If {x^m}{y^n} = {(x + y)^{m + n}},then {left. | vedclass

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(B) Given the equation: ${x^m}{y^n} = {(x + y)^{m + n}}$Taking the natural logarithm on both sides:$m \ln x + n \ln y = (m + n) \ln(x + y)$Differentia

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$2$

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(B) Given the equation: ${x^m}{y^n} = {(x + y)^{m + n}}$Taking the natural logarithm on both sides:$m \ln x + n \ln y = (m + n) \ln(x + y)$Differentiating both sides with respect to $x$:$\frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = (m + n) \frac{1}{x + y} \left( 1 + \frac{dy}{dx} \right)$Rearranging the terms to isolate $\frac{dy}{dx}$:$\frac{dy}{dx} \left( \frac{n}{y} - \frac{m + n}{x + y} \right) = \frac{m + n}{x + y} - \frac{m}{x}$$\frac{dy}{dx} \left( \frac{nx + ny - my - ny}{y(x + y)} \right) = \frac{mx + nx - mx - my}{x(x + y)}$$\frac{dy}{dx} \left( \frac{nx - my}{y(x + y)} \right) = \frac{nx - my}{x(x + y)}$$\frac{dy}{dx} = \frac{y}{x}$At $x = 1$ and $y = 2$,we have:${\left. {\frac{dy}{dx}} \right|_{x = 1, y = 2}} = \frac{2}{1} = 2$.

If ${x^m}{y^n} = {(x + y)^{m + n}}$,then ${\left. {\frac{dy}{dx}} \right|_{x = 1, y = 2}}$ is equal to:

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