-,~. Implicit Differentiation. 1. r + T = X2 - y2 = x3-xy+y2=4. ry' + 2xy + T + 2yxy' = 0 (r + 2xy)y'= _(y2+ 2xy)

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1 Section.5 mplicit Differentiation f(x) = x nx, f(l) = 0 f'(x) = + nx, f(l) = r(x) = -, x r() = Pl(X) =f(l) + f'(l)(x - ) = x -, P(l) = 0 -,~. 3 - P(x)= f() + f'(l)(x - ) + ~r(l)(x - ) = (x- )+ (x - ), P(l) = 0 P'(x) =, P'(l) = P/(x) = + (x - ) = x, P/(l) = P/'(x) = x, P"()= The values off, P, P' and their first derivatives agree at x =.The values of the second derivatives offandp agree atx =. 57. False. fy = ( - X)/,theny' =!(l - X)-/(_). 58. False. ff(x) = sinx, thenf'(x) = (sin x)( cos x). 59. True Section.5 mplicit Differentiation. r + T = 6. X - y = 6 x + yy' = 0 -x y'=y x - yy' = 0 X y'= ȳ 3. Xl/ + yl/ = 9 4. x3+y3=8 -X-l/ + _y-l/y' = 0 y'= _X-l/ = - y-l/ -V; i 3r + 3yy' = 0 r y' = - y 5. x3-xy+y=4 3r - xy'- y + yy'= 0 (y- x)y' = y - 3r y - 3r y'=- y- x 6. ry + yx= - ry' + xy + T + yxy' = 0 (r + xy)y'= _(y+ xy), -y(y + x) y = x(x + y) 7. xey- lox+ 3y = 0 8. exy+r-y= 0 xeydy + ey dy = 0 dy (xey+ 3) = 0 - ey dy_lo-ey -xey+3 (x: + y)exy+ x - y: = 0 dy (xexy- y) = -yexy- x dy= yexy+x xexy- y

2 46 Chapter Differentiation 9. x3- xy+ 3xy= 38 3x - xy' - 4xy + 6xyy' + 3y = 0 x(3y - x)y' = 4xy - 3x- 3y y' = 4xy - 3X -3y x(3y - x) 0. sinxcosy = [sin x( - sin y)y' + COgy(cos x)] = 0, cosxcosy y = sinxsiny ==cot x coty. sin x + cos y = cosx - 4(siny)y' = 0 cosx y'=4siny. (sin TTX + COg 7Ty) = (sin 7TX + COS7TY)[7T COg TTX - 7T(sin7Ty)y = 0 7T COg TTX - 7T(sin7TY)y'= 0 COg TTX y' = sin TTY 3. sinx = x(l + tany) COgx = X(secy)y' + (l + tany)(l), cosx-tany- y = xsecy 4. coty=x-y (-cscy)y' = - y' y'=-cscy =~ -cot Y = -tany 5. y = sin(xy) y' = [xy' + y] cos(xy) y' - xcos(xy)y' = ycos(xy), - y cos(xy) y - - x cos(xy) 6. x=,sec- y y' = --sec-tany y Y y' = sec(l/~~~(l/y) = _y cos(~)cot(~) 7. x3y3_y_x=0 3x3yZy'+ 3xl - y' - = 0 (3x3y- l)y' = - 3xy3, - - 3xl y-3x3y- 8. (xy)/- X + y = 0 ~xy)-/(xy' + y) - + y' = 0 --y' + -L- - + y' = 0 Fxy Fxy xy' + Y - Fxy + 4Fxyy' = 0, Fxy - y y = 4Fxy + x 9. x - 3 n y + y = 0 3dy x---+y-=0 y dy x =:(~ - Y) dy- x - xy - (3/y) - y y 0. n(xy)+ 5x = 30 nx + ny + 5x = dy + 5 = 0 x y --=---5 dy y x dy = _l - 5y = - ( y + 5XY ) x x

