If 3^x y^x = x^{3y}, then the value of frac{d | vedclass

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(D) Given the equation: $3^x y^x = x^{3y}$.Taking the natural logarithm on both sides:$\ln(3^x y^x) = \ln(x^{3y})$$x \ln 3 + x \ln y = 3y \ln x$.Diffe

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$\frac{1}{3}$

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(D) Given the equation: $3^x y^x = x^{3y}$.Taking the natural logarithm on both sides:$\ln(3^x y^x) = \ln(x^{3y})$$x \ln 3 + x \ln y = 3y \ln x$.Differentiating both sides with respect to $x$:$\frac{d}{dx}(x \ln 3) + \frac{d}{dx}(x \ln y) = \frac{d}{dx}(3y \ln x)$$\ln 3 + (\ln y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx}) = 3 \cdot \frac{dy}{dx} \cdot \ln x + 3y \cdot \frac{1}{x}$.At $x = 1$, we first find $y$ from the original equation:$3^1 y^1 = 1^{3y} \implies 3y = 1 \implies y = \frac{1}{3}$.Substituting $x = 1$ and $y = \frac{1}{3}$ into the differentiated equation:$\ln 3 + \ln(\frac{1}{3}) + 1 \cdot \frac{1}{1/3} \cdot \frac{dy}{dx} = 3 \cdot \frac{dy}{dx} \cdot \ln 1 + 3(\frac{1}{3}) \cdot \frac{1}{1}$.Since $\ln(\frac{1}{3}) = -\ln 3$ and $\ln 1 = 0$:$\ln 3 - \ln 3 + 3 \frac{dy}{dx} = 0 + 1$.$3 \frac{dy}{dx} = 1$.$\frac{dy}{dx} = \frac{1}{3}$.

If $3^x y^x = x^{3y}$, then the value of $\frac{dy}{dx}$ at $x = 1$ is:

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