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

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

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independent of both $p$ and $q$

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(D) Given the equation $x^p \cdot y^q = (x + y)^{p + q}$.Taking the natural logarithm on both sides:$\ln(x^p \cdot y^q) = \ln((x + y)^{p + q})$$p \ln x + q \ln y = (p + q) \ln(x + y)$Differentiating both sides with respect to $x$:$\frac{p}{x} + \frac{q}{y} \frac{dy}{dx} = (p + q) \frac{1}{x + y} (1 + \frac{dy}{dx})$$\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)}$$\frac{dy}{dx} \left( \frac{qx - py}{y(x + y)} \right) = \frac{qx - py}{x(x + y)}$$\frac{dy}{dx} = \frac{y}{x}$Since $\frac{dy}{dx} = \frac{y}{x}$,it is independent of both $p$ and $q$.

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

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