Show that the operation \ is associative ii. find the identity of under \*.

The distance squared formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $(x_2-x_1)^2 + (y_2-y_1)^2$.

$$SM^2 = (x-1)^2 + (y-0)^2 = (x-1)^2 + y^2$$ $$SN^2 = (x-2)^2 + (y-(-3))^2 = (x-2)^2 + (y+3)^2$$

$$(x^2 - 2x + 1) + y^2 - [(x^2 - 4x + 4) + (y^2 + 6y + 9)] = 1$$ $$x^2 - 2x + 1 + y^2 - x^2 + 4x - 4 - y^2 - 6y - 9 = 1$$ $$2x - 6y - 12 = 1$$ $$2x - 6y = 13$$

The equation $2x - 6y = 13$ is the equation of a straight line.
The constant properties are its gradient and y-intercept.
Rearrange to slope-intercept form $y = mx + c$:

$$-6y = -2x + 13$$ $$y = \frac{-2}{-6}x + \frac{13}{-6}$$ $$y = \frac{1}{3}x - \frac{13}{6}$$

The gradient is $\frac{1}{3}$ and the y-intercept is $-\frac{13}{6}$.

The equation of the locus of S is $\boxed{\text{2x - 6y = 13}}$.
Its constant properties are: gradient $m = \frac{1}{3}$ and y-intercept $c = -\frac{13}{6}$.

$$-6y = 13$$ $$y = -\frac{13}{6}$$

The point where the locus meets the y-axis is $(0, -\frac{13}{6})$.

The locus meets the y-axis at \boxed{\text{(0, -\frac{13}{6})}}.

$$3y^2 = 16x$$ $$\frac{d}{dx}(3y^2) = \frac{d}{dx}(16x)$$ $$6y \frac{dy}{dx} = 16$$ $$\frac{dy}{dx} = \frac{16}{6y} = \frac{8}{3y}$$

At $(3,4)$, $y=4$.

$$m = \frac{dy}{dx}\Big|_{(3,4)} = \frac{8}{3(4)} = \frac{8}{12} = \frac{2}{3}$$

Using the point $(x_1, y_1) = (3,4)$ and gradient $m = \frac{2}{3}$:

$$y - 4 = \frac{2}{3}(x - 3)$$ $$3(y - 4) = 2(x - 3)$$ $$3y - 12 = 2x - 6$$ $$3y = 2x + 6$$ $$2x - 3y + 6 = 0$$

The equation of the tangent is $\boxed{\text{2x - 3y + 6 = 0}}$.

$$f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \times \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}$$ $$f'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}$$ $$f'(x) = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$$

$$f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$$ $$f'(x) = \frac{1}{\sqrt{x+0} + \sqrt{x}}$$ $$f'(x) = \frac{1}{\sqrt{x} + \sqrt{x}}$$ $$f'(x) = \frac{1}{2\sqrt{x}}$$

The derivative of $f(x) = \sqrt{x}$ from first principles is \boxed{\text{\frac{1}{2\sqrt{x}}}}.

$$y = (x-2)(2x^2 - 3x - 5)$$ $$y = x(2x^2 - 3x - 5) - 2(2x^2 - 3x - 5)$$ $$y = 2x^3 - 3x^2 - 5x - 4x^2 + 6x + 10$$ $$y = 2x^3 - 7x^2 + x + 10$$

$$\frac{dy}{dx} = \frac{d}{dx}(2x^3 - 7x^2 + x + 10)$$ $$\frac{dy}{dx} = 2(3x^{3-1}) - 7(2x^{2-1}) + 1x^{1-1} + 0$$ $$\frac{dy}{dx} = 6x^2 - 14x + 1$$

The gradient $\frac{dy}{dx}$ is \boxed{\text{6x^2 - 14x + 1}}}.