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AP Calculus Name___________________Pd.___ Implicit Differentiation Derivatives (2) Day 1 1-10: Find dy/dx by implicit differentiation. 1. 3 3 1 x y + = 2. 2 2 4 x xy y + = 3. 3 2 3 2 2 x xy xy + = 4. y xe x y = 5. 2 2 cos y x x y = + 6. cos( ) 1 sin xy y = + 7. 4 cos sin 1 x y = 8. sin y e x x xy = + 9. x y e x y = 10. 2 2 1 x y xy + = + ¥K7tEx¥y7=¥kD 2xtx¥yTtyx3 - 2yd¥=0 2xtxd¥tyU ) 2yd£×=0 2ydf×-Xdf×=2xty Exley - x ) - - Zxty ¥=%t £ . - - Exch Exod x. date 't e' ¥63 - - I II , x. e 's .¥CyTteKD - l - de DX Xe 's test - dy d. X x :÷¥:¥i÷i .ie : ¥×CcosCxyD=g¥gi¥[ sing ] COSY # txsinlxyadf-ysinlxyj-sinlxyl.ae#y=coscygdgxcyyITkosyt- xsincxyfysinlxy ) ' si :* :* :±÷÷÷¥÷÷÷÷¥÷ii¥÷ - xsincxyldy ydd DX DX :i .ms?::i:::i;ie:.:i :¥¥s÷t¥¥ ekosxtsinxebgkf-ltxdzytysinxeYII-xdduj-lty-ekosxf-xkxi.gl "7=¥¥¥④④ chain ' zcxtyjk .dz#tyT--x?gtxCy7t-5adzEx7ztyF.C1tdfxJ--x2l2y)ddIxty42x ) zx¥ytz¥yd¥ - - 2×311×+2×5 zf.FI#-2x2ydfz-- 2×5 - Lay [x¥ 2×5111--2×5 ;¥g ÷÷÷÷÷÷i÷

# Xe .ie :±÷÷÷¥÷÷÷÷¥÷ii¥÷...dy y xy dx x xy y = 7) sin sin tan tan cos cos dy x y xy dx x y == 8) 1cos sin y y dy y e x dx xe x + = 9) 2 2 x y x y dy y y e dx yxe = 10)

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### Text of Xe .ie :±÷÷÷¥÷÷÷÷¥÷ii¥÷...dy y xy dx x xy y = 7) sin sin tan tan cos cos dy x y xy dx x...

• AP Calculus Name___________________Pd.___ Implicit Differentiation Derivatives (2) Day 1 1-10: Find dy/dx by implicit differentiation. 1. 3 3 1x y+ =

2. 2 2 4x xy y+ =

3. 3 2 32 2x x y xy+ =

4. yxe x y=

5. 2 2cosy x x y= +

6. cos( ) 1 sinxy y= +

7. 4cos sin 1x y =

8. sinye x x xy= +

9. xye x y=

10. 2 21x y x y+ = +

¥K7tEx¥y7=¥kD2xtx¥yTtyx3 - 2yd¥=02xtxd¥tyU) - 2yd£×=02ydf×-Xdf×=2xtyExley-x) -- Zxty ¥=%t£. -- Exch -Exodx. date'te'¥63 -- I - II

,

x. e's .¥CyTteKD .- l - de

DX

Xe's test - dy

d.X

x:÷¥:¥i÷i.ie:¥×CcosCxyD=g¥gi¥[sing]

- Lay[x¥ - 2×5111--2×5 ;¥g

÷÷÷÷÷÷i÷

• AP Calculus Name___________________Pd.___ Implicit Differentiation Derivatives (2) Day 1 11-12: Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

11. sin 2 cos2 , ,2 4

x x y=

12. 2 22 2, (1,2)x xy y x+ + =

13-14: Find b im lici diffe en ia i n. 13. 2 29 9x y+ =

14. 4 4 4x y a+ = 4 constanta =

① Point

② Slope #hit,

#(cost2yDtCostly)fzCx][cos (2x)= Xl - sinky)) Gy]tcosC2yXl)2cosC2xl= - Xsinczyl -2¥ t costly)Zxsinlzy)# = costly) - ZCOSCZX)

died = C0SC2y)-2x)

2xsinC2y )=cos( Iz) - 2COSTa¥h¥i¥j¥¥ ' - ¥2. Iz sin (2 'II )

¥44]t¥Cy4]=¥[a4]M°4×3+49314=0

I

453daL, =- 4×3 n 11×2=-3×44--3×6. >

g- ydy = - 4×3 IIDX Ty3 I

f÷4z= - 3xYy4t×4 ]¥x=- Toriginal problem÷¥÷¥÷¥¥÷s¥*÷÷÷÷÷.dd¥=y3t3x7tyf3yff, dIIz=-3ay!dd÷z= - 3x2y3t3x3y2¥YT42×4-2=-3×43+3×354-yx÷)-

y6

42×7=-3×19--331yyb

• AP Calculus Name___________________Pd.___ Implicit Differentiation Derivatives (2) Day 1 Review

15. 3

2lim

3x x+ 16.

2lim

3x x 17.

2

( ) 5 '( )xg x find g x=

18. ( ) ( )2 5( ) 3 2 2 5 '( )h x x x find h x= +

19. 2

( ) '( )3

f x find f xx

= 20. ( )3 2( ) sin '( )k x x find k x=

2

dy xdx y

= 2) 2 22 2

dy x y x ydx x y y x

+= = 3)

3 2

2 2

2 63

dy y xy xdx x xy

=

4) 11

y

y

dy edx xe

=+

5) 2 sincos 2

dy x y xdx x y

+= 6) sin( )

sin( ) cosdy y xydx x xy y

=

7) sin sin tan tancos cos

dy x yx y

dx x y= = 8) 1 cos

sin

y

y

dy y e xdx x e x

+= 9) 2

2

xy

xy

dy y y edx

y x e

=

10) 22

4 1

1 4

xy x ydydx x y x y

+=

+ 11) 2

4 2y x= 12) ( )72 1

2y x=

13)

2

2 3

9

81

dy xdx y

d ydx y

=

=

14) 33

2 2 4

2 7

3

dy xdx y

=

=

15)

16) 0 17) 2'( ) ln5 2 5xg x x= 18) 4'( ) 2(3 2)(2 5) (21 5)h x x x x= + + 19)

2

2'( )

( 3)f x

x= 20) ( ) ( )2 2 2'( ) 6 sin cosk x x x x=

(J- Lo)EB

0--00

" '"43×-254×[12×+55]taxi- 55,1×[13×-25]KIX)=( since))3

"H)=l3x-27512*551.242×+55.213×-2343) K'(X)=3( sin (XZ))?¥§in(X2)h'K7- 1013×-2512×+55't612×+5513×-27 K' (X)=3sin2(X4(cos(X2Dh' (X)-- 213×-2712×+5141513×-2)t3(2Xt5D"(x) -- 213×-2712×+57445×-101-6×1-15) K' ( X) -_3Sin4x2)COS(x2)l2X)

flint KCH=6Xsin4x4co

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