If x y neq 0, x+y neq 0 and x^m y^n=(x+y)^{m+ | vedclass

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

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$\frac{y}{x}$

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(A) Given, $x^m y^n = (x+y)^{m+n}$.Taking the natural logarithm on both sides, we get:$m \ln x + n \ln y = (m+n) \ln (x+y)$.Differentiating both sides with respect to $x$:$\frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = \frac{m+n}{x+y} \left(1 + \frac{dy}{dx}ight)$.Rearranging the terms to isolate $\frac{dy}{dx}$:$\frac{m}{x} - \frac{m+n}{x+y} = \frac{dy}{dx} \left( \frac{m+n}{x+y} - \frac{n}{y} ight)$.Simplifying both sides:$\frac{m(x+y) - x(m+n)}{x(x+y)} = \frac{dy}{dx} \left( \frac{y(m+n) - n(x+y)}{y(x+y)} ight)$.$\frac{mx + my - mx - nx}{x(x+y)} = \frac{dy}{dx} \left( \frac{my + ny - nx - ny}{y(x+y)} ight)$.$\frac{my - nx}{x(x+y)} = \frac{dy}{dx} \left( \frac{my - nx}{y(x+y)} ight)$.Since $my - nx 
eq 0$ (as $x, y$ are variables), we can cancel the terms:$\frac{1}{x} = \frac{dy}{dx} \left( \frac{1}{y} ight)$.Therefore, $\frac{dy}{dx} = \frac{y}{x}$.

If $x y \neq 0, x+y \neq 0$ and $x^m y^n=(x+y)^{m+n}$, where $m, n \notin N$, then $\frac{d y}{d x}$ is equal to

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