 Hello friends, welcome to the session, I am Malka, I am going to help you solve the question that is differentiate with respect to x, the function in exercise is 1 to 11 that is sin inverse x root x, where 0 is less than equal to x is less than equal to 1. So, let us start with the solution hit y equal to sin inverse x root x. Now, we will differentiate both sides with respect to x, we get dy by dx equal to 1 upon square root of 1 minus x square which is x into square root of x square into dy dx of x into square root of x. We get dy by dx equal to 1 upon square root of 1 minus x cube into, now here we will apply the product rule which will give us x into dy dx of x square plus square root of x plus square root of x into dy dx of x, this is equal to 1 upon square root of 1 minus x cube into x into 1 upon 2 root x plus root x into 1, this is equal to 1 upon square root of 1 minus x cube into, now we can write x equal to square root of x into square root of x upon 2 square root of x plus square root of x, square root of x and square root of x cancel out, so this is equal to 1 upon square root of 1 minus x cube into, now we will take 2 as LCM we get square root of x plus 2 square root of x, this is equal to 1 upon square root of 1 minus x cube into 3 square root of x upon 2, this implies dy by dx equal to 1 upon square root of 1 minus x cube into 3 square root of x upon 2, this implies dy by dx equal to 3 by x upon 2 into x upon 1 minus x cube where 0 is less than equal to x is less than equal to 1, so hope you understood the solution and enjoyed the session goodbye and take care.