 Hello friends, welcome to the session I am Malka. Let's discuss the question that is differentiate with respect to x the function in exercises 1 to 11. Our given function is x to the power x square minus 3 plus x minus 3 to the power x square for x greater than 3. So let's start with the solution. Let y equal to x to the power x square minus 3 plus x minus 3 to the power x square. Now again let us assume that that u equal to x to the power x square minus 3 and v equal to x minus 3 to the power x square. Now taking log of both sides of u and v, we get log u equal to x square minus 3 log x log v equal to x square log x minus 3. Now differentiate both sides with respect to x we get 1 upon u du over dx equal to x square minus 3 into 1 upon x plus log x into 2x. This implies du upon dx equal to u into x square minus 3 upon x plus 2x into log x. Now this implies du upon dx equal to, we will substitute the value of u which is x to the power x square minus 3 into x square minus 3 upon x plus 2x log x. This is our first equation. Similarly differentiate that is log v equal to x square on differentiating this we get 1 upon v into du upon dx equal to x square into 1 upon x minus 3 into upon dx equal to v into x square upon x minus 3 plus x minus 3. Now we substitute the value of v. Therefore we get dv upon dx equal to x minus 3 to the power x square into x square upon x minus 3 plus 2x log x minus 3. So this is our second equation. Now we are given y equal to x to the power x square minus 3 plus x minus 3 to the power x square. This implies y equal to u plus v. Now this implies dy by dx equal to du upon dx plus dv upon dx on differentiating both the sides with respect to x. This implies on substituting the value of du upon dx and dv upon dx we get dy upon dx equal to x to the power x square minus 3 into x square minus 3 upon x plus 2x log x plus x minus 3 to the power x square into x square upon x minus 3 plus 2x log x minus 3. dy upon dx equal to x to the power x square minus 3 into x square minus 3 upon x plus 2x log x plus x minus 3 to the power x square into x square upon x minus 3 plus 2x log x minus 3 which is a required solution. Hope you understood it and enjoyed the session. Goodbye and take care.