Find the co-ordinates of the point on curve `y=(x^2-1)/(x^2+1), (x>0)` where the gradient of the tangent to the curve is maximum.

Explanation

Here, equaton of the curve,

`y = (x^2-1)/(x^2+1)`

`:. dy/dx = ((x^2+1)(2x)-(x^2-1)(2x))/(x^2+1)^2`

`=>dy/dx = (4x)/(x^2+1)^2`

We have to find a point, where slope is maximum.

`:. (d^2y)/dx^2` should be `0`.

`=>(d^2y)/dx^2 = ((x^2+1)^2 4-4x(2(x^2+1))(2x) )/(x^2+1)^4 `

`=>((x^2+1)^2 4-4x(2(x^2+1))(2x) )/(x^2+1)^4 =0`

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