Consider the following for the two (02) items that follow: Let (x+y)^ p+q = x^p y^q , where p , q are positive integers. If p + q = 10 , then what is dy dx equal to?

Step-by-step solution

Given (x+y)^ p+q = x^p y^q Taking natural logarithm on both sides: (p+q) (x+y) = p x + q y Differentiating both sides with respect to x : p+q x+y (1 + dy dx ) = p x + q y dy dx Rearranging the terms to solve for dy dx : p+q x+y - p x = ( q y - p+q x+y ) dy dx x(p+q) - p(x+y) x(x+y) = ( q(x+y) - y(p+q) y(x+y) ) dy dx qx - py x(x+y) = ( qx - py y(x+y) ) dy dx dy dx = y x The result is independent of the value of p+q . Answer: y x