Use the limit definition of the derivative to prove that the derivative of 2x^3 - 1 is 6x^2.

Step-by-step solution

1. 1

   Write down the definition. For $f(x)=2x^3-1$ the derivative is $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$. The whole proof is a matter of making that quotient simple enough for $h\to 0$ to become a plain substitution.

2. 2

   Expand $f(x+h)$. Using $(x+h)^3=x^3+3x^2h+3xh^2+h^3$, we get $f(x+h)=2x^3+6x^2h+6xh^2+2h^3-1$.

3. 3

   Subtract $f(x)$. The $2x^3$ and the $-1$ cancel, leaving $f(x+h)-f(x)=6x^2h+6xh^2+2h^3$. Every surviving term carries at least one factor of $h$, which is what keeps the quotient finite.

4. 4

   Divide by $h$. Factoring $h$ out gives $\frac{h\left(6x^2+6xh+2h^2\right)}{h}=6x^2+6xh+2h^2$, valid for every $h\neq 0$ — and that is all the limit ever needs.

5. 5

   Take the limit. The remaining expression is a polynomial in $h$, so substitution is legal: $\lim_{h\to 0}\left(6x^2+6xh+2h^2\right)=6x^2$.

6. 6

   State the conclusion. Therefore $\frac{d}{dx}(2x^3-1)=6x^2$, matching the power rule and confirming that the additive constant $-1$ contributes nothing.