 Hello friends welcome to the session I am Alka, let us discuss the question which is differentiate with respect to x, the function in exercise is 1 to 11, given function is x to the power x plus x to the power a plus a to the power x plus a to the power a for some fixed a greater than 0 and x greater than 0, so let's start with the solution let equal to x to the power x plus x to the power a plus a to the power x plus a to the power a, now again let us assume that let u equal to x to the power x and v equal to x to the power x, now we will take log of both the sides of u and v both, we get log u equal to x log x and here log v equal to x log a now we will differentiate both sides with respect to x we get upon u into du by dx equal to x into 1 upon x plus log x into 1, here we get 1 upon v into dv by dx equal to log a into 1, this implies du by dx equal to u xx cancel out into 1 plus log x and here we get dv by dx equal to v into log x and here we get dv by dx equal to v into log a, this implies du by dx equal to v substituted value of u which is x to the power x into 1 plus log x and dv by dx equal to we substitute the value of v which is a to the power x into log a, now let this be our first equation, now we are given that y equal to x to the power x plus x to the power a plus a to the power x plus a to the power a, differentiate sides with respect to x we get dy by dx equal to, now since we know that x to the power x is du so it can be written as dy dx of u plus dy dx of x to the power a is ax a minus 1 and we know that a to the power x is v so it will be written as dy dx of v plus 0 since v by dx of any constant is 0, this implies dy by dx equal to now we will substitute the value of du by dx which is x to the power x into 1 plus log x plus ax a minus 1 plus we will substitute the value of v by dx which is ax log therefore dy by dx equal to x to the power x 1 plus log x x a minus 1 plus a to the power x log a hope you understood the solution and enjoyed the session goodbye and take care