Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 21

Find the derivative of $\displaystyle f(x) = x^3 - 3x + 5$ using the definition and the domain of its derivative.

Using the definition of derivative


$
\begin{equation}
\begin{aligned}

\qquad f'(x) &= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
&&
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{(x + h)^3 - 3(x + h) + 5 - (x^3 - 3x + 5)}{h}
&& \text{Substitute $f(x + h)$ and $f(x)$}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{\cancel{x^3} + 3x^2 h + 3xh^2 + h^3 - \cancel{3x} - 3h + \cancel{5} - \cancel{x^3} + \cancel{3x} - \cancel{5}} {h}
&& \text{Expand and combine like terms}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{3x^2 h + 3xh ^2 + h^3 - h}{h}
&& \text{Factor the numerator}
\\
\\
\qquad f'(x) &= \lim_{h \to 0} \frac{\cancel{h}(3x^2 + 3xh + h^2 - 3)}{\cancel{h}}
&& \text{Cancel out like terms}
\\
\\
\qquad f'(t) &= \lim_{h \to 0} (3x^2 + 3xh + h^2 - 3) = 3x^2 + 3x(0) + (0)^2 - 3
&& \text{Evaluate the limit}

\end{aligned}
\end{equation}
$


$\qquad \fbox{$f'(x) = 3x^2 - 3$}$

Both $f(x)$ and $f'(x)$ are polynomiale functions that are continuous on every number. Therefore, their domain is $(-\infty, \infty)$