 Hi, and welcome to the session. I am Deepika here. Let's discuss a question which says differentiate the following function with respect to x. x raised to power x cos x plus x squared plus 1 upon x squared minus 1. Let's start the solution. Let y is our given function. y is equal to x raised to power x cos x plus x squared plus 1 upon x squared minus 1. Again put u is equal to x raised to power x cos x and v is equal to x squared plus 1 upon x squared minus 1. Then y is equal to u plus v. Now on differentiating both sides with respect to x we have dy by dx is equal to du by dx plus dv by dx. Let us give this equation as number 1. Let us first find out du by dx. So now u is equal to x raised to power x cos x. By taking logarithm on both the sides we have log u is equal to x cos x log x. Now on differentiating both the sides with respect to x we have 1 over u into du by dx is equal to here we will apply the product root. Now x cos x into derivative of log x which is 1 by x plus x log x into derivative of cos x which is minus sin x plus cos x log x into derivative of x which is 1. Therefore du by dx is equal to u into cos x minus x log x plus cos x log x. Now on substituting the value of u here we get du by dx is equal to x raised to power x cos x into cos x minus x sin x log x plus cos x log x. Now let us find out dv by dx. Now v is equal to x square plus 1 upon x square minus 1. Now we know that derivative of u upon v is equal to derivative of u into v minus u into derivative of v upon v square where v is not equal to 0 and this is known as r. Therefore dv by dx is equal to now here we will apply the quotient rule which says derivative of u upon v is equal to derivative of u into v minus u into derivative of v upon v square. And this is known as our quotient rule. So here we have u is our x square plus 1 and v is our x square minus 1. So we have d by dx of x square plus 1 into x square minus 1 minus x square plus 1 into d by dx of x square minus 1 upon x square minus 1 whole square. Therefore dv by dx is equal to 2x into x square minus 1 minus x square plus 1 into 2x upon x square minus 1 whole square and this is equal to 2x. Let us take it common x square minus 1 minus x square minus 1 upon x square minus 1 whole square and this is equal to minus 4x upon x square minus 1 whole square. Now on substituting the value of dv by dx and du by dx which we have solved above in equation 1 which was dy by dx is equal to du by dx plus dv by dx. So we have dy by dx is equal to x raise to power x cos x into cos x minus x sin x log x plus cos x log x minus 4x upon x square minus 1 whole square. Hence the derivative of the above function is x raise to power x cos x into cos x into 1 plus log x minus x sin x log x minus 4x upon x square minus 1 whole square. And this is our answer for the above question. I hope the solution is clear to you. Bye and take care.