e^x+y+cosxy=2,求y的导数|x=0
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直接两边积分，过程如图
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e^x+y+cosxy=2
x=0
e^0 +y(0) + 1 = 2
y(0) =0
(x,y)=(0,0)
//
e^x+y+cosxy=2
d/dx (e^x+y+cosxy )=0
e^x + y' -sin(xy) .(xy)' =0
e^x + y' -sin(xy) .(xy' + y) =0
[ xsin(xy) -1] y' = e^x -ysin(xy)
y' = [e^x -ysin(xy)]/[ xsin(xy) -1]
y'|(x,y)=(0,0)
=[1 -0]/[ 0 -1]
=-1
x=0
e^0 +y(0) + 1 = 2
y(0) =0
(x,y)=(0,0)
//
e^x+y+cosxy=2
d/dx (e^x+y+cosxy )=0
e^x + y' -sin(xy) .(xy)' =0
e^x + y' -sin(xy) .(xy' + y) =0
[ xsin(xy) -1] y' = e^x -ysin(xy)
y' = [e^x -ysin(xy)]/[ xsin(xy) -1]
y'|(x,y)=(0,0)
=[1 -0]/[ 0 -1]
=-1
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