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

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

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$y / x$

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(A) Given,$x^{p} y^{q}=(x+y)^{p+q}$.Taking the natural logarithm on both sides,we get:$p \ln x + q \ln y = (p+q) \ln (x+y)$.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 isolate $\frac{dy}{dx}$:$\frac{q}{y} \frac{dy}{dx} - \frac{p+q}{x+y} \frac{dy}{dx} = \frac{p+q}{x+y} - \frac{p}{x}$.$\frac{dy}{dx} \left( \frac{q(x+y) - y(p+q)}{y(x+y)} \right) = \frac{x(p+q) - p(x+y)}{x(x+y)}$.$\frac{dy}{dx} \left( \frac{qx + qy - py - qy}{y(x+y)} \right) = \frac{px + qx - px - py}{x(x+y)}$.$\frac{dy}{dx} \left( \frac{qx - py}{y(x+y)} \right) = \frac{qx - py}{x(x+y)}$.Canceling the common term $(qx - py)$ and $(x+y)$ from both sides:$\frac{dy}{dx} = \frac{y}{x}$.

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

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