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

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

Taking the 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 $\dfrac{dy}{dx}$:

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

Simplifying the left side:

$\dfrac{x(p+q) - p(x+y)}{x(x+y)} = \dfrac{px + qx - px - py}{x(x+y)} = \dfrac{qx - py}{x(x+y)}$

Simplifying the right side:

$\dfrac{q(x+y) - y(p+q)}{y(x+y)} = \dfrac{qx + qy - py - qy}{y(x+y)} = \dfrac{qx - py}{y(x+y)}$

Equating the two sides:

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

Canceling the common terms from both sides:

$\dfrac{1}{x} = \dfrac{1}{y} \dfrac{dy}{dx}$

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

The derivative $\dfrac{dy}{dx}$ is $\dfrac{y}{x}$, which is independent of both $p$ and $q$.

Answer: is independent of both $p$ and $q$