chapter_04

# chapter_04 - January 27, 2005 11:44 L24-ch04 Sheet number 1...

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January 27, 2005 11:44 L24-ch04 Sheet number 1 Page number 127 black 127 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 2. dy/dx = 1 3 £ 2 + tan( x 2 ) ¤ 2 / 3 sec 2 ( x 2 )(2 x )= 2 3 x sec 2 ( x 2 ) £ 2 + tan( x 2 ) ¤ 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 4. dy/dx = 1 2 · x 2 x 2 5 ¸ 1 / 2 d dx · x 2 x 2 5 ¸ = 1 2 · x 2 x 2 5 ¸ 1 / 2 12 x ( x 2 5) 2 = 6 x ( x 2 5) 3 / 2 x 2 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) 6. dy/dx = 3 2 x 1 x 2 + 1 x 2 3(2 x 1) 2 / 3 = 4 x +3 3 x 2 (2 x 1) 2 / 3 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 8. dy/dx = 1 2 £ cos( x 3 ) ¤ 3 / 2 £ sin( x 3 ) ¤ (3 x 2 3 2 x 2 sin( x 3 ) £ cos( x 3 ) ¤ 3 / 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 10. (a) 1 2 y 1 / 2 dy dx cos x =0or dy dx =2 y cos x (b) y = (2 + sin x ) 2 =4+4sin x + sin 2 x so dy dx = 4 cos x + 2 sin x cos x (c) from Part (a), dy dx y cos x = 2 cos x (2 + sin x )=4cos x + 2 sin x cos x 11. 2 x y dy dx =0so dy dx = x y 12. 3 x 2 y 2 dy dx =3 y 2 +6 xy dy dx , dy dx = 3 y 2 3 x 3 y 2 6 xy = y 2 x 2 y 2 2 xy 13. x 2 dy dx xy 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

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January 27, 2005 11:44 L24-ch04 Sheet number 2 Page number 128 black 128 Chapter 4 14. x 3 (2 y ) dy dx +3 x 2 y 2 5 x 2 dy dx 10 xy +1=0 (2 x 3 y 5 x 2 ) dy dx =10 xy 3 x 2 y 2 1so dy dx = 10 xy 3 x 2 y 2 1 2 x 3 y 5 x 2 15. 1 2 x 3 / 2 dy dx 2 y 3 / 2 =0 , dy dx = y 3 / 2 x 3 / 2 16. 2 x = ( x y )(1 + dy/dx ) ( x + y )(1 dy/dx ) ( x y ) 2 , 2 x ( x y ) 2 = 2 y +2 x dy dx so dy dx = x ( x y ) 2 + y x 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 ) 18. sin( xy 2 ) · y 2 xy dy dx ¸ = dy dx , dy dx = y 2 sin( xy 2 ) 2 xy sin( xy 2 )+1 19. 3 tan 2 ( xy 2 + y ) sec 2 ( xy 2 + y ) µ 2 xy dy dx + y 2 + dy dx =1 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 ) 20. (1 + sec y )[3 xy 2 ( dy/dx )+ y 3 ] xy 3 (sec y tan y )( dy/dx ) (1 + sec y ) 2 =4 y 3 dy dx , multiply through by (1 + sec y ) 2 and solve for dy dx to get dy dx = y (1 + sec y ) 4 y (1 + sec y ) 2 3 x (1 + sec y xy sec y tan 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 22. dy dx = x 2 y 2 , d 2 y dx 2 = y 2 (2 x ) x 2 (2 ydy/dx ) y 4 = 2 xy 2 2 x 2 y ( x 2 /y 2 ) y 4 = 2 x ( y 3 + x 3 ) y 5 , but x 3 + y 3 =1so d 2 y dx 2 = 2 x y 5 23.
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## This note was uploaded on 06/10/2010 for the course MATH 200-177 taught by Professor Richardwhite during the Spring '10 term at Drexel.

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chapter_04 - January 27, 2005 11:44 L24-ch04 Sheet number 1...

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