2.6
1. 4xy+2x+3x2 =
(a) Find y' by implicit differentiation.
x x dx
x dy dx
xdy
y 2 6 y
0 6
2+ = ⇒ =- - - +
+
⇒
(b) Solve the equation explicitly for y and differentiate to get y’ in terms of x.
x x y x
x x xy
2 4 2 3 2
4 3
2 - -
=
⇒
= + +
2
2 4
3 6
2
x x x
y x dx
dy - - - - -
=
=
(c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for y into your solution for part (a).
2 2 2
2 1 3 4
3 4 ' 3 2 3 4
2 4
x x x
y x x x
x
y x - -
- - -
- -
- = - ⇒ = - =
=
7. 4x2y2+xsiny= Find dx
dy by implicit differentiation.
y x y x
y y xy
y xy y yy x
xy 2 cos
sin ' 2
0 cos ' sin ' 2
2 2 2 2 2
= +
⇒
= +
+ +
⇒ - -
13. xy =1+x2y Find dx
dy by implicit differentiation.
xy x x
y xy y xy
xy y x xy xy xy y y x xy xy
xy y
2 2
2
2 ' 4 '
2 4
' '
2 2 '
-
= -
⇒ +
= +
⇒ +
+ =
⇒
15. If f(x)+x2[f(x)]3 =10 and f(1)=2, find f'(1). 0
) ( ' )]
( [ 3 )]
( [ 2 ) (
' + 3+ 2 2 =
⇒ f x x f x x f x f x 將x=1代入上式，
13 (1) 16 ' 0 (1) ' (1)]
3[
(1)]
2[
(1)
' + f 3+ f 2 f = ⇒ f =-
f
17. 3x2+xy+y2 = (1,1) Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
y x
y y x
yy xy y
x 2
' 2 0 ' 2 '
2 + + + = ⇒ = +
⇒ - -
將(1,1)代入上式y'(1,1) =-1
其切線方程式為y-1=-(x-1)，即x+ -y 2=0
21. )2(x2+y2)2 =25(x2-y2 (3,1) Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
) ' 2 25(2 )
' 2 )(2
4(x2+y2 x+ yy = x- yy
⇒
將(3,1)代入上式，
13 ' 9 ) ' 1 2 3 25(2 )
' 1 2 3 )(2 1
4(32 + 2 ⋅ + ⋅ ⋅y = ⋅ - ⋅ ⋅y ⇒ y=-
其切線方程式為 ( 3)
13 1 - 9 -
- x
y =
25. 1x3+ y3 = Find y" by implicit differentiation.
2 2 2 2 2
2 3 ' 0 '
3 - -x y-
y y x y
y
x + = ⇒ = =
⇒
5 5
3 5 3
4 2 2
2 3 2 2 3
2
2 2 ( ) 2
2 2
2 2
' 2
2 y
x y
y x y x
x xy y
x y x xy y
y x xy
y" - -
- -
- -
- - + - = - - - = - - = + =
=
42. Find equations of both the tangent lines to the ellipse x2+ y4 2 =36 that pass through the point (12,3).
y y x
yy
x 8 ' 0 ' 4
2 + = ⇒ =-
⇒
設切點(a,b)，則其切線方程式為 ( )
4 x a b b a
y- =- - 通過(12,3)，即
) 4 (12
3 a
b
b - a -
- = 得a2 +4b2 =12(a+b) 與原式a2+ b4 2 =36相交，即
⎪⎩
⎪⎨
⎧
=
=
=
⇒ =
= +
⎩ ⇒
⎨⎧
= +
+
= +
5 , 9 5 240, 3 36 3
4
) ( 12 4
2 2
2 2
b - a
b a b
b a a
b a b
a
其切線方程式為y=3或 )
5 ( 24 3 2 5
9 x-
y+ =
4.7
5. If y=x3+2x and =5 dt
dx , find dt
dy when x=2.
dt x dx dt
dx dt x dx dt
dy =3 2 +2 =(3 2+2)
⇒
70 5 2) 2 (3 2
5 2
=
⋅ +
⋅
=
=
= dt ,dx
dt x
dy
9. If snowball melts so that its surface area decreases at a rate of 1cm2/min, find the rate at which the diameter decreases when the diameter is 10 cm.
設雪球直徑 D cm，表面積為 A cm2， 2
2
4 D2 πD π
A ⎟ =
⎠
⎜ ⎞
⎝
= ⎛
dt πDdD dt
DdD dt π
πD dA
A= 2⎯d/dt⎯→ = (2 )=2
dt dA πD dt
dD = ⋅
⇒ 2
1
由題目可知要求
π π
dt dD
dt ,dA
D 20
) 1 1 10 ( 2
1
1 10
- -
-
=
⋅ ⋅
=
=
=
(cm/min)
21. Two carts, A and B, are connected by a rope 39ft long that passes over a pulley P. The point Q is on the floor 12 ft directly beneath P and between the carts. Cart A is being pulled away from Q at a speed of 2 ft/s. How fast is cart B moving toward Q at the instant when cart A is 5 ft from Q?
