If {x^p}{y^q} = {(x + y)^{p + q}}, then frac{ | vedclass

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(A) Given the equation: ${x^p}{y^q} = {(x + y)^{p + q}}$Taking the natural logarithm on both sides:$\ln({x^p}{y^q}) = \ln({(x + y)^{p + q}})$$p\ln x +

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$\frac{y}{x}$

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(A) Given the equation: ${x^p}{y^q} = {(x + y)^{p + q}}$Taking the natural logarithm on both sides:$\ln({x^p}{y^q}) = \ln({(x + y)^{p + q}})$$p\ln x + q\ln y = (p + q)\ln(x + y)$Differentiating both sides with respect to $x$:$\frac{d}{dx}(p\ln x + q\ln y) = \frac{d}{dx}((p + q)\ln(x + y))$$\frac{p}{x} + \frac{q}{y}\frac{dy}{dx} = \frac{p + q}{x + y}(1 + \frac{dy}{dx})$Multiply by $xy(x + y)$ to clear denominators:$py(x + y) + qx(x + y)\frac{dy}{dx} = x(p + q)y + x(p + q)x\frac{dy}{dx}$$pxy + py^2 + qx^2\frac{dy}{dx} + qxy\frac{dy}{dx} = pxy + qxy + (px^2 + qx^2)\frac{dy}{dx}$Rearranging terms to isolate $\frac{dy}{dx}$:$\frac{dy}{dx}(qx^2 + qxy - px^2 - qx^2) = pxy + qxy - pxy - py^2$$\frac{dy}{dx}(qxy - px^2) = qxy - py^2$$\frac{dy}{dx} = \frac{qxy - py^2}{qx^2 - pxy} = \frac{y(qx - py)}{x(qx - py)}$$\frac{dy}{dx} = \frac{y}{x}$

If ${x^p}{y^q} = {(x + y)^{p + q}},$ then $\frac{dy}{dx} = $

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