 Hi and welcome to the session, I am Deepika here. Let's discuss the question different shades of following function with respect to x x plus 1 by x raised to power x plus x raised to power 1 plus 1 over x So let's start the solution solution Let y is our given function y is equal to x plus 1 by x raised to power x plus x raised to power 1 plus 1 over x now put u is equal to x plus 1 by x raised to power x and v is equal to x raised to power 1 plus 1 over x taking logarithm on both the sides of u and v we get log u is equal to x plus 1 by x and log v is equal to 1 plus 1 over x log x Now differentiating both the sides with respect to x we get have Let us first take log u is equal to x log x plus 1 by x Now here we will apply the product rule. So we have 1 by u into du by dx is equal to x into Variative of log x plus 1 by x So this is equal to 1 over derivative of log x is 1 over x so 1 over x plus 1 by x Into dy dx of x plus 1 by x plus log x plus 1 by x Into derivative of x that is 1 So this is equal to 1 by u du by dx is equal to x over x plus 1 by x Now d by d by dx of x plus 1 by x is 1 Minus derivative of 1 by x is 1 minus 1 over x square plus log x plus 1 by x So this is equal to 1 by u du by dx is equal to now here Take the answer make square plus 1 and this is x into x x square Here also take the answer make square minus 1 over x square plus log x plus 1 by x So this is equal to x square minus 1 over x square plus 1 Plus log x plus 1 by x This is equal to 1 by u into du by dx So this implies du by dx is equal to U into x square minus 1 over x square plus 1 plus log x plus 1 by x on Substituting value of u here value of u we get The u by dx is equal to u is our x plus 1 by x raise to power x Into x square minus 1 over x square plus 1 plus log x plus 1 by x So let us take this as number one now consider log v So let us consider log v is equal to 1 plus 1 over x into log x Again differentiating both the sides Here we will also use the product rule with respect to x We have 1 by v into dv by dx is equal to 1 plus 1 by x into dy dx of log x plus log x into dy dx of 1 plus 1 by x This is equal to 1 by v into dv by dx is equal to 1 plus 1 by x Into 1 over x plus log x into d by dx of 1 plus 1 by x is Derivative of 1 is 0 and 1 by x is minus 1 over x square. So we get This implies dv by dx is equal to v into this is x plus 1 upon x square minus log x upon x square Let us take this as number two now given y is equal to x plus 1 by x raise to power x plus x raise to power 1 plus 1 by x and we have put y is equal to u plus v So this implies dy by dx is equal to du by dx plus dv by dx substituting values of du by dx dv by dx From one and two we get we get d y by dx is equal to so d y by dx is equal to The u by dx that is x plus 1 by x raise to power x Into x square minus 1 over x square plus 1 plus log x plus 1 by x So this was a value of the u by dx plus From 2 we have dv by dx is equal to x raise to power 1 plus 1 by x into x plus 1 by x square minus log x over x square hence we have differentiate the given function and our answer is x plus 1 by x raise to power x into x square minus 1 over x square plus 1 plus log x plus 1 by x raise to power 1 plus 1 by x into x plus 1 minus log x upon x square So this is our answer. I hope the question is clear to you. Bye and have a nice day