The question asks to differentiate each function with respect to $x$. Let’s go through each of the four cases step by step:

1. $y = 2x^3 - 3x + 1$

Using basic differentiation rules: $\frac{dy}{dx} = \frac{d}{dx}(2x^3) - \frac{d}{dx}(3x) + \frac{d}{dx}(1)$ $\frac{dy}{dx} = 6x^2 - 3 + 0$ $\boxed{\frac{dy}{dx} = 6x^2 - 3}$

2. $y = \frac{1}{x}$

Rewrite $\frac{1}{x}$ as $x^{-1}$, then differentiate: $\frac{dy}{dx} = \frac{d}{dx}(x^{-1})$ $\frac{dy}{dx} = -x^{-2}$ $\boxed{\frac{dy}{dx} = -\frac{1}{x^2}}$

3. $y = \frac{3}{x^2}$

Rewrite $\frac{3}{x^2}$ as $3x^{-2}$, then differentiate: $\frac{dy}{dx} = \frac{d}{dx}(3x^{-2})$ $\frac{dy}{dx} = 3(-2)x^{-3}$ $\frac{dy}{dx} = -\frac{6}{x^3}$ $\boxed{\frac{dy}{dx} = -\frac{6}{x^3}}$

4. $y = \frac{1}{x} + \frac{2}{x^3}$

Rewrite the terms as $y = x^{-1} + 2x^{-3}$, then differentiate term by term: $\frac{dy}{dx} = \frac{d}{dx}(x^{-1}) + \frac{d}{dx}(2x^{-3})$ $\frac{dy}{dx} = -x^{-2} + 2(-3)x^{-4}$ $\frac{dy}{dx} = -\frac{1}{x^2} - \frac{6}{x^4}$ $\boxed{\frac{dy}{dx} = -\frac{1}{x^2} - \frac{6}{x^4}}$

Final Results:

1. $\frac{dy}{dx} = 6x^2 - 3$
2. $\frac{dy}{dx} = -\frac{1}{x^2}$
3. $\frac{dy}{dx} = -\frac{6}{x^3}$
4. $\frac{dy}{dx} = -\frac{1}{x^2} - \frac{6}{x^4}$

Would you like further clarification or examples?

5 Related Questions:

1. How do you apply the power rule in differentiation?
2. What are the general rules for differentiating polynomial functions?
3. How does the chain rule apply to more complex functions?
4. Can you explain higher-order derivatives of these functions?
5. How do we find critical points using derivatives?

Tip:

Always rewrite terms like $\frac{1}{x^n}$ as $x^{-n}$ before differentiating—it simplifies the process!