设y=y(x)由方程cos(x+y)+y=1,求dy/dx我做到-sin(x+y)*(x+y)'+y'=0
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设y=y(x)由方程cos(x+y)+y=1,求dy/dx我做到-sin(x+y)*(x+y)'+y'=0
设y=y(x)由方程cos(x+y)+y=1,求dy/dx
我做到
-sin(x+y)*(x+y)'+y'=0
设y=y(x)由方程cos(x+y)+y=1,求dy/dx我做到-sin(x+y)*(x+y)'+y'=0
cos(x+y)+y=1 两边对 x 求导:
=> - sin(x+y) * (1 + y ') + y ' = 0
=> y ' = sin(x+y) / [1 - sin(x+y)]
dy/dx = sin(x+y) / [1 - sin(x+y)]
设y=y(x)由方程cos(x+y)+y=1,求dy/dx我做到-sin(x+y)*(x+y)'+y'=0
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