 Hello and welcome to the session. My name is Mansi and I'm going to help you with the following question. The question says find the derivative of x raise to power n minus a raise to power n divided by x minus a for some constant a. Let us start with the solution to this question. First of all let us see the quotient rule that is d by dx of fx x by dx is equal to dx into d by dx of fx minus fx into d by dx of dx divided by dx the whole square. So if we take fx to be equal to x raise to power n minus a raise to power n and dx to be equal to x minus a then we get d by dx of x raise to power n minus a raise to power n divided by x minus a is equal to gx that is x minus a into d by dx of x raise to power n minus a raise to power n minus fx that is x raise to power n minus a raise to power n into d by dx of gx that is x minus a and this whole divided by x minus a the whole square. This is equal to x minus a into, now d by dx of x raise to power n minus a raise to power n is equal to, d by dx of x raise to power n minus d by dx of a raise to power n that will be, n into x raise to power n minus 1 minus 0 because a is a constant, so derivative with respect to x of a constant is 0 minus x raise to power n minus a raise to power n into, d by dx of x minus a will be 1 divided by x minus a the whole square. This will be equal to x minus a into, n into x raise to power n minus 1 minus x raise to power n minus a raise to power n divided by x minus a the whole square. Now on opening the brackets in the numerator we have, n into x raise to power n minus 1 into x minus a into n into x raise to power n minus 1 minus x raise to power n plus a raise to power n divided by x minus a the whole square. Now this can be written as, n into x raise to power n minus 1 plus 1 becomes x raise to power n minus a n x raise to power n minus 1 minus x raise to power n plus a raise to power n divided by x minus a the whole square. So this is our answer to the question. I hope that you understood the question and enjoyed the session. Have a good day.