 Hello friends welcome to the session I am Malka, let us discuss the question that is differentiate with respect to x, the function in exercise is 1 to 11. Our given function is log x to the power log x where x is always more than 1. So let us start with the solution let y equal to log x to the power log x. Now taking log of both sides we get log y equal to log x log of log x will differentiate both the sides with respect to x. We get 1 upon y into dy by dx equal to 1 upon x into log of log x plus log x into 1 upon log x into 1 upon x this implies dy by dx equal to y into log of log x upon x plus log x upon x log x where log x will cancel out with log x we get this implies dy by dx equal to y into log of log x upon x plus 1 upon x this implies dy by dx equal to now we will substitute the value of y which is log x to the power log x into I am taking x and we get log of log x plus 1 upon x this implies dy by dx equal to log x to the power log x into log of log x plus 1 upon x which is the required solution where x is always greater than 1. So hope you understood it and enjoyed the session goodbye and take care.