3 Section.5 mplicit Differentiation 47. xy = 4 xy' + y() = 0 xy' = -y y'= -y X At(-4,-): y'= -4'. xl-y3=o x - 3yy' = 0 x y' = 3y At (,): y' = 3" 3. xl - 9 y = xl + 9 ' - (X + 9)(x) - (xl - 9)x yy - (xl + 9), 8x y = (X+ 9)y At (3,0): y' is undefined. 4. (x + y)3 = x3 + y3 x3 + 3xy + 3xy + y3 = x3 + y3 3xy + 3xy = 0 xy + xy = 0 xy' + xy +.xyy' + y = 0 (xl + xy)y' = _(y + xy), y(y + x) y = - x(x + y) At(-l, ): y'=-. 5. X/3 + y/3 = 5 6. x3+y3_xy=0 -X-/3 + ::;y-/3y' = X-/3 y' = Y y-/3 - ~ 3xl + 3yy'-.xy'- y = 0 At (8,): y' = -:" At (, ): y' = -. (3y- x)y' = y - 3xl, y - 3xl y = 3y- x 7. tan(x + y) = x 8. xcosy = ( + y~ sec(x+ y) = At (0, 0): y' = O., - sec(x+ y) Y = sec(x+ y) - -tan(x + y) - tan(x+ y) + xl --xl+ x[-y' sin y] + cosy = 0 cosy y'= xsiny - cot Y =-coty- x x At (,~): y' = ~' 9. ex}' - x = 0, (,0) 30. y = nx, (e,l) ex}'[xy'+ y] - = 0 At (,0): ex}'xy'= - yex)' y'=4', - - yex)' y - xex)' yy' = ~ y' = xy At(e,): y'=e'

4 48 Chapter Differentiation 3. Jx+Jy=3 -x-/ + -y-/y' = 0 y' = (/)X-/ Jy (/)y-/ = - Jx At (4, ): y' = -" Tangent line: y - = --(x - 4) x + y - 6 = 0 -~7 5 ~ - y = --x x- 3. Y - X +, (x + )() - (x - )(x) yy = (x+ ) x+-r+x (x + ) +x-x y' = Y(X+ ) ( y"s, ~ At, 5 ): y = [(y"s)/s](4+ )- Oy"S. Tangent line: y"s y - - = -(x - ) S Oy"S loy"sy - 0 = x - x - loy"sy + 8 = 0 ~ -~=r (X + 4)y = 8 (X + 4)y' + y(x) = 0 '- -xy y-x+4 - _x[8/(x + 4)] - x+4-6x = (x + 4) -3 At (,): y' = M = (4 - X)y= X3 (4 - x)(yy~ + y(_) = 3x At (,): y'=. 3X+y y' = y(4 - x) (or, you could just solve for y: y = X~ 4) 35. (x + y) = 4xy (x + y)(x + yy~ = 4xy'+ y(8x) 4X3+ 4xyy' + 4xy + 4yV = 4xy' + 8xy 4xyy' + 4yV 4y'(xy + y3 - X) - 4xy' = 8xy - 4x3-4xy = 4(xy- x3- xy),_xy-x3-xy Y - xy+y3-x 36. X3+y3_6xy=O 3X+ 3yy' - 6xy' - 6y = 0 y'(3y - 6x) = 6y - 3x 6y -3x y -x y' y - 6x - y - x ( 4, (6/3) - (6/9) 3 4 At 3' 3 8) : y = (64/9) - (8/3) = 40 = S. At (,): y' = O.

5 Section.5 mplicit Differentiation (a) x + y= 6 y= 6 - r (b) y = :t~6 - r (c) Explicitly: dy = :tl.(6 - r)-/(-x) - +=x - - x = -x - ~6 - X- :t~6 - r y (d) mplicitly: -6 x+yy'=o x y'= -;; 38. (a) (X- 4x + 4) + (y + 6y + 9) = (c) Explicitly: (b) - -3 (x - ) + (y + 3) = 4 (Circle) (y + 3)= 4 - (x - ) Y = -3:t--/4 - (x - ) dy = :t 4 - (x - )]-/(-)(x- ) +=(x- ) (~4 - (x - ) -(x - ) :t~4 - (x - ) -(x-) -3 :t ~4 - (x - )+ 3 = - (x - ) y+3 (d) mplicitly: x + yy' y' = 0 (y + 6)y' = -(x - ),_-(x-) y y:t3 39. (a) 6y= 44-9X 9 y = -(44-9X)= -(6 - X) 6 6 (c) Explicitly: dy 3 - = :t-(6 - r)-/(-x) 8 3 y = :t4~6 - X 3x -3x -9x = +=4~6 - r = 4(4/3)y = 6y (b) (d) mplicitly: -6 8x + 3yy' = 0-9x y' = 6y

