Big-O Notation and Calculus?
an algorithm linear according to Big-O notation reduces the size of the problem by a constant amount at each step


I don't think that's really true. It seems to me that all you're doing here is observing that a discrete linear function changes by a constant amount at each step and wondering if that's connected to the fact that the derivative of a continuous linear function is constant. That doesn't have much to do with what linear-time algorithms necessarily do.

Since algorithms are intrinsically discrete and calculus is intrinsically about continuous properties, I'm not aware of any real relationships between them. One might occasionally use calculus, e.g., to bound a sum by an integral or in tricks like
$$
\begin{align*}
\sum_{i=1}^k in^i &amp;= \sum_{i=1}^k (i+1)n^i - \sum_{i=1}^k n^i\\
 &amp;= \sum_{i=1}^k \frac{\mathrm{d}}{\mathrm{d}n}n^{i+1} - \sum_{i=1}^k n^i\\
 &amp;= \frac{\mathrm{d}}{\mathrm{d}n}\sum_{i=1}^k n^{i+1} - \sum_{i=1}^k n^i\,,
\end{align*}$$
but I doubt that's the sort of application you had in mind.