 Hello friends, welcome to the session I am Alka. Let us solve the question which is differentiated with respect to x the function in exercise is 1 to 11 that is cot inverse square root of 1 plus sin x plus square root of 1 minus sin x upon square root of 1 plus sin x minus square root of 1 minus sin x where 0 is less than x is less than 1 by 2. So, let us start with the solution let y equal to cot inverse square root of 1 plus sin x plus square root of 1 minus sin x upon square root of 1 plus sin x minus square root of 1 minus sin x. Now, we see that y equal to cot inverse into 9 phase of 1 we will write sin square x by 2 plus cos square x by 2 and for sin x we will write 2 sin x by 2 cos x by 2 plus square root of sin square x by 2 plus cos square x by 2 minus 2 sin x by 2 cos x by 2 and similarly we will write in the denominator that is for 1 we will write sin square x by 2 plus cos square x by 2 plus 2 sin x by 2 cos x by 2 minus sin square x by 2 plus cos square x by 2 minus 2 sin x by 2 cos x by 2 by this implies y equal to cot inverse square root of sin x by 2 plus cos x by 2 whole square plus square root of cos x by 2 minus sin x by 2 whole square upon square root of sin x by 2 plus cos x by 2 whole square minus square root of cos x by 2 minus sin x by 2 whole square we get y equal to cot inverse sin x by 2 plus cos x by 2 plus cos x by 2 minus sin x by 2 upon sin x by 2 plus cos x by 2 minus cos x by 2 plus sin x by 2 this implies y equal to cot inverse sin x by 2 cancel out with sin x by 2 cos x by 2 cancel out with minus cos x by 2 we get cot inverse into 2 cos x by 2 upon 2 sin x by 2 to do cancel out this implies y equal to cot inverse cot x by 2 this implies y equal to cot cot cancel out y equal to x by 2 now differentiate both sides with respect to x we get dy by dx equal to 1 by 2 therefore dy by dx equal to 1 by 2 is the required solution hope you understood it and enjoyed the session goodbye and take care