This is an exercise from Morris Kline's "Calculus: An Intuitive and Physical Approach":
The total cost $C$ of producing $x$ units of some item is a function of $x$. (Of course physically $x$ takes on only positive integral values, but it is convenient to think of it as taking on all real values in some domain.) Economists use the term marginal cost for the rate of change of $C$ with respect to $x$. Suppose that $C = 5x^2 + 15x + 200$, what is the marginal cost when x = 15? Would this marginal cost be the cost of the 16th unit?
Since the question asks for the rate of change at a specific value of $x$, we compute $C' = 10x + 15$. Then the marginal cost when $x = 15$ is equal to $165$.
I am still trying to understand how this would be the cost of the 16th unit though. Is it that since the rate of change of $C$ at $x = 15$ is 165, moving from 15 units to 16 units adds that rate to the overall cost? Similar to how if an object is moving at 165 mph, then in one hour it would travel 165 additional miles. Can this even be compared to an object in motion though? The object could easily change speeds within the hour, but there can be no fractional units produced.
Using the method of increments on the total cost formula yields \begin{align} C + \Delta C &= 5(x +\Delta x)^2 + 15(x + \Delta x) + 200 \\ &= 5x^2 + 10x \Delta x + 5(\Delta x)^2 + 15x + 15 \Delta x + 200 \\ \end{align}
Then \begin{align} \Delta C = C + \Delta C - C&= 10x \Delta x + 5(\Delta x )^2 + 15 \Delta x \\ \end{align}
Setting $x = 15$ and $ \Delta x = 1$, we have \begin{align} \Delta C = 10* 15 * 1 + 5 * 1^2 + 15* 1 = 170 \end{align}
Why is this value of $\Delta C$ not the cost of the 16th unit?
What you're calling the method of increments is the slope of the chord from 15 to 16, $(C(15+\Delta)-C(15))/\Delta $ with $\Delta=1$, and the derivative is what happens as $\Delta \rightarrow 0$, giving you the slope of the tangent line to the graph at 15. They're not going to be the same unless the function is constant.
When calculus gets used in economics it is usually just a story. It's not meant to be taken literally. Most results have "non-calculus" versions that come from linear programming or lattice arguments, but the arguments are more tedious and abstract, despite making more practical sense.