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y=sin(x+y)的隐函数的二阶导数.
人气:197 ℃ 时间:2020-07-05 18:28:01
解答
y'=cos(x+y)*(1+y')y'=1/[1/cos(x+y)-1]y"= -sin(x+y)(1+y')/{cos(x+y)^2[1/cos(x+y)-1]^2}=-sin(x+y)[1+1/[1/cos(x+y)-1]/[1-cos(x+y)]^2=-sin(x+y)/[1-cos(x+y)]^3
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