If 3x2 − 5y3 + xy − 1 = 0, find (dy)/(dx) at the point (5, -2).

Remember to use the chain rule for terms involving $y$ and the product rule for $xy$.

$$\frac{d}{dx}(3x^2) - \frac{d}{dx}(5y^3) + \frac{d}{dx}(xy) - \frac{d}{dx}(1) = \frac{d}{dx}(0)$$ $$6x - 15y^2 \frac{dy}{dx} + \left(1 \cdot y + x \cdot \frac{dy}{dx}\right) - 0 = 0$$ $$6x - 15y^2 \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$$

$$\frac{dy}{dx}(x - 15y^2) = -6x - y$$ $$\frac{dy}{dx} = \frac{-6x - y}{x - 15y^2}$$

$$\frac{dy}{dx} = \frac{-6(5) - (-2)}{5 - 15(-2)^2}$$ $$\frac{dy}{dx} = \frac{-30 + 2}{5 - 15(4)}$$ $$\frac{dy}{dx} = \frac{-28}{5 - 60}$$ $$\frac{dy}{dx} = \frac{-28}{-55}$$ $$\frac{dy}{dx} = \frac{28}{55}$$

The value of $\frac{dy}{dx}$ at the point $(5, -2)$ is $\boxed{\frac{28}{55}}$.

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Given that $A = \begin{pmatrix} x^2 & 3 \\ 1 & 3x \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 6 \\ 2 & x \end{pmatrix}$, find the possible values of $x$ such that $AB = BA$.

$$AB = \begin{pmatrix} x^2 & 3 \\ 1 & 3x \end{pmatrix} \begin{pmatrix} 3 & 6 \\ 2 & x \end{pmatrix}$$ $$AB = \begin{pmatrix} (x^2)(3) + (3)(2) & (x^2)(6) + (3)(x) \\ (1)(3) + (3x)(2) & (1)(6) + (3x)(x) \end{pmatrix}$$ $$AB = \begin{pmatrix} 3x^2 + 6 & 6x^2 + 3x \\ 3 + 6x & 6 + 3x^2 \end{pmatrix}$$

$$BA = \begin{pmatrix} 3 & 6 \\ 2 & x \end{pmatrix} \begin{pmatrix} x^2 & 3 \\ 1 & 3x \end{pmatrix}$$ $$BA = \begin{pmatrix} (3)(x^2) + (6)(1) & (3)(3) + (6)(3x) \\ (2)(x^2) + (x)(1) & (2)(3) + (x)(3x) \end{pmatrix}$$ $$BA = \begin{pmatrix} 3x^2 + 6 & 9 + 18x \\ 2x^2 + x & 6 + 3x^2 \end{pmatrix}$$

$$\begin{pmatrix} 3x^2 + 6 & 6x^2 + 3x \\ 3 + 6x & 6 + 3x^2 \end{pmatrix} = \begin{pmatrix} 3x^2 + 6 & 9 + 18x \\ 2x^2 + x & 6 + 3x^2 \end{pmatrix}$$

Equating the (1,2) elements:

$$6x^2 + 3x = 9 + 18x$$ $$6x^2 - 15x - 9 = 0$$

Dividing by 3:

$$2x^2 - 5x - 3 = 0$$

Equating the (2,1) elements:

$$3 + 6x = 2x^2 + x$$ $$0 = 2x^2 + x - 6x - 3$$ $$2x^2 - 5x - 3 = 0$$

Both equations lead to the same quadratic equation.

We can factor the quadratic equation:

$$(2x+1)(x-3) = 0$$

This gives two possible values for $x$:

$$2x+1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}$$ $$x-3 = 0 \implies x = 3$$

The possible values of $x$ are $\boxed{-\frac{1}{2}, 3}$.