Yes. In fact, there are an infinite number of Calculus!
Define the generalized mean of a set of number by,
$$(1) \quad \mu_{F}=F^{-1}\left({1 \over N} \cdot \sum_{k=1}^{N} F(x_k)\right)$$
Where $F(x_k)$ is a function over the number. Check that $F(x)=x$ yields the arithmetic mean, $F(x)=\log(x)$ yields the geometric mean, and $F(x)={1 \over x}$ yields the harmonic mean. To define the derivative and integral, we just need a moment of insight to see how manipulating $(1)$ yields standard calculus, given a proper $F$. We find that,
$$(2) \quad {{df} \over {dx}}_F=\lim_{\epsilon \to 0^+} \ F^{-1} \left( {1 \over {\epsilon}} \cdot \left[F(f(x+\epsilon)-F(f(x)) \right] \right)$$
and
$$(3) \quad \int_a^b f(x) \ dx_F=\lim_{N \to {\infty}} \ F^{-1}\left({{b-a} \over N} \cdot \sum_{k=1}^{N} F\left(f\left(a+(b-a) \cdot {k \over N}\right)\right)\right)$$
Where $(2)$ is the $F$ derivative and $(3)$ is the $F$ integral. Setting $F(x)=x$ yields standard calculus, $F(x)=\log(x)$ yields product calculus, and $F(x)={1 \over x}$ yields harmonic calculus. At this point speaking of "interpretations" for $(2)$ and $(3)$ given an arbitrary $F(x)$ is naïve. It's best to think in terms of the way the derivative approximates a $f(x)$ around a point. We can manipulate $(2)$ into,
$$(4) \quad f(x+dx)=F^{-1} \left(F(f(x))+F \left({{df} \over {dx}} \right) \cdot dx \right)$$
Once again, regular calculus yields the best linear approximation, product calculus yields the best exponential approximation, and harmonic calculus yields the best rational function approximation.
I should reiterate that this formalism creates an infinite number of calculus, each with their own unique properties and uses.