Question

What is the second derivative of y= (1/x)^x    [ ie, y = (1/x) whole power x]? 

Answer 1

take y=(1/x)^x
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)

Answer 2

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.