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+ = +

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x. e's .¥CyTteKD .- l - de

DX

Xe's test - dy

d.X

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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,

5¥=EHIxlscnnaiLnxD-fxCOxcosnai@nyDcosl2x7adxl2xJ-x.d

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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

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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=

Answers: 1) 2

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

d y x adx 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

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flint KCH=6Xsin4x4co