If `y=(x-1)log(x-1)-(x+1)log(x+1),"p r o v et h a t"(dy)/(dx)=log((x-1)/(1+x))`

Explanation

We have, `y=(x-1) log (x-1)-(x+1) log (x+1)`

`frac{d y}{d x}=frac{d}{d x}[(x-1) log (x-1)-(x+1) log (x+1)] `

`=[(x-1) frac{d}{d x} log (x-1)+log (x-1) frac{d}{d x}(x-1)]-[(x+1) frac{d}{d x} log (x+1)+log (x+1) frac{d}{d x}(x+1)] `

`=[(x-1) times frac{1}{(x-1)} frac{d}{d x}(x-1)+log (x-1) times(1)]-[(x+1) times frac{1}{(x+1)} times frac{d}{d x}(x+1)+log (x+1)(1)] `

`=[1+log (x-1)]-[1+log (x+1)] `

`=log (x-1)-log (x+1) `

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

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