Is there an alternative way (instead of taking derivative) to prove that a function is monotonic? We know that to prove that a function is monotonic (increasing or decreasing), a very strong method is using the first derivative. My question is that if we have a very complicated function, is there an alternative way to prove that a function is monotonic?
 A: As alternative way may be this helps: we know, that function $f$ is (strictly) concave on an interval if and only if its derivative function $f^{'} = g$ is (strictly) monotonically decreasing. So, if you take integral from your function and check if it is (strictly) concave, then you'll have answer.
A: Try to check if the function has any extrema within the region of interest, not on the boundary. This can be done using some derivative-free approach (e.g., Nelder-Mead). If there are any minima or maxima, then the function is not monotone.
Alternatively, you can try to solve numerically the problems $f(x_1,\dotsc,x_n)=z$ for different values of $z$ within the range and with different initial guesses for  $x_1,\dotsc,x_n$. If there will be more than one solution for some $z$, this would indicate that the function bends somewhere.
Obviously, these methods cannot guarantee that your function is monotone indeed, but they can be performed in an automated way for very general forms of the function.
A: As noted in the comments, you haven't given us an actual function. For example, we could be talking about singular functions such as Cantor function that is locally constant almost everywhere (derivative is zero for all $x \in [0,1]$ save for a countable subset).
Given that, you have to go back to basics:
Definition:
A function $f$ is monotonic when $x\leq y \implies f(x) \leq f(y)$
So, rather unhelpfully, if you prove this then voila, it's monotonic.
If you know the actual function, you can get more specific.
For example, you may be able to derive a difference where it is easy to prove the function is always non-negative (ok, kinda cheating, but not strictly a derivative):
$$f(x+\delta) - f(x) \geq 0 \;\forall \delta \geq 0$$
This may be no easier than simply proving it meets the definition.
As with a lot of things, the devil is in the details.
