 Hi and welcome to the session. I am Deepika here. Let's discuss the question, differentiate the following function with respect to x. x cos x raised to power x plus x sin x raised to power 1 by x. So let's start the solution. That y is our given function, y is equal to x cos x raised to power x plus x sin x raised to power 1 by x. Again put u is equal to x cos x raised to power x and v is equal to x sin x raised to power 1 by x. So our y will be u plus v. Differentiate both sides with respect to x. We have d y by dx is equal to d u by dx plus d v by dx. So let us take this equation as number 1. Now we will find the u by dx from u is equal to x cos x raised to power x and d v by dx from v is equal to x sin x raised to power 1 by x. Then we will substitute the values of d u by dx and d v by dx in this equation to get the derivative of our given function. So now consider u is equal to now u is equal to x cos x raised to power x. Again taking logarithm on both sides we get log u is equal to log of x cos x. Now method is same. We will find, we will differentiate with respect to x and on the right hand side we will apply the product rule. We have 1 by u into d u by dx is equal to now apply the product rule here x into derivative of log of x cos x. We know derivative of log x is 1 over x so derivative of log of x cos x is 1 over x cos x into d by dx of x cos x. For this term also we will apply the product rule which is now derivative of x cos x is x into derivative of cos x that is minus sin x plus cos x into derivative of x. Now so again x into derivative of log of x cos x plus log of x cos x into derivative of x that is 1. I hope this step is clear to you. Now 1 by u d u by dx is equal to so this is minus x tan x plus 1 plus log of x cos x. So this implies d u by dx is equal to u into 1 minus x tan x. plus log of x cos x. So substitute the value of u here we get d u by dx is equal to x cos x raise to power x into 1 minus x tan x plus log of x cos x. Now we got the derivative of u now let us consider v so now v is equal to raise to power 1 by x again taking logarithm on both sides. We get log v is equal to 1 by x into log of x sin x. Now again process is same differentiate above both sides with respect to x. Now we have 1 by v into dv by dx is equal to now here apply the product rule here so 1 by x into derivative of log of x sin x. So this is equal to 1 over x sin x into derivative of x sin x. So for derivative of x sin x we will again apply the product rule which is equal to x into derivative of sin x that is cos x plus sin x into derivative of x this is 1. So again now 1 by x into derivative of log of x sin x plus log of x sin x into derivative of 1 by x which is minus 1 over x square. So this is equal to 1 by v into dv by dx is equal to here we get 1 by x cot x is equal to 1 so this is 1 over x square plus minus 1 by x square log of x sin x. So this implies dv by dx is equal to into 1 by x cot x plus 1 by x square minus 1 by x square log of x sin x substitute the value of v here we get dv by dx is equal to now v is our x sin x is to power 1 by x. Now inside the bracket take x square as I will say we get x cot x plus 1 minus log of x sin x so this is the value of dv by dx now substitute the value of the values of u by dx and dv by dx. In equation 1 which was dv by dx is equal to dv by dx plus dv by dx so we get dv by dx is equal to now dv by dx is x cot x is to power x into 1 minus x tan x plus log of dv by dx which is plus is to power 1 by x with our differentiation of our given function so our answer is cos x is to power x into 1 minus x plus log of x cos x is to power 1 by x x cot x plus 1 minus log of for the given question I hope the question is clear to you bye and have a nice day.