Find dy/dx at (5,-2) if 3x2−5y3+xy−1=0. Also find possible values of x for AB=BA given A=[[x2,3],[1,3x]] and B=[[3,6],[2,x]].

$$\frac{d}{dx}(3x^2) - \frac{d}{dx}(5y^3) + \frac{d}{dx}(xy) - \frac{d}{dx}(1) = \frac{d}{dx}(0)$$

Applying the power rule, chain rule, and product rule:

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

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

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

$$\boxed{\frac{28}{55}}$$

: 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 top-left elements: $3x^2+6 = 3x^2+6$ (This is always true).
Equating the bottom-right elements: $6+3x^2 = 6+3x^2$ (This is always true).

Equating the top-right elements:

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

Divide by 3:

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

Equating the bottom-left elements:

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

Both equations yield the same quadratic equation.

We can factor the quadratic equation:

$$2x^2 - 6x + x - 3 = 0$$ $$2x(x - 3) + 1(x - 3) = 0$$ $$(2x + 1)(x - 3) = 0$$

Set each factor to zero to find the values of $x$:

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

The possible values of $x$ are:

$$\boxed{x = -\frac{1}{2} \text{ or } x = 3}$$