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Page 1 of 11 Answer Key to Math Session for CHEM 3615 Fall 2008 Exercise for Section 2: Use the product and chain rule to derive the formula for dx ) v / u ( d dx v udv vdu dx dv v 1 u dx du v 1 dx ) v / 1 ( d u dx du v 1 dx ) v / u ( d 2 2 = + = + = , hence ( )( ) 2 v dx / dv u dx / du v dx ) v / u ( d = (1) y = 4x 2 – 8x + 1, then 8 x 8 0 ) 1 ( 8 ) x 2 ( 4 ) 1 ( dx d ) x ( dx d 8 ) x ( dx d 4 ) 1 x 8 x 4 ( dx d dx dy 2 2 = + = + = + = (2) y = (3x - 1) (2x + 5) OR () ( ) [] 5 x 2 1 x 3 dx d dx dy + = = 6x 2 + 15x -2x – 5 = 6x 2 + 13x – 5 + + + = ) 1 x 3 ( dx d ) 5 x 2 ( ) 5 x 2 ( dx d ) 1 x 3 ( 13 x 12 0 ) 1 ( 13 ) x 2 ( 6 ) 5 x 13 x 6 ( dx d dx dy 2 + = + = + = ) 0 3 )( 5 x 2 ( ) 0 2 )( 1 x 3 ( + + + = 13 x 12 + = (3) y = t 2 and x 1 x t = , hence 2 2 2 2 ) x 1 ( x x 1 x t y = = = [ ] ) 1 ( x 1 2 x ) x 2 ( ) x 1 ( ) x 1 ( dx d x ) x ( dx d ) x 1 ( ) x 1 ( x dx d dx dy 3 2 2 2 2 2 2 2 2 + = + = = 3 3 2 2 3 2 3 2 2 ) x 1 ( x 2 ) x 1 ( x 2 x 2 x 2 ) x 1 ( x 2 ) x 1 ( x 2 ) x 1 ( 2 x ) x 1 ( x 2 = + = + = + = OR + = = = 1 2 x 1 dx d x ) x ( dx d x 1 1 ) t 2 ( x 1 x dx d ) t ( dt d dx dt dt dy dx dy 3 2 2 2 ) x 1 ( x 2 ) x 1 ( x x 1 x 1 x 2 ) x 1 ( x x 1 1 ) t 2 ( ) x 1 ( dx d ) x 1 ( 1 x x 1 1 ) t 2 ( = + = + = + = (4) 2 / 1 2 2 / 1 2 ) x 1 ( ) x 1 ( x 1 x 1 y + = + = 2 / 1 2 2 / 1 2 / 1 2 / 1 2 2 / 1 2 2 / 1 ) x 1 ( dx d ) x 1 ( ) x 1 ( dx d ) x 1 ( ) x 1 ( ) x 1 ( dx d dx dy + + + = + =

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Page 2 of 11 () ) x 1 ( dx d ) x 1 ( 2 / 1 ) x 1 ( ) x 1 ( dx d ) x 1 ( 2 / 1 ) x 1 ( 2 2 / 3 2 2 / 1 2 / 1 2 / 1 2 + + + + = ( ) ) x 2 ( ) x 1 ( 2 / 1 ) x 1 ( ) 1 ( ) x 1 ( 2 / 1 ) x 1 ( 2 / 3 2 2 / 1 2 / 1 2 / 1 2 + + + = 2 / 1 2 / 3 2 2 / 1 2 / 1 2 2 / 3 2 2 / 1 2 / 1 2 / 1 2 ) x 1 ( ) x 1 ( 2 ) x 1 ( ) x 1 ( x 2 ) x 1 ( ) x 1 ( ) x 1 ( x ) x 1 ( ) x 1 ( 2 1 + + = + + = 2 / 1 2 / 3 2 2 2 / 1 2 / 3 2 2 2 2 / 1 2 / 3 2 2 ) x 1 ( ) x 1 ( 2 1 x 2 x ) x 1 ( ) x 1 ( 2 x 2 x 2 x 1 ) x 1 ( ) x 1 ( 2 ) x 1 ( x 2 x 1 + = + + = + = Another method: define u = (1-x) 1/2 , and v = (1+x 2 ) 1/2 v u y = ; 2 / 1 2 / 1 ) x 1 ( 2 1 ) 1 ( ) x 1 ( 2 1 dx du = = ; 2 / 1 2 2 / 1 2 ) x 1 ( x ) x 2 ( ) x 1 ( 2 1 dx dv + = + = [] ) x 1 ( ) x 1 )( x ( ) x 1 ( ) x 1 )( 2 / 1 ( ) x 1 ( v dx dv v dx du v dx v u d dx dy 2 2 / 1 2 2 / 1 2 / 1 2 / 1 2 2 + + + = = = 2 / 1 2 / 3 2 2 2 / 1 2 / 3 2 2 ) x 1 ( ) x 1 ( 2 1 x 2 x ) x 1 ( ) x 1 ( 2 x ) x 1 ( 2 ) 1 )( x 1 ( + = + + = (5) 2 x e y = , then 2 2 2 x 2 x x xe 2 x dx d e e dx d dx dy = = = (6) + = x 1 x 1 ln y ) x 1 )( x 1 ( 2 ) x 1 ( x 1 x 1 ) x 1 ( ) x 1 ( ) x 1 ( 1 ) x 1 ( ) 1 )( 1 )( x 1 ( ) x 1 ( ) x 1 ( ) 1 ( ) x 1 ( 1 ) x 1 ( dx d ) x 1 ( 1 ) x 1 ( ) x 1 ( ) x 1 ( ) x 1 ( dx d ) x 1 ( 1 x 1 1 dx d ) x 1 ( ) x 1 ( ) x 1 ( x 1 x 1 dx d x 1 x 1 1 x 1 x 1 ln dx d dx dy 2 2 2 + = + + + = + + + = + + + = + + + + = + + = + =
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