 Hello and welcome to the session. The question says find the derivative of the following functions. It is to be understood that A, B, C, D, P, Q, R and S affects non-zero constants and M and N are integers. So the fourth question is A x plus B into C x plus D whole square. Let's start with the solution and let us denote the given function by Y. So Y is equal to A x plus B and C x plus D whole square. Now let A x plus B is equal to U and C x plus D is equal to V. Now we have to find the derivative of Y that is dy upon dx we have to find out. So we have d dx of UV and by the product rule we have U into d dx of V plus x of U into V. So with the help of this product rule we shall find the derivative of the given function. Now let us substitute the functions U and V. So U is A x plus B into d dx of function V which is C x plus D whole square plus d dx of function U which is A x plus B into the function V which is C x plus D whole square. This is further equal to A x plus B and derivative of C x plus D whole square is 2 into C x plus D into d dx of C x plus D plus and derivative of A x plus B is derivative of constant is U so derivative of B is 0 and here derivative of x with respect to x is 1 so we have from here we have A into C x plus D whole square this is further equal to 2 times of A x plus B into C x plus D and derivative of C x plus D is C plus A into C x plus D whole square. This is equal to 2 C into A x plus B into C x plus D plus A into C x plus D whole square thus on finding the derivative of the given function A x plus B into C x plus D whole square we get 2 C into A x plus B into C x plus D plus A into C x plus D whole square. This completes the session. Take care and have a good day.