calculate the derivative by the limit definition: f(x) = 6x^3 + 2

To calculate the derivative of the function $$\(f(x) = 6x^3 + 2\)$$ using the limit definition of a derivative, you can use the following formula: $$\[f'(x) = \lim_{{h \to 0}} \frac{f(x + h) - f(x)}{h}\]$$ In this case, $$\(f(x) = 6x^3 + 2\)$$, and we want to find f'(x). Now, plug this into the limit definition formula: $$\[f'(x) = \lim_{{h \to 0}} \frac{(6(x + h)^3 + 2) - (6x^3 + 2)}{h}\]$$ Let's simplify the expression inside the limit: $$\[f'(x) = \lim_{{h \to 0}} \frac{6(x^3 + 3x^2h + 3xh^2 + h^3) + 2 - 6x^3 - 2}{h}\]$$ Now, we can cancel out the terms that will simplify: $$\[f'(x) = \lim_{{h \to 0}} \frac{6(3x^2h + 3xh^2 + h^3)}{h}\]$$ Next, factor out an \(h\) from the numerator: $$\[f'(x) = \lim_{{h \to 0}} \frac{h(18x^2 + 18xh + 6h^2)}{h}\]$$ Now, cancel out the common factor of \(h\) in the numerator and denominator: $$\[f'(x) = \lim_{{h \to 0}} 18x^2 + 18xh + 6h^2\]$$ Now, we can calculate the limit as \(h\) approaches 0: $$\[f'(x) = 18x^2 + 18x(0) + 6(0)^2\]$$ $$\[f'(x) = 18x^2\]$$ So, the derivative of $$\(f(x) = 6x^3 + 2\)$$ with respect to x is $$\(f'(x) = 18x^2\)$$.