To evaluate (dy)/(dx) if y = (8 − 3x6)/(8x5):

Here are the solutions to the differentiation problems.

1.1
To evaluate $\frac{dy}{dx}$ if $y = \frac{8 - 3x^6}{8x^5}$:

Step 1: Rewrite the function by separating the terms and expressing them with negative exponents.

$$y = \frac{8}{8x^5} - \frac{3x^6}{8x^5} = x^{-5} - \frac{3}{8}x^{6-5} = x^{-5} - \frac{3}{8}x$$

Step 2: Differentiate $y$ with respect to $x$ using the power rule, $\frac{d}{dx}(x^n) = nx^{n-1}$.

$$\frac{dy}{dx} = -5x^{-5-1} - \frac{3}{8}(1)x^{1-1} = -5x^{-6} - \frac{3}{8}$$

Step 3: Express the answer with positive exponents.

$$\frac{dy}{dx} = -\frac{5}{x^6} - \frac{3}{8}$$

The final answer is $\boxed{-\frac{5}{x^6} - \frac{3}{8}}$.

1.2
To evaluate $\frac{dy}{dx}$ if $y = \frac{2\sqrt{x} + 1}{x^2}$:

Step 1: Rewrite the function by separating the terms and expressing them with fractional and negative exponents.

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

Step 2: Differentiate $y$ with respect to $x$ using the power rule.

$$\frac{dy}{dx} = 2\left(-\frac{3}{2}\right)x^{-3/2 - 1} + (-2)x^{-2 - 1} = -3x^{-5/2} - 2x^{-3}$$

Step 3: Express the answer with positive exponents.

$$\frac{dy}{dx} = -\frac{3}{x^{5/2}} - \frac{2}{x^3}$$

The final answer is $\boxed{-\frac{3}{x^{5/2}} - \frac{2}{x^3}}$.

1.3
To evaluate $\frac{dy}{dw}$ if $y = 4(\sqrt[3]{x^2})$ and $x = w^{-3}$:

Step 1: Rewrite $y$ in terms of $x$ with fractional exponents and find $\frac{dy}{dx}$.

$$y = 4x^{2/3}$$ $$\frac{dy}{dx} = 4 \cdot \frac{2}{3}x^{2/3 - 1} = \frac{8}{3}x^{-1/3}$$

Step 2: Find $\frac{dx}{dw}$.

$$x = w^{-3}$$ $$\frac{dx}{dw} = -3w^{-3-1} = -3w^{-4}$$

Step 3: Use the chain rule $\frac{dy}{dw} = \frac{dy}{dx} \cdot \frac{dx}{dw}$. Substitute $x = w^{-3}$ into $\frac{dy}{dx}$.

$$\frac{dy}{dx} = \frac{8}{3}(w^{-3})^{-1/3} = \frac{8}{3}w^{(-3)(-1/3)} = \frac{8}{3}w^1 = \frac{8}{3}w$$

Step 4: Multiply $\frac{dy}{dx}$ and $\frac{dx}{dw}$.

$$\frac{dy}{dw} = \left(\frac{8}{3}w\right)(-3w^{-4}) = -8w^{1-4} = -8w^{-3}$$

Step 5: Express the answer with positive exponents.

$$\frac{dy}{dw} = -\frac{8}{w^3}$$

The final answer is $\boxed{-\frac{8}{w^3}}$.

1.4
To evaluate $f'(x)$ if $f(x) = \frac{x^3 - 5x^2 + 4x}{x-4}$:

Step 1: Factor the numerator of $f(x)$.

$$x^3 - 5x^2 + 4x = x(x^2 - 5x + 4) = x(x-1)(x-4)$$

Step 2: Simplify $f(x)$ by canceling common factors.

$$f(x) = \frac{x(x-1)(x-4)}{x-4} = x(x-1) = x^2 - x \quad \text{for } x \neq 4$$

Step 3: Differentiate $f(x)$ with respect to $x$ using the power rule.

$$f'(x) = \frac{d}{dx}(x^2 - x) = 2x - 1$$

The final answer is $\boxed{2x - 1}$.

1.5
To evaluate $f'(x)$ if $f(x) = \left(x - \frac{3}{x}\right)^2$:

Step 1: Rewrite the function with negative exponents and expand the square.

$$f(x) = (x - 3x^{-1})^2 = x^2 - 2(x)(3x^{-1}) + (3x^{-1})^2$$ $$f(x) = x^2 - 6x^{1-1} + 9x^{-2} = x^2 - 6 + 9x^{-2}$$

Step 2: Differentiate $f(x)$ with respect to $x$ using the power rule.

$$f'(x) = \frac{d}{dx}(x^2 - 6 + 9x^{-2}) = 2x - 0 + 9(-2)x^{-2-1} = 2x - 18x^{-3}$$

Step 3: Express the answer with positive exponents.

$$f'(x) = 2x - \frac{18}{x^3}$$

The final answer is $\boxed{2x - \frac{18}{x^3}}$.

1.6
To evaluate $\frac{dy}{dx}$ if $x^2 - 3 = \sqrt[3]{y}$:

Step 1: Isolate $y$ by cubing both sides of the equation.

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