6 50 Chapter Differentiation 40. (a) y = + X = r (b) Y = -z-..}r + 4 (c) Explicitly: dy = +.!. (r + 4)-/(x) -4 -zx =../r zx _x - 4(/)..}r + 4-4y (d) mplicitly: yy' - -x = 0 X y' = 4y 4. r+xy=5 x + xy' + Y = 0 + xy" + y' + y' = 0 -(x+y) y'=~ xy"= -( +y " - [ - (x + y)/x] (x + y) Y = x =~ " (x + y) x 0 y= r.:;=x3 (Note: Youcouldwritey = (5 - r)/x and calculate y' and y" directly.) 4. ry-x=3 xyy' + xy- = 0 ryy' + xy - = 0 l-xy y'- -~ xyy'+r(y + xyy"+ xyy'+ y= 0 4xyy' + r(y + xyy"+ y = 0 4-4xy ( - xy) xyy"+ y= 0 X xy 4xy - 4ry4 + - xy + ry4 + x4y3y" + H = 0 x4y3y" = xy4- xy - x,.4- xy - y" =.f x4y3

7 Section.5 mplicit Differentiation X - y = 6 x - yy' = 0 X y'= ȳ x - yy' = 0 - yy" - (y ~ = 0 ( X ),, -yy-y =0 y - y3y" = X 44.-xy=x-y y-xy=x-l x- y=-=- - y' = 0 y"= 0 x y" = y - xl = -6 y3 y3 45. y=x3 46. y = 4x yy' = 3X, 3X 3x xy 3y x3 3y y y - y xy - x y - x " x(3y~ - 3y() y = 4xl - x[3. (3y/x)] - 6y - 4X yy' = 4, y =- Y - y" = -y-y' = [.~ Y Y y3 ] -ll. - 3x - 4X - 4y 47. X + y = 5 -x Y'=y At (4, 3): -4 Tangent line: y - 3 = 3(x - 4) ~ 4x + 3y - 5 = 0 ~ 6 Normal line: y - 3 = ~(x - 4) ~ 3x - 4y = O. At (- 3,4): Tangent line: y - 4 = ~(x + 3) ~ 3x - 4y + 5 = 0-4 Normal line: y - 4 = 3(x + 3) ~ 4x + 3y = O. -9lEZJ 9-6 ~ 6-9 [SJ] 9-6

8 5 Chapter Differentiation 48. x + y = 9 -x Y'=y At (0, 3): 4 EZELJ Tangent line: y = 3-6[TJ6 Normal line: x = O. -4 At (,,)5): - Tangentline: y -.,/5 =.,/5 (x - ) => x +.,/5y - 9 = 0 Normalline: y -.,/5 = ~(x - ) =>.,/5x - y = O. UE3-6 LBJ x + y = r x + yy' = 0 -x y' = - y = slopeof tangentline l = slope of normal line x Let (xo'y~ be a point on the circle. f Xo= 0, then the tangent line is horizontal, the normal line is vertical and, hence, passes through the origin. f Xo*- 0, then the equation of the normal line is y - y = Yo(x - x ) 0 Xo 0 Yo y=-x Xo which passes through the origin. 50. y = 4x yy' = 4 y'=-= y lat(,) Equation of normal at (, ) is y - = - (x - ), Y = 3 - x. The centers of the circles must be on the normal and at a distance of 4 units from (,). Therefore, (x - )+ [(3 - x) - ]= 6 (x - )= 6 x = ::f:j. Centersof thecircles: ( + J, - J) and( - J, + J) Equations:(x - - J) + (y - + J) = 6 (x - + J) + (y - - J) = 6

9 Section.5 mplicit Differentiation x + 6y+ 00x - 6Oy+ 400 = 0 5Ox + 3yy' Oy' = 0 Horizontal tangents occur when x = -4:, Ox y = 60-3y 5(6)+ 6y+ 00(-4) - 6Oy+ 400 = 0 Horizontaltangents: (- 4,0); (- 4, 0). Vertical tangents occur when y = 5: 5r x = 0 Verticaltangents: (0,5), (- 8, 5). y(y - 0)= 0 ~ y = 0,0 5x(x + 8) = O~x = 0,-8 (-8.5) r + y- 8x + 4y + 4 = 0 8x + yy' y' = 0 8-8x 4-4x y' = y + 4 = y + Horizontal tangents occur when x = : 4() + y - 8() + 4y + 4 = (.0) """ (0,-)l (.-) -3-4 (,:'4) -s y + 4y = y(y + 4) = 0 ~ y = 0, -4 Horizontal tangents: (,0), (, -4). Vertical tangents occur when y = - : 4r + (- )- 8x + 4(- ) + 4 = 0 4r - 8x= 4x(x- ) = 0 ~ x = 0, Verticaltangents: (0, -), (, -). 53. y = xr-=t ny = nx + Zn(r - ) Hz) = ~ + X ~ ~- [ x- ] _x- - Y x(r - ) - R-=t 54. y = ~(x - l)(x - )(x - 3) ny = Z[n(x- ) + n(x - ) + n(x - 3)] ~(z) = Ux ~ + x ~ + x ~ 3] [ 3X- x + = Z (x - )(x - )(x - 3)] dy - 3r - x+ - y 3r - x+ ~(x - l)(x - )(x - 3)

