To determine the derivative of $f(x) = -2x$ from first principles, we start by using the definition of the derivative:

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

Given $f(x) = -2x$, we substitute into the definition:

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

Simplify the expression inside the limit:

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

Notice that the $-2x$ and $2x$ terms cancel each other out:

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

Now, cancel the $h$ in the numerator and the denominator:

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

Since $-2$ is a constant and independent of $h$, the limit is simply:

$f'(x) = -2$

Answer to the Given Question:

The correct expression obtained after canceling $h$ in the numerator and denominator is $-2$.

Would you like to go into more detail about any step? Here are five related questions to expand on this topic:

1. How does the derivative from first principles apply to more complex functions?
2. Can you explain the geometric interpretation of the derivative?
3. How does the derivative of $f(x) = -2x$ compare to the derivative of other linear functions?
4. What happens if we have a higher-degree polynomial? How does first principles differ?
5. How do you find the derivative of a function that is not linear using first principles?

Tip: When simplifying expressions during differentiation from first principles, always ensure that any terms involving $h$ can be canceled before taking the limit as $h$ approaches zero.