設 A 車距 Q 點 x ft，B 車距 Q 點 y ft，
則AP+BP= 144+x2 + 144+y2 =39
144 0 2
2 144
2 2
2
2 =
+ +
⎯→ +
⎯ dt
dy y y dt
dx x
d/dt x
dt dx x y
y x
dt dy
2 2
144 144
+
= - +
當x=5時，y=2 133，且 =2 dt dx
由題目可知要求
133 133 10 133 2 10
5 144 133 2
) 133 (2 144 5
2 2 2
5
- -
- ⋅ = =
+
= +
=
= dt ,dx
dt x
dy (ft/s)
25. Gravel is being dumped from a conveyor belt at a rate of 30 ft2/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 10 ft high?
設圓錐高為 x ft，則底半徑為 2
xft，體積
12 2
3
1 2 πx3
x x π
V ⎟ =
⎠
⎜ ⎞
⎝
= ⎛ ft3
dt dV πx dt dx dt dx πx dt
d/dt dV
2
2 4
4 ⇒ =
=
⎯→
⎯
由題目可知要求
π π
dt dx
dt ,dV
x 5
30 6 10 4
2 30
10
=
⋅ ⋅
=
=
=
(ft/min)
35. A plane flying with a constant speed of 300 km/h passes over a ground radar station at an altitude of 1 km and climbs at an angle of 30 . At what rate is the distance from the plane to the o radar station increasing a minute later?
設飛機飛行路徑為AB=x，塔台位於O 點，且OA=1km
欲求OB的時變率，即
dt dy
由餘弦定理可知
x x x
x
y2 =12 + 2-2 cos120o =1+ 2 +
dt x dx dt
dx dt xdx dt ydy
d/dt⎯→2 =2 + =(2 +1)
⎯
dt dx y x dt dy
2 1 2 +
=
⇒
一分鐘後，x=5，y= 31
由題目可知要求
31 300 1650 31
2 1 5 2
300 5,
= + ⋅
= ⋅
=
= dt x dx
dt
dy (km/hr)
2.8
3. Find the linearization L(x) of the function at a. f(x)=cosx, 2
=π a )
)(
( ) ( : )
(x y f a f' a x a
L - = -
⇒ y - x- -x ) 2
( π2 =π
=
11-14. Use a linear approximation (or differentials) to estimate the given number.
11. (2.001) 5
令y= f(x)=x5，則當切點為2 時的切線方程式為 128 80 )
2 80(
32 )
2 )(
2 ( ) 2
( - - - -
- f f' x y x y x
y = ⇒ = ⇒ =
32.08(2.001)5 ≈80×2.001-128=
13. (8.06) 2/3
令y= f(x)=x2/3，則當切點為8 時的切線方程式為 4 8) 1
1( 4 8)
(8)(
(8)= f' x ⇒ y = x ⇒ y= x+ f
y- - - -
A
O
B
3 4.02 8.06 4 3
(8.06)2/3 ≈1× + =
17. Find the differential of each function.
(a) y=x2sin2x
x x x x dy x x x x dx
dx
d/dx⎯→⎯ dy =2 sin2 + 22cos2 ⇒ =2 (sin2 + cos2 )
⎯
(b) y= 4+5x
dx
dy x dx x
d/dx dy
5 4 2
5 5
4 2
5
= + + ⇒
=
⎯
⎯→
⎯
19. Let y tan= x.
(a) Find the differential dy .
x dy xdx
dx
d/dx⎯→⎯ dy =sec2 ⇒ =sec2
⎯
(b) Evaluate dy and yΔ if 4
x= and π dx -= 0.1.
( 01) 02
sec2 4
1 0
4 . .
dyx π/ ,dx - . ⎟ - =-
⎠
⎜ ⎞
⎝
= ⎛
=
=
π
) 0.18237
(4 4 0.1)
( ) ( ) (
Δ = + - = π - - π ≈-
f f
x f dx x f y
21. The edge of a cube was found to be 30cm with a possible error in measurement of 0.1cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing (a) the volume of the cube and (b) the surface area of the cube.
(a) 設邊長 x cm，則正方體體積為 V cm3 V =x3
dx x dV dx x
dV 2 2
3
3 ⇒ =
=
當x=30，dx=0.1 時，dV x=30,dx=0.1 =3(30)2(0.1)=270(最大可能誤差)
相對誤差(relative error)為 0.01
30 270
3 =
= Δ ≈
V dV V
V
百分誤差(percentage error) 為Δ ×100%≈ ×100%=1%
V dV V
V
(b) 設邊長 x cm，則正方體表面積為 A cm2 6x2
A=
xdx dA
dx x
dA=12 ⇒ =12
當x=30，dx=0.1 時，dAx=30,dx=0.1=12(30)(0.1)=36(最大可能誤差)
相對誤差(relative error)為 0.0067
150 1 30 6
36
2 = ≈
= ⋅
≈ A dA A ΔA
百分誤差(percentage error) 為 100% 0.67%
A 100% A A
A× ≈ × =
Δ d
