If `y=log(x+1/x),` prove that `(dy)/(dx)=(x-1)/(2x(x+1))`

Explanation

`y=log (sqrt{x}+frac{1}{sqrt{x}})`

`frac{d y}{d x}=frac{d}{d x} log (sqrt{x}+frac{1}{sqrt{x}}) `

`=frac{1}{sqrt{x}+frac{1}{sqrt{x}}} frac{d}{d x}(sqrt{x}+frac{1}{sqrt{x}}) `

`=frac{sqrt{x}}{x+1}(frac{1}{2 sqrt{x}}-frac{1}{2 x sqrt{x}}) `

`=frac{1}{2} frac{sqrt{x}}{x+1}(frac{x-1}{x sqrt{x}}) `

`=frac{x-1}{2 x(x+1)} `

So, ` frac{d y}{d x}=frac{x-1}{2 x(x+1)}`