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

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

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$\frac{y}{x}$

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(A) Given the equation: $x^py^q=(x+y)^{p+q}$ 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 isolate $\frac{dy}{dx}$: $\frac{q}{y} \frac{dy}{dx} - \frac{p+q}{x+y} \frac{dy}{dx} = \frac{p+q}{x+y} - \frac{p}{x}$ $\frac{dy}{dx} \left( \frac{q(x+y) - y(p+q)}{y(x+y)} \right) = \frac{x(p+q) - p(x+y)}{x(x+y)}$ $\frac{dy}{dx} \left( \frac{qx + qy - py - qy}{y(x+y)} \right) = \frac{px + qx - px - py}{x(x+y)}$ $\frac{dy}{dx} \left( \frac{qx - py}{y(x+y)} \right) = \frac{qx - py}{x(x+y)}$ Dividing both sides by $\frac{qx - py}{x+y}$: $\frac{dy}{dx} = \frac{y}{x}$

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

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