求由方程sin(x+y)=cosxlny所解定的隐函数的导数dy/dx？
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sin(x + y) = cosx lny
dsin(x + y)/d(x + y) · d(x + y)/dx = lny d(cosx)/dx + cosx d(lny)/dy · dy/dx
cos(x + y) · (1 + dy/dx) = lny · (- sinx) + cosx · 1/y · dy/dx
cos(x + y) + cos(x + y) dy/dx = - sinx lny + 1/y cosx dy/dx
[cos(x + y) - 1/y cosx] dy/dx = - sinx lny - cos(x + y)
dy/dx = [sinx lny + cos(x + y)]/[(1/y)cosx - cos(x + y)],1,
dsin(x + y)/d(x + y) · d(x + y)/dx = lny d(cosx)/dx + cosx d(lny)/dy · dy/dx
cos(x + y) · (1 + dy/dx) = lny · (- sinx) + cosx · 1/y · dy/dx
cos(x + y) + cos(x + y) dy/dx = - sinx lny + 1/y cosx dy/dx
[cos(x + y) - 1/y cosx] dy/dx = - sinx lny - cos(x + y)
dy/dx = [sinx lny + cos(x + y)]/[(1/y)cosx - cos(x + y)],1,
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