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求由方程所确定的隐函数的导数dy/dx? y=cos(x+y)
$(acontent)
人气:103 ℃ 时间:2019-08-19 07:34:14
解答
y=coa(x+y)
dy/dx=-sin(x+y)·(1+dy/dx)
dy/dx=-sin(x+y)-sin(x+y)·dy/dx
[1+sin(x+y)]dy/dx=-sin(x+y)
dy/dx=-sin(x+y)/[1+sin(x+y)]
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