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Y=(1/x)^x i.e; y=x^-x taking log on bothsides, logy=log(x)^-x differentiating w.r.t. x we get, (1/y)(dy/dx)=-x.(1/x)+logx.(-1) (dy/dx)=y(-1-logx) taking second derivative (d^2y/dx^2)=(-dy/dx)-y(1/x)-(dy/dx)logx.