If x^py^q=(x+y)^p+q, then prove that dy/dx=y/x.

Given : xp * yq =(x+y)p+q Take log on both side log xp + log yq = log(x+y)p+q => plog x + qlogy = (p+q)*log (x+y) Now differentiate with respect to x p/x + (q/y)* dy/dx = {(p+q)/(x + y)}*(1 + dy/dx) => p/x + (q/y)* dy/dx = {(p+q)/(x + y) + {(p+q)/(x + y)}*(dy/dx) => (q/y)* dy/dx - {(p+q)/(x + y)}*(dy/dx) = (p+q)/(x + y) - p/x => {(q/y) - (p+q)/(x + y)}*(dy/dx) = (p+q)/(x + y) - p/x => [(qx + qy - py - qy)/{y*(x + y)}]*(dy/dx) = (px + qx - px - py)/{x*(x+y)} => {(qx - py)/y}*(dy/dx) = (qx - py)/x => (dy/dx)/y = 1/x => dy/dx = y/x