Let `y=tan^(-1){(sqrt(1+x^(2))-1)/(x)}`
Putting `x=tan theta`, we get
`y=tan^(-1){(sqrt(1+tan^(2)theta-1))/(tan theta)}=tan^(-1){(sec theta-1)/(tan theta)}`
`=tan^(-1){(((1)/(cos theta)-1))/(sin theta).cos theta}=tan^(=1)((1-cos theta)/(sin theta))`
`=tan^(-1){(2sin^(2)(theta//2))/(2sin(theta//2)cos(theta//2))}=tan^(-1){"tan"(theta)/(2)}`
`=(theta)/(2)=(1)/(2)tan^(-1)x.`
`therefore y=(1)/(2)tan^(-1)x`
`rArr(dy)/(dx)=(1)/(2).(d)/(dx)(tan^(-1)x)=(1)/(2(1+x^(2))).`
