MATH
prodquotpracsol

# prodquotpracsol - Solutions to derivatives f(x = x2 10x 100...

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Solutions to derivatives: f ( x ) = x 2 - 10 x + 100 g ( x ) = x 100 + 50 x + 1 V ( r ) = 4 3 πr 3 f ( x ) = 2 x - 10 g ( x ) = 100 x 99 + 50 V ( r ) = 4 πr 2 s ( t ) = t 8 + 6 t 7 - 18 t 2 + 2 t F ( x ) = (16 x ) 3 G ( y ) = ( y 2 + 1)(2 y - 7) s ( t ) = 8 t 7 + 42 t 6 - 36 t + 2 F ( x ) = 12288 x 2 G ( y ) = 2 y (2 y - 7) + ( y 2 + 1)2 Y ( t ) = bt - 9 R ( x ) = 10 x 7 g ( x ) = x 2 + 1 x Y ( t ) = - 9 bt - 10 R ( x ) = - 7 10 x - 8 g ( x ) = 2 x - 1 x 2 f ( t ) = t - 1 t h ( x ) = x + 2 x - 1 f ( u ) = 1 - u 2 1 + u 2 f ( t ) = 1 2 t - 1 / 2 + 1 2 t - 3 / 2 h ( x ) = ( x - 1) - ( x +2) ( x - 1) 2 f ( u ) = (1+ u 2 )( - 2 u ) - (1 - u 2 )(2 u ) (1+ u 2 ) 2 G ( s ) = ( s 2 + s + 1)( s 2 + 2) H ( t ) = 3 t ( t + 2) y = x 2 + 4 x + 3 x G ( s ) = 4 s 3 + 3 s 2 + 6 s + 2 H ( t ) = 1 3 t - 2 / 3 ( t + 2) + t 1 / 3 y = x (2 x +4) - ( x 2 +4 x +3)(1 / 2) x - 1 / 2 x y = x - 1 x + 1 y = 5 x y = x 4 / 3 - x 2 / 3 y = ( x +1)( 1 2 x - 1 / 2 ) - ( x - 1)( 1 2 x - 1 / 2 ) ( x +1) 2 y = 5 1 2 x - 1 / 2 y = 4 3 x 1 / 3 - 2 3 x - 1 / 3 y = 1 x 4 + x 2 + 1 y = x 2 + x + x - 1 + x - 2 y = ax 2 + bx + c y = - (4 x 3 +2 x ) ( x 4 + x 2 +1) 2 y = 2 x + 1 - 1 x 2 - 2 x 3 y = 2 ax + b y = A + B x + C x 2 y = 3 t - 7 t 2 + 5 t - 4 y = 4 t + 5 2 - 3 t y = - B x 2 - 2 C x 3 y = ( t 2 +5 t - 4)(3) - (3 t - 7)(2 t +5) ( t 2 +5 t - 4) 2 y = 4(2 - 3 t ) - (4 t +5)( - 3) (2 - 3 t ) 2 y = x + 5 x 2 y = x 4 - 4
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• Spring '08
• Johnson

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