To solve this problem using the definition of the derivative, we apply the formula:

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

The given function is $f(x) = 2x^2 - 1$.

Step-by-Step Solution:

1. Substitute $f(x)$ and $f(x+h)$ into the definition: $f'(x) = \lim_{h \to 0} \frac{(2(x+h)^2 - 1) - (2x^2 - 1)}{h}$

2. Expand $(x+h)^2$: $(x+h)^2 = x^2 + 2xh + h^2$ Substituting this into $2(x+h)^2 - 1$: $f(x+h) = 2(x^2 + 2xh + h^2) - 1 = 2x^2 + 4xh + 2h^2 - 1$

3. Simplify $f(x+h) - f(x)$: $f(x+h) - f(x) = \big(2x^2 + 4xh + 2h^2 - 1\big) - \big(2x^2 - 1\big)$ Simplify terms: $f(x+h) - f(x) = 4xh + 2h^2$

4. Divide by $h$: $\frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2}{h}$ Factor $h$ from the numerator: $\frac{4xh + 2h^2}{h} = 4x + 2h$

5. Take the limit as $h \to 0$: $\lim_{h \to 0} (4x + 2h) = 4x$

Final Answer:

$f'(x) = 4x$

Let me know if you’d like a more detailed explanation or further help.

Related Questions to Explore:

1. How does the definition of the derivative help in finding the slope of a curve?
2. What happens if you apply this process to a linear function like $f(x) = 3x + 2$?
3. How would you find the derivative using standard differentiation rules instead?
4. Can you explain the significance of the $h \to 0$ limit in this calculation?
5. How does this process change if the function has higher powers of $x$?

Tip: Always double-check the algebra when expanding and simplifying terms, as small errors can lead to incorrect results.