10 54 Chapter Differentiation 55. X -= y = (x - ) n y = n x + n(3x - ) - ln(x - ) ( dy ) 3 Y = ~ + (3x - ) - x - dy [ 3x - 5x + 8 = Y x(3x- )(x - ) ] 3x3-5r + 8x - (x - )3~3x ~ + y= x- ny = 3[n(x + ) -n(x + ) -n(x - )] H) = HX ~ - x ~ - x ~ ] dy - -4xy - - 4x -vir + - 3(x4- ) - 3(x4- ) r x - 3(r + )/3(X- )4/3 57. x(x - )3/ Y = v"x+l 3 n y = n x + ln(x - ) - ln(x + ) ( dy ) 3( ) ( ) y =~+x- -x+ =~[~+ x~ - x~ ] = [ 4X + 4x - = (x + x - )~ x(x - ) ] (x + )3/ 58. (x + )(x + ) y = (x - )(x - ) ny = n(x+ ) + n(x+ ) - n(x- ) - n(x- ) H)=x~l+x~-x~l-x~ dy = Y X - + r - 4 = Y (r - )(r - 4) [ - -4 ] [ -6X+ ] - (x + )(x + ). - 6(X - ) - (x - )(x - ) (x + )(x - )(x + )(x - ) 6(X- ) (x - )(x - ) 59. y = X/x ny=-nx x. ( dy ) = ~ (. ) + nx ( -~ ) = ~( - n x) y xx X X dy = y X ( - n x) = x(/x)-( - n x) 60. y =XX- n y = (x - l)(n x) ~() = (x - )(;) + nx dy X - [ =y~+lnx ] = xx- (x - + xlnx) 6. y = (x - )X+ ny = (x + )ln(x - ) ~ () = (x + )C ~ ) + n(x - ) dy X + [ x- - = y '-- + n(x- ) ] = (x X + - )X+ [ - x- + n(x - ) ] 6. y = ( + X)l/x ny = -n( + x) x. ( dy ) =. ( ~ ) + n( + x) ( -- y xl+x X) dy = y.. [ ~ - n(x + ) xx+ x ] = ( + X)l/x [ ~ - n(x+ ) x x+ x ]

11 Section.5 mplicit Differentiation Find the points of intersection by letting y = 4x in the equation,x + y = 6.,x+ 4x = 6 and (x + 3)(x - ) = 0 The curves intersect at (, :t). Ellipse: 4x + yy' = 0 x y'= -- Y Parabola: yy' = 4 y' =- y -6 At (, ), the slopes are: y' = - y' =. At (, -), the slopes are: y' = y' = -. Tangents are perpendicular. 64. Find the points of intersection by letting y = x3 in the equation,x + 3y = 5. x + 3X3= 5 and 3x3+x_5=0 ntersect when x =. Points of intersection: (,:t ) y = X3: yy' = 3X '- 3X Y -- y At (, ), the slopes are: 3 y' =- At (, -), the slopes are: 3 y'= - Tangents are perpendicular. x + 3y = 5: 4x + 6yy' = 0 x y' = - 3y y' = -'3' y' = '3' i=x3 65. y = -x and x = siny Point of intersection: (0, 0) y = -x: x=siny: y' = - = y'cosy y'=secy -6 6 At (0, 0), the slopes are: y' = - y' =. Tangents are perpendicular.

