If x^y cdot y^x = 16,then frac{dy}{dx} at (2, | vedclass

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(D) Given equation: $x^y \cdot y^x = 16$. Taking natural logarithm on both sides: $\log(x^y \cdot y^x) = \log(16)$. Using logarithmic properties: $y \

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-$1$

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(D) Given equation: $x^y \cdot y^x = 16$. Taking natural logarithm on both sides: $\log(x^y \cdot y^x) = \log(16)$. Using logarithmic properties: $y \log x + x \log y = \log(16)$. Differentiating both sides with respect to $x$ using the product rule: $\frac{dy}{dx} \log x + y \cdot \frac{1}{x} + \log y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} = 0$. Substitute the point $(2, 2)$ into the derivative equation: $\frac{dy}{dx} \log(2) + 2 \cdot \frac{1}{2} + \log(2) + 2 \cdot \frac{1}{2} \cdot \frac{dy}{dx} = 0$. $\frac{dy}{dx} \log(2) + 1 + \log(2) + \frac{dy}{dx} = 0$. $\frac{dy}{dx} (\log(2) + 1) = -(1 + \log(2))$. $\frac{dy}{dx} = -\frac{1 + \log(2)}{1 + \log(2)} = -1$.

If $x^y \cdot y^x = 16$,then $\frac{dy}{dx}$ at $(2, 2)$ is

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