 Hello friends, welcome to the session. I am Mulkand and this is the question which is find dy upon dx if y equal to sin inverse x plus sin inverse square root of 1 minus x square where minus 1 is less than equal to x is less than equal to 1. Let us start with the solution. We are given y equal to sin inverse x plus sin inverse square root of 1 minus x square. Now differentiate both sides with respect to x we get dy upon dx equal to d upon dx of sin inverse x plus d upon dx of sin inverse square root of 1 minus x square. This implies dy upon dx equal to 1 upon square root of 1 minus x square plus d upon dx of sin inverse square root of 1 minus x square is 1 upon square root of 1 minus square root of 1 minus x square whole square into d upon dx of square root of 1 minus x square. This implies dy upon dx equal to 1 upon square root of 1 minus x square plus 1 upon square root of 1 minus 1 minus plus x square into 1 upon 2 into square root of 1 minus x square into minus 2 x. This implies dy upon dx equal to 1 upon square root of 1 minus x square plus minus 2 x upon x into square root of 1 minus x square. This implies dy upon dx equal to 1 upon square root of 1 minus x square. Here 2 to cancel out, xx will also cancel out we get minus 1 upon square root of 1 minus x square. This implies dy upon dx equal to 0. Hope you understood the solution and enjoyed the session. Goodbye and take care.