find dy/dx by implicit differentiation:
1 + x = sin ( xy^2 )
Urgent!! Math Question differentiation (15)
2008-10-09 8:56 pm
回答 (2)
2008-10-13 9:09 pm
✔ 最佳答案
總結一下,就是這樣:1 = cos (xy^2)[ d(xy^2)/dx]
1 = cos(xy^2)[ 2xyy' + y^2]
1 = 2xycos(xy^2)y' + y^2cos(xy^2)
y' = [1- y^2cos(xy^2)]/[ 2xycos(xy^2)].
1 + x = sin ( xy^2 )
0+1 = cos(xy^2)(x2y(dy/dx)+y^2(1))
x2y(dy/dx)+y^2 = 1/cos(xy^2)
x2y(dy/dx) = 1/cos(xy^2)-y^2
x2y(dy/dx) = (1 - y^2 cos(xy^2))/cos(xy^2)
ok?
希望幫到你
2008-10-09 9:41 pm
1 = cos (xy^2)[ d(xy^2)/dx]
1 = cos(xy^2)[ 2xyy' + y^2]
1 = 2xycos(xy^2)y' + y^2cos(xy^2)
y' = [1- y^2cos(xy^2)]/[ 2xycos(xy^2)].
1 = cos(xy^2)[ 2xyy' + y^2]
1 = 2xycos(xy^2)y' + y^2cos(xy^2)
y' = [1- y^2cos(xy^2)]/[ 2xycos(xy^2)].
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