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## Answers

take log on both sides

log y=-xlogx

diff w.r.t x

(1/y)(dy/dx)=-(log x+1)

dy/dx=y(log x +1)=(1/x)^x(log x +1)

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.