x के सापेक्ष `tan^(-1)[(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))]` (w.r.t.x) को अवकलित करें ।

Explanation

माना कि `y=tan^(-1)[(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))]`
`=tan^(-1)[(sqrt(1+costheta)-sqrt(1-costheta))/(sqrt(1+costheta)+sqrt(1-costheta))] " " (x=costheta` रखने पर)
`=tan^(-1)[(sqrt(2"cos"^(2)(theta)/(2))-sqrt(2"sin"^(2)(theta)/(2)))/(sqrt(2"cos"^(2)(theta)/(2))+sqrt(2"sin"^(2)(theta)/(2)))]=tan^(-1)[("cos"(theta)/(2)-"sin"(theta)/(2))/("cos"(theta)/(2)+"sin"(theta)/(2))]`
`=tan^(-1)[(1-"tan"(theta)/(2))/(1+"tan"(theta)/(2))]=tan^(-1)[tan((pi)/(4)-(theta)/(2))]=(pi)/(4)-(theta)/(2)`
`thereforey=(pi)/(4)-(1)/(2) "cos"^(-1)x therefore(dy)/(dx)=-(1)/(2)(-(1)/(sqrt(1-x^(2))))=(1)/(2sqrt(1-x^(2)))`