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. The derivative of y with respect to x

Step-by-step solution

Given (x+y)^ p+q = x^p y^q Taking the 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 Simplifying the left side: x(p+q) - p(x+y) x(x+y) = px + qx - px - py x(x+y) = qx - py x(x+y) Simplifying the right side: q(x+y) - y(p+q) y(x+y) = qx + qy - py - qy y(x+y) = qx - py y(x+y) Equating the two sides: qx - py x(x+y) = qx - py y(x+y) dy dx Cance