Chapter 4 Solutions Manual

# Chapter 4 Solutions Manual - 11:31 L38-ch04 Sheet number 1...

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January 28, 2005 11:31 L38-ch04 Sheet number 1 Page number 67 black 67 CHAPTER 4 Derivatives of Logarithmic, Exponential, and Inverse Trigonometric Functions EXERCISE SET 4.1 1. y =(2 x 5) 1 / 3 ; dy/dx = 2 3 (2 x 5) 2 / 3 3. dy/dx = 2 3 µ x +1 x 2 1 / 3 x 2 ( x +1) ( x 2) 2 = 2 ( x 1 / 3 ( x 2) 5 / 3 5. dy/dx = x 3 µ 2 3 (5 x 2 5 / 3 (10 x )+3 x 2 (5 x 2 2 / 3 = 1 3 x 2 (5 x 2 5 / 3 (25 x 2 +9) 7. dy/dx = 5 2 [sin(3 /x )] 3 / 2 [cos(3 /x )]( 3 /x 2 )= 15[sin(3 /x )] 3 / 2 cos(3 /x ) 2 x 2 9. (a) 1+ y + x dy dx 6 x 2 =0 , dy dx = 6 x 2 y 1 x (b) y = 2+2 x 3 x x = 2 x +2 x 2 1 , dy dx = 2 x 2 +4 x (c) From Part (a), dy dx =6 x 1 x 1 x y x 1 x 1 x µ 2 x x 2 1 =4 x 2 x 2 11. 2 x y dy dx =0so dy dx = x y 13. x 2 dy dx xy +3 x (3 y 2 ) dy dx y 3 1=0 ( x 2 +9 xy 2 ) dy dx =1 2 xy 3 y 3 so dy dx = 1 2 xy 3 y 3 x 2 xy 2 15. 1 2 x 3 / 2 dy dx 2 y 3 / 2 , dy dx = y 3 / 2 x 3 / 2 17. cos( x 2 y 2 ) · x 2 (2 y ) dy dx xy 2 ¸ =1, dy dx = 1 2 xy 2 cos( x 2 y 2 ) 2 x 2 y cos( x 2 y 2 ) 19. 3 tan 2 ( xy 2 + y ) sec 2 ( xy 2 + y ) µ 2 xy dy dx + y 2 + dy dx so dy dx = 1 3 y 2 tan 2 ( xy 2 + y ) sec 2 ( xy 2 + y ) 3(2 xy + 1) tan 2 ( xy 2 + y ) sec 2 ( xy 2 + y ) 21. 4 x 6 y dy dx , dy dx = 2 x 3 y , 4 6 µ dy dx 2 6 y d 2 y dx 2 , d 2 y dx 2 = 3 ³ dy dx ´ 2 2 3 y = 2(3 y 2 2 x 2 ) 9 y 3 = 8 9 y 3 23. dy dx = y x , d 2 y dx 2 = x ( dy/dx ) y (1) x 2 = x ( y/x ) y x 2 = 2 y x 2

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January 28, 2005 11:31 L38-ch04 Sheet number 2 Page number 68 black 68 Chapter 4 25. dy dx = (1 + cos y ) 1 , d 2 y dx 2 = (1 + cos y ) 2 ( sin y ) dy dx = sin y (1 + cos y ) 3 27. By implicit diferentiation, 2 x +2 y ( dy/dx )=0 , dy dx = x y ;a t( 1 / 2 , 3 / 2), dy dx = 3 / 3; at (1 / 2 , 3 / 2), dy dx =+ 3 / 3. Directly, at the upper point y = 1 x 2 , dy dx = x 1 x 2 = 1 / 2 p 3 / 4 = 1 / 3 and at the lower point y = 1 x 2 , dy dx = x 1 x 2 =+1 / 3. 29. 4 x 3 +4 y 3 dy dx =0,so dy dx = x 3 y 3 = 1 15 3 / 4 ≈− 0 . 1312. 31. 4( x 2 + y 2 ) µ 2 x y dy dx =25 µ 2 x 2 y dy dx , dy dx = x [25 4( x 2 + y 2 )] y [25+4( x 2 + y 2 )] ;at(3 , 1) dy dx = 9 / 13 33. 4 a 3 da dt 4 t 3 =6 µ a 2 at da dt , solve For da dt to get da dt = 2 t 3 +3 a 2 2 a 3 6 at 35. 2 a 2 ω b 2 λ =0so = b 2 λ a 2 ω 37. (a) –4 4 –2 2 x y (b) Implicit diferentiation oF the equation oF the curve yields (4 y 3 y ) dy dx =2 x 1so dy dx =0 only iF x =1 / 2 but y 4 + y 2 0, so x / 2 is impossible. (c) x 2 x ( y 4 + y 2 ) = 0, so by the Quadratic ±ormula x = 1 ± p 1+4 y 2 y 4 2 =1+ y 2 , y 2 which gives the parabolas x y 2 ,x = y 2 . 39. (a) y x 2 –2 2 –2 (b) x ≈± 1 . 1547 (c) Implicit diferentiation yields 2 x x dy dx y y dy dx = 0. Solve For dy dx = y 2 x 2 y x .I F dy dx then y 2 x =0or y x .Thu s4= x 2 xy + y 2 = x 2 2 x 2 x 2 =3 x 2 , x = ± 2 3 .
January 28, 2005 11:31 L38-ch04 Sheet number 3 Page number 69 black Exercise Set 4.1 69 41. Solve the simultaneous equations y = x,x 2 xy + y 2 = 4 to get x 2 x 2 + x 2 =4 ,x = ± 2 ,y = x = ± 2, so the points of intersection are (2 , 2) and ( 2 , 2).

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Chapter 4 Solutions Manual - 11:31 L38-ch04 Sheet number 1...

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