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x+y+cosy=0确定的隐函数y=y(x)的二阶导
人气:205 ℃ 时间:2020-10-01 21:49:30
解答
x+y+cosy=0
x'+y'+(cosy)'=0'
1+y'-siny*y'=0
(siny-1)y'=1
y'=1/(siny-1)
y"=-(siny-1)'/(siny-1)^2
=-cosy*y'/(siny-1)^2
=-cosy*(1/(siny-1))/(siny-1)^2
=-cosy/(siny-1)^3
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