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

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(A) Given the equation $x^y \cdot y^x = 16$. Taking the natural logarithm on both sides,we get: $y \ln x + x \ln y = \ln 16$. Differentiating both sid

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-$1$

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(A) Given the equation $x^y \cdot y^x = 16$. Taking the natural logarithm on both sides,we get: $y \ln x + x \ln y = \ln 16$. Differentiating both sides with respect to $x$ using the product rule: $\frac{d}{dx}(y \ln x) + \frac{d}{dx}(x \ln y) = \frac{d}{dx}(\ln 16)$. $\left( \frac{dy}{dx} \ln x + y \cdot \frac{1}{x} \right) + \left( 1 \cdot \ln y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} \right) = 0$. Rearranging the terms to isolate $\frac{dy}{dx}$: $\frac{dy}{dx} (\ln x + \frac{x}{y}) = -(\frac{y}{x} + \ln y)$. $\frac{dy}{dx} = -\frac{\frac{y}{x} + \ln y}{\ln x + \frac{x}{y}}$. Now,substitute the point $(2, 2)$ into the expression: $\left( \frac{dy}{dx} \right)_{(2, 2)} = -\frac{\frac{2}{2} + \ln 2}{\ln 2 + \frac{2}{2}} = -\frac{1 + \ln 2}{1 + \ln 2} = -1$.

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

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