Need help using the alternative formula to find the derivative at $x = c$ Don't know if I'm simplifying wrong or if it is a simple mistake but I cannot get the answer of 4. Please help and include exact step by step details. Thank you.
$$f(x)=x^3 + 2x^2 + 1,\;\;\;c= -2$$

Using the alternative definition of the derivative, given what you posted in a comment, we'll start with the approximation of the derivative given by $\lim_{x \to -2}\dfrac{f(x) - f(c)}{x - c}$
\begin{align*}\require{cancel}
\lim_{x \to -2} \dfrac{f(x) - f(c)}{x - c}
&=\lim_{x\to -2} \dfrac{x^3 + 2x^2 + 1 - ((-2)^3 + 2(-2)^2 + 1)}{x - (-2)}\\
&=\lim_{x \to -2}  \dfrac{x^3 + 2x^2 + 1 -(-8 + 2\cdot 4 + 1)}{x + 2}\\
&=\lim_{x \to -2} \dfrac{x^3 + 2x^2 + 1 - (1)}{x + 2}\\
&= \lim_{x \to -2}  \dfrac{x^2({x + 2)}}{{x+2}}\\
&= \lim_{x \to -2}  \dfrac{x^2\cancel{(x + 2)}}{\cancel{x+2}} \\
&= \lim_{x \to -2} x^2 = (-2)^2 = 4
\end{align*}
