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標題:

Differentiation

發問:

23)' If xsin(2x + y) = y^2 Find dy/dx.

最佳解答:

xsin(2x+y)=y^2 x{d[sin(2x+y)]/dx]+sin(2x+y)[(dx/dx)]=2y(dy/dx) (x)[cos(2x+y)]d(2x+y)/dx+sin(2x+y)=2y(dy/dx) (x)[cos(2x+y)(2+dy/dx)+sin(2x+y)=2y(dy/dx) 2xcos(2x+y)+xcos(2x+y)(dy/dx)+sin(2x+y)=2y(dy/dx) [xcos(2x+y)-2y](dy/dx)=-sin(2x+y)-2xcos(2x+y) dy/dx=[-sin(2x+y)-2xcos(2x+y)]/[xcos(2x+y)-2y] 2010-06-23 14:44:53 補充： 補充得好好 不過兩方答案都得.....

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• 我想做兼職
• L 之終章.最後的23天 的圖片邊度有得搵-[10點]@1@
• 合‘23’問題

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其他解答:

分別係 個負響上面定下面.....|||||補充： Simplified dy/dx=[-sin(2x+y)-2xcos(2x+y)]/[xcos(2x+y)-2y] dy/dx=-[sin(2x+y)+2xcos(2x+y)]/[xcos(2x+y)-2y] dy/dx=[sin(2x+y)+2xcos(2x+y)]/-[xcos(2x+y)-2y] dy/dx=[sin(2x+y)+2xcos(2x+y)]/[-xcos(2x+y)+2y] dy/dx=[sin(2x+y)+2xcos(2x+y)]/[2y-xcos(2x+y)]