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

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(A) Given the equation: ${x^p}{y^q} = {(x + y)^{p + q}}$Taking the natural logarithm on both sides:$p \ln x + q \ln y = (p + q) \ln (x + y)$Differentiating both sides with respect to $x$:$\frac{p}{x} + \frac{q}{y} \frac{dy}{dx} = \frac{p + q}{x + y} \left( 1 + \frac{dy}{dx} \right)$Rearranging the terms to solve for $\frac{dy}{dx}$:$\frac{dy}{dx} \left( \frac{q}{y} - \frac{p + q}{x + y} \right) = \frac{p + q}{x + y} - \frac{p}{x}$$\frac{dy}{dx} \left( \frac{q(x + y) - y(p + q)}{y(x + y)} \right) = \frac{x(p + q) - p(x + y)}{x(x + y)}$$\frac{dy}{dx} \left( \frac{qx + qy - py - qy}{y(x + y)} \right) = \frac{px + qx - px - py}{x(x + y)}$$\frac{dy}{dx} \left( \frac{qx - py}{y(x + y)} \right) = \frac{qx - py}{x(x + y)}$$\frac{dy}{dx} = \frac{y}{x}$Differentiating again with respect to $x$ using the quotient rule:$\frac{d^2y}{dx^2} = \frac{x \frac{dy}{dx} - y}{x^2}$Substitute $\frac{dy}{dx} = \frac{y}{x}$:$\frac{d^2y}{dx^2} = \frac{x(\frac{y}{x}) - y}{x^2} = \frac{y - y}{x^2} = 0$

If ${x^p}{y^q} = {(x + y)^{p + q}}$,then $\frac{{{d^2}y}}{{d{x^2}}} = $

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