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