If {x^m}{y^n} = 2{(x + y)^{m + n}}, the value | vedclass

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(C) Given the equation: ${x^m}{y^n} = 2{(x + y)^{m + n}}$Taking the natural logarithm on both sides:$\ln({x^m}{y^n}) = \ln(2{(x + y)^{m + n}})$$m\ln x

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$y/x$

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(C) Given the equation: ${x^m}{y^n} = 2{(x + y)^{m + n}}$Taking the natural logarithm on both sides:$\ln({x^m}{y^n}) = \ln(2{(x + y)^{m + n}})$$m\ln x + n\ln y = \ln 2 + (m + n)\ln(x + y)$Now,differentiate both sides with respect to $x$:$\frac{d}{dx}(m\ln x + n\ln y) = \frac{d}{dx}(\ln 2 + (m + n)\ln(x + y))$$\frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = 0 + \frac{m + n}{x + y}(1 + \frac{dy}{dx})$$\frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = \frac{m + n}{x + y} + \frac{m + n}{x + y}\frac{dy}{dx}$Rearranging the terms to isolate $\frac{dy}{dx}$:$\frac{dy}{dx}(\frac{n}{y} - \frac{m + n}{x + y}) = \frac{m + n}{x + y} - \frac{m}{x}$$\frac{dy}{dx}(\frac{nx + ny - my - ny}{y(x + y)}) = \frac{mx + nx - mx - my}{x(x + y)}$$\frac{dy}{dx}(\frac{nx - my}{y(x + y)}) = \frac{nx - my}{x(x + y)}$$\frac{dy}{dx} = \frac{nx - my}{x(x + y)} 	imes \frac{y(x + y)}{nx - my}$$\frac{dy}{dx} = \frac{y}{x}$

If ${x^m}{y^n} = 2{(x + y)^{m + n}},$ the value of $\frac{dy}{dx}$ is

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