Question

x+y+cosy=0确定的隐函数y=y(x)的二阶导

Answer 1

x+y+cosy=0x'+y'+(cosy)'=0'1+y'-siny*y'=0(siny-1)y'=1y'=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