If x^{1/2} y^{1/3} = (x + y)^n and x frac{dy} | vedclass

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(C) Given $x^{1/2} y^{1/3} = (x + y)^n$. Taking natural logarithm on both sides: $\frac{1}{2} \ln x + \frac{1}{3} \ln y = n \ln(x + y)$. Differentiati

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$\frac{5}{6}$

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(C) Given $x^{1/2} y^{1/3} = (x + y)^n$. Taking natural logarithm on both sides: $\frac{1}{2} \ln x + \frac{1}{3} \ln y = n \ln(x + y)$. Differentiating with respect to $x$: $\frac{1}{2x} + \frac{1}{3y} \frac{dy}{dx} = \frac{n}{x + y} (1 + \frac{dy}{dx})$. Given $x \frac{dy}{dx} - y = 0 \implies \frac{dy}{dx} = \frac{y}{x}$. Substituting $\frac{dy}{dx} = \frac{y}{x}$ into the differentiated equation: $\frac{1}{2x} + \frac{1}{3y} (\frac{y}{x}) = \frac{n}{x + y} (1 + \frac{y}{x})$. $\frac{1}{2x} + \frac{1}{3x} = \frac{n}{x + y} (\frac{x + y}{x})$. $\frac{3 + 2}{6x} = \frac{n}{x}$. $\frac{5}{6x} = \frac{n}{x} \implies n = \frac{5}{6}$.

If $x^{1/2} y^{1/3} = (x + y)^n$ and $x \frac{dy}{dx} - y = 0$, then $n =$

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