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求隐函数导数ln(xy)=e^(x+y)所确定的隐函数y=y(x)的导数dy/dx.
刚才问题写错,是这个
人气:208 ℃ 时间:2019-08-15 08:27:02
解答
[ln(xy)]' = [e^(x+y)]'
(xy)'/(xy) = e^(x+y) * (x+y)'
(y + xy')/(xy) = e^(x+y) *(1 + y')
y' = y[e^(x+y) -1]/[x(1 - ye^(x+y)]
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