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

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

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$\frac{y}{x}$

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(D) Given the equation: $x^{p} + y^{q} = (x + y)^{p + q}$.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 $x$:$\frac{p}{x} + \frac{q}{y} \cdot \frac{dy}{dx} = \frac{p + q}{x + y} \left( 1 + \frac{dy}{dx} \right)$.Rearranging the terms to isolate $\frac{dy}{dx}$:$\frac{p}{x} - \frac{p + q}{x + y} = \left( \frac{p + q}{x + y} - \frac{q}{y} \right) \frac{dy}{dx}$.Simplifying both sides:$\frac{p(x + y) - x(p + q)}{x(x + y)} = \left( \frac{y(p + q) - q(x + y)}{y(x + y)} \right) \frac{dy}{dx}$.$\frac{px + py - px - qx}{x(x + y)} = \left( \frac{py + qy - qx - qy}{y(x + y)} \right) \frac{dy}{dx}$.$\frac{py - qx}{x(x + y)} = \frac{py - qx}{y(x + y)} \cdot \frac{dy}{dx}$.Canceling the common term $(py - qx)$ and $(x + y)$ from both sides,we get:$\frac{1}{x} = \frac{1}{y} \cdot \frac{dy}{dx}$.Therefore,$\frac{dy}{dx} = \frac{y}{x}$.

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

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