xy+sin(x+y)=1,两边求导数
y+xy'+cos(x+y)*(1+y')=0
xy'+cos(x+y)y'=-[cos(x+y)+y]
∴y'[cos(x+y)+x]=-[cos(x+y)+y]
∴y'=-[y+cos(x+y)]/[x+cos(x+y)]不好意思呀,回复晚了。这一步有点不明白:cos(x+y)*(1+y')。谢谢了sin(x+y) 是复合函数,求导时,先求外导,再乘上内导∴[sin(x+y)]'=cos(x+y)*(x+y)'=cos(x+y)*(x'+y')=cos(x+y)*(1+y')
求下列所确定的隐函数方程y=y(x)的导数.
求下列所确定的隐函数方程y=y(x)的导数.
xy+sin(x+y)=1
xy+sin(x+y)=1
数学人气:869 ℃时间:2019-11-08 15:27:08
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