Consider the following for the two (02) items that follow: Let...

Given $(x+y)^{p+q} = x^p y^q$

Taking natural logarithm on both sides:

$(p+q) \ln(x+y) = p \ln x + q \ln y$

Differentiating both sides with respect to $x$:

$\dfrac{p+q}{x+y} \left(1 + \dfrac{dy}{dx}\right) = \dfrac{p}{x} + \dfrac{q}{y} \dfrac{dy}{dx}$

Rearranging the terms to solve for $\dfrac{dy}{dx}$:

$\dfrac{p+q}{x+y} - \dfrac{p}{x} = \left( \dfrac{q}{y} - \dfrac{p+q}{x+y} \right) \dfrac{dy}{dx}$

$\dfrac{x(p+q) - p(x+y)}{x(x+y)} = \left( \dfrac{q(x+y) - y(p+q)}{y(x+y)} \right) \dfrac{dy}{dx}$

$\dfrac{qx - py}{x(x+y)} = \left( \dfrac{qx - py}{y(x+y)} \right) \dfrac{dy}{dx}$

$\dfrac{dy}{dx} = \dfrac{y}{x}$

The result is independent of the value of $p+q$.

Answer: $\dfrac{y}{x}$