If 2x^y + 3y^x = 20,then frac{dy}{dx} at (2, | vedclass

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(B) Given the equation $2x^y + 3y^x = 20$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(2x^y) + \frac{d}{dx}(3y^x) = 0$. Using the fo

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$-\left(\frac{2+\log_e 8}{3+\log_e 4}\right)$

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(B) Given the equation $2x^y + 3y^x = 20$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(2x^y) + \frac{d}{dx}(3y^x) = 0$. Using the formula $\frac{d}{dx}(a^b) = a^b \frac{d}{dx}(b \ln a)$,we get: $2x^y \left(\frac{y}{x} + \ln x \cdot \frac{dy}{dx}\right) + 3y^x \left(\frac{x}{y} \cdot \frac{dy}{dx} + \ln y\right) = 0$. At the point $(2, 2)$,$x=2$ and $y=2$: $2(2^2) \left(\frac{2}{2} + \ln 2 \cdot \frac{dy}{dx}\right) + 3(2^2) \left(\frac{2}{2} \cdot \frac{dy}{dx} + \ln 2\right) = 0$. $8(1 + \ln 2 \cdot \frac{dy}{dx}) + 12(\frac{dy}{dx} + \ln 2) = 0$. $8 + 8 \ln 2 \cdot \frac{dy}{dx} + 12 \frac{dy}{dx} + 12 \ln 2 = 0$. $\frac{dy}{dx} (12 + 8 \ln 2) = -(8 + 12 \ln 2)$. $\frac{dy}{dx} = -\frac{8 + 12 \ln 2}{12 + 8 \ln 2} = -\frac{2(4 + 6 \ln 2)}{4(3 + 2 \ln 2)} = -\frac{2 + 3 \ln 2}{3 + 2 \ln 2}$. Since $3 \ln 2 = \ln 2^3 = \ln 8$ and $2 \ln 2 = \ln 2^2 = \ln 4$,we have: $\frac{dy}{dx} = -\left(\frac{2 + \log_e 8}{3 + \log_e 4}\right)$.

If $2x^y + 3y^x = 20$,then $\frac{dy}{dx}$ at $(2, 2)$ is equal to

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