12 56 Chapter Differentiation 66. Rewriting each equation and differentiating, x3 = 3(y - ) x(3y - 9) = 3 x3 y=-+ 3 y = ~(~ + 9) y' = X y' = - X' -3 For each value of x, the derivatives are negative reciprocals of each other, Thus, the tangent lines are orthogonal at both points of intersection, 67. xy=c x_y=k xy' + Y = 0 x - yy' = 0 X y'= _l X y'= - Y At any point of intersection (x, y) the product of the slopes is (-y/x)(x/y) = -. The curves are orthogonal. -,~"c:s=:rn, ~ bl:tj x+y=c y = Kx x + yy' = 0 y'= K X y'= - ȳ At the point of intersection (x,y) the product of the slopes is (- x/y)(k) = (- x/kx)(k) = -. The curvesareorthogonal..tm3..~, y - 3x4 = 0 (a) 4yy' - x3= 0 4yy'= x3, x3 3x3 Y y - y dy (b) 4ydt - x3- dt = 0 ydy dt = 3X3 dt 70. x - 3xy + y3 = 0 (a) x - 3y - 6xyy' + 3yy' = 0 (-6xy + 3y)y' = 3y - x y'= 3y. - x 3y - 6xy dy dy (b) x - - 3y- - 6xy - + 3y- = 0 dt dt dt dt (x - 3y)= (6xy - 3y)dy. dt dt 7. cos TTY- 3 sin TTX= (a) - 7Tsin (77)')y' - 37T CDSTTX = 0, - 3 cos TTX Y = sin TTY (b) -7Tsin(7Ty)dy dt - 37Tcos(7Tx) dt = 0 dy -sin(7tytdt = 3 COS(7TX7t 7. 4 sinx cos y = (a) 4 sin x(-sin y)y' + 4cosxcosy = 0, cos x cos y y = sin x sin y (b) 4sinx(-siny)! + 4COSX~ cosy = 0.. dy cos x cos Y dt = sm x sm y dt

13 Section.5 mplicit Differentiation (a).0 = 4(4x - y) 0 4y = 6x -.0 y = 4X y =j::: - ~.0 -"~,, -0 (b) y = 3 ~ 9 = 4X x+ 36 = 0 36 = 6x -.0 X = 6 ~56-44 = 8 + "8 -~ O Notethatx = 8 h8 = 8 fi = ( fi). Hence, there are four values of x: - - fi, - fi, -+ fi, + fi fi d h ' ' - x(8 - X) To t e s ope, yy - x - x ~ y - (3). For x = - - fi, y' = Hfi + 7),andthelineis Yl = Hfi + 7)(x+ + fi) + 3 = t[(fi + 7)x+ 8fi + 3]. Forx = - fi, y' = Hfi - 7),andthelineis yz = Hfi - 7)(x - + fi) + 3 = H(fi - 7)x + 3-8fi]. Forx = - + fi,y'= -Hfi - 7),andthelineis Y3= -Hfi - 7)(x + - fi) + 3 = -t[( fi - 7)x - (3-8fi)]. Forx = + fi,y' = -Hfi + 7),andthelineis Y4= -Hfi + 7)(x - - fi) + 3 = -t[( fi + 7)x - (8fi + 3)]. (c) Equating Y3 and Y4' --(fi - 7)(x + - fi) + 3 = --(fi + 7)(x - - fi) (fi - 7)(x+ - fi) = (fi + 7)(x- - fi) fix + fi - 7-7x fi = fix - fi x - 7-7fi 6fi = 4x 8fi x=7 f x = 8f, theny = 5 andthe linesintersectat (8f, 5).

14 58 Chapter Differentiation 74. tany=x y'secy =, ~ ~ y = - -- < y < - secy' secy = + tany = + X y'= +x 75. Letf(x) = x" = xp/q, where p and q are nonzero integers and q > O.First consider the case where p = f(x) = Xl/q is given by ~[Xl/q] = lim f(x + /lx) - f(x) = lim f(t) - f(x) ~-->o /lx '-->x t - x where t = x + /lx. Observe that f(t) - f(x) - t/q - Xl/q - t/q - Xl/q t - x - t - x - (tl/q)q - (Xl/q)q (tl/q - Xl/q)(tl-(l/q) t/q - Xl/q + tl-(/q)xl/q tl/qxl-(/q)+ Xl-(l/q») - tl-(l/q) + tl-(/q)xl/q tl/qxl-(/q) + X-(/q)' As t~x, the denominator approaches qxl-(l/q). That is, d _[Xl/q] = = -X(l/q)-l. qxl-(l/q) q Now consider f(x) = :xf'/q= (:xf')/q. From the Chain Rule, f'(x) =.!.(:xf')(l/q)-l~:xf'] q =.!.(:xf')(l/q)-lp:xf'-l =!!..;,<p/q)-p]+(p-) =!!..X(p/q)-bnx"- ( n =!!..). q q q q 76. ~+Jy=Jc ~+~dy=o ~ Jy dy Jy - ~ Tangent line at (xo' Yo): y - Yo = - ~ (x - xo) yxo x-intercept:(xo + ~-So, 0) y-intercept:(0, Yo+ ~-So) Sum of intercepts: (xo+ ~-50) + (Yo+ ~-50) = Xo+ ~-50 + Yo= (~ + -50) = (Jc) = C.