X + x+ 1 X - x I ' ( ) 3x - 1 fter si lifi. grve X = r ==j== = = a er simp I cation. 4 x - X + 1 ( 4 X +x + 1) 3 x 4 + X + 1 (b)1'(x) = O <=? 3x 4-1 = 0 <=? x= ± iff ~± (c) Yes. l'(x) = 0 where f has horizontal tangents. r has two max ima and one minimum where f has inflection points (a) If f is even, then f (x ) = f( - x ). Using the Chain Rule to differentiate this equation, we get 1'(x) = 1' (- x ) d~ (- x) = - 1' (- x ). Thus, 1' (-x) = - 1' (x ), so l' is odd. (b) If f is odd, then f (x ) = - f (- x ). Differentiating this equation, we get i ' (x ) = -1'(- x)(- 1) = l'(-x), so l' is even. 90. [ ~~: ~ r = {J(x ) [g(x )]- l }' = 1' (x ) [g (X)t1+ (- 1) [g(x)]- g'( x)f(x) 1'(x) f(x)g'(x) 1'(x)g(x) - f(x)g'(x ) g(x ) - [g(x)] [g( X)] This is an alternative derivation of thejormula in the Quotient Rule. But part ofthe purp ose ofthe Quotient Rule is to show that if f and 9 are differentiable, so is f / g. The proof in Section 3. does that; this one doesn't. 91. (a) d~ (sin" x cosnx ) = nsin n - 1 x cos x cos n x + sin" x (-nsin n x) [Product Rule] = n sin n - 1 x (cosnx cos x - sin n x sin x ) [factor out n sin n-1x ] = n sin n - 1 x cos(n x + x ) [Additi on Formula for cosine] = n sin n-1 xcos[(n + l )x ] [factor out x] (b) d~ (cos" x cos nx) = n cosn- 1 x (- sinx) cos nx + cos" x (-n sinnx) [Product Rule] = - n cosn- 1 x (cosnx sin x + sin nx cos x ) [factor out - n cosn- 1 x ] = - n cosn- 1 x sin(nx + x ) [Addition Formula for sine] = - n cosn- 1 xsin [(n + l )x ] [factor out x] 9. "The rate ofchange of y5 with respect to x is eighty times the rate of change of y with respect to x" <=? 5y 4 dy = 80 dy <=? 5y4 = 80 (Note that dy / dx =1= 0 since the curve never has a dx dx horizontal tangent) <=? y4 = 16 <=? Y = (since y > 0 for all x ) 93. Since tl? = C ; o)b rad, we have :B (sin BO) = :B (Sin 1;0B) = 1;0 cos l;o B = 1;0 cosb o. f is not differentiable at x = o.

12 SECTION 3.4 THECHAIN RULE (b) f( x) = [sin z] = Vsin x =} j' (x ) = ~(s i n x)-1/ sinx cos x = I s ~ n x i cos x smx COS X if sin x > 0 - { - cos x if sin x < 0 f is not differentiable when x = mr, n an integer. (c) g(x) = sin [z ] = sin H =} y if x > 0 x x {cos x g'(x) = cos Ixl ' -I I = -I I cos x = x x - cos x if x < 0 g is not differentiable at O. g g'. dy dy du 95. The Cham Ru le says that dx = du dx ' so [Product Rule] _ [.!!.. ( d y) 3udy du] (du) + du du dy + [.!!.. ( dy) du] ( d U) + d - du du? dx dx dx dx du du du dx dx dx 3 du