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

Given: $$2x^y + 3y^x = 20$$. We need $$\frac{dy}{dx}$$ at $$(2, 2)$$.

Verify: At $$(2,2)$$: $$2(2^2) + 3(2^2) = 8 + 12 = 20$$ ✓

Differentiate implicitly. For $$x^y$$, take log: $$\ln(x^y) = y\ln x$$.

$$\frac{d}{dx}(x^y) = x^y\left(\frac{y}{x} + y'\ln x\right)$$

For $$y^x$$: $$\ln(y^x) = x\ln y$$.

$$\frac{d}{dx}(y^x) = y^x\left(\ln y + \frac{xy'}{y}\right)$$

Differentiating the equation:

$$ 2x^y\left(\frac{y}{x} + y'\ln x\right) + 3y^x\left(\ln y + \frac{xy'}{y}\right) = 0 $$

At $$(2, 2)$$: $$x^y = 4$$, $$y^x = 4$$:

$$ 2(4)\left(1 + y'\ln 2\right) + 3(4)\left(\ln 2 + y'\right) = 0 $$

$$ 8 + 8y'\ln 2 + 12\ln 2 + 12y' = 0 $$

$$ y'(8\ln 2 + 12) = -(8 + 12\ln 2) $$

$$ y' = -\frac{8 + 12\ln 2}{12 + 8\ln 2} = -\frac{4(2 + 3\ln 2)}{4(3 + 2\ln 2)} = -\frac{2 + \ln 8}{3 + \ln 4} $$

The correct answer is $$-\dfrac{2 + \log_e 8}{3 + \log_e 4}$$.