यदि ` y= sin ^(-1) { (sqrt (1+x)-sqrt (1-x))/(2)}` तब ` (dy)/(dx)` का मान ज्ञात कीजिए |

Explanation

`y= sin ^(-1) ""{( sqrt(1+x)-sqrt(1-x))/(2)}`
माना ` x= cos 2theta ,` तब
` " "y=sin ^(-1) { (sqrt (1+cos 2theta) -sqrt( 1-cos 2theta))/(2)}`
` sin ^(-1) {(sqrt( 2cos ^(2)theta )-sqrt (1-cos 2theta ))/(2)}`
` =sin ^(-1) {(sqrt ( 2)cos theta - sqrt (2) sin theta )/(2)}`
` = sin ^(-1) { (1)/(sqrt(2) )cos theta -(1)/(sqrt(2))sin theta }`
` " "=sin ^(-1) {sin ""(pi)/(4) cos theta - cos ""(pi)/(4) sin theta }`
` =sin ^(-1) {sin ((pi)/(4) -theta )}`
`" "((pi)/(4)-theta ) =((pi)/(4) -(1)/(2) cos ^(-1) x ),` जहाँ `theta =(1)/(2) cos ^(-1)x `
` y= (pi)/(4) -(1)/(2) cos ^(-1) x `
` therefore " "(dy)/(dx) =(d)/(dx) ((pi)/(4) -(1)/(2)cos ^(-1)x )`
` =(d)/(dx) ((pi)/(4) )-(d)/(dx) ((1)/(2) cos ^(-1) x )`
` " "= (0-(1)/(2) *((-1))/(sqrt (1-x^(2))) ) =(1)/(2sqrt(1-x^(2)))`