Find the derivatives of the following functions from first principles or by using differentiation rules: f(x) = -2x2, f(x) = 3x2 − 6x, and y = 2*sqrt(x) - 8/x.

$$f(x+h) - f(x) = (-2x^2 - 4xh - 2h^2) - (-2x^2)$$ $$f(x+h) - f(x) = -2x^2 - 4xh - 2h^2 + 2x^2$$ $$f(x+h) - f(x) = -4xh - 2h^2$$

$$f'(x) = \lim_{h \to 0} (-4x - 2h) = -4x - 2(0) = -4x$$

The derivative is $\boxed{\text{-4x}}$.

6.2.1

To differentiate $f(x) = 3x^2 - 6x$ with respect to $x$, we apply the power rule $\frac{d}{dx}(ax^n) = nax^{n-1}$.

$$\frac{d}{dx}(3x^2) = 3 \cdot 2x^{2-1} = 6x$$ $$\frac{d}{dx}(6x) = 6 \cdot 1x^{1-1} = 6x^0 = 6$$

$$f'(x) = 6x - 6$$

The derivative is $\boxed{\text{6x - 6}}$.

6.2.2

To determine $\frac{dy}{dx}$ if $y = 2\sqrt{x} - \frac{8}{x}$, we first rewrite $y$ using exponent notation.

$$\frac{dy}{dx} = \frac{d}{dx}(2x^{1/2}) - \frac{d}{dx}(8x^{-1})$$ $$\frac{d}{dx}(2x^{1/2}) = 2 \cdot \frac{1}{2} x^{1/2 - 1} = x^{-1/2}$$ $$\frac{d}{dx}(8x^{-1}) = 8 \cdot (-1) x^{-1 - 1} = -8x^{-2}$$

$$\frac{dy}{dx} = x^{-1/2} - (-8x^{-2}) = x^{-1/2} + 8x^{-2}$$

The derivative is \boxed{\text{x^{-1/2} + 8x^{-2}}}.

6.3

To determine $\frac{dy}{dx}$ if $xy - 5 = \sqrt{x^3}$, we first solve for $y$ explicitly.

$$xy = \sqrt{x^3} + 5$$ $$xy = x^{3/2} + 5$$ $$y = \frac{x^{3/2} + 5}{x}$$ $$y = \frac{x^{3/2}}{x} + \frac{5}{x}$$ $$y = x^{3/2 - 1} + 5x^{-1}$$ $$y = x^{1/2} + 5x^{-1}$$

$$\frac{dy}{dx} = \frac{d}{dx}(x^{1/2}) + \frac{d}{dx}(5x^{-1})$$ $$\frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2}$$ $$\frac{d}{dx}(5x^{-1}) = 5 \cdot (-1)x^{-1 - 1} = -5x^{-2}$$

$$\frac{dy}{dx} = \frac{1}{2}x^{-1/2} - 5x^{-2}$$

The derivative is \boxed{\text{\frac{1}{2}x^{-1/2} - 5x^{-2}}}.

6.4

To determine the equation of the tangent to the graph of $f(x) = 10x - x^2$ at the point where $f'(x) = -2$.

$$f(x) = 10x - x^2$$ $$f'(x) = \frac{d}{dx}(10x) - \frac{d}{dx}(x^2) = 10 - 2x$$

$$10 - 2x = -2$$ $$-2x = -2 - 10$$ $$-2x = -12$$ $$x = 6$$

$$f(6) = 10(6) - (6)^2$$ $$f(6) = 60 - 36$$ $$f(6) = 24$$

The point of tangency is $(6, 24)$. The slope of the tangent ($m$) is given as $-2$.

$$y - 24 = -2(x - 6)$$ $$y - 24 = -2x + 12$$ $$y = -2x + 12 + 24$$ $$y = -2x + 36$$

The equation of the tangent is $\boxed{\text{y = -2x + 36}}$.