Derivations of Newton's laws? I feel convinced that the mathematics behind newtons laws can be derived from simpler priciples. The fact that displacement s can be described by a cartesian coordinate system with a parameter t and that the laws are constant with both ds and dt, we could somehow (though I'm not certain how) arrive to the definition that dE = Fds and dP = Fdt. F is a function applied to a closed system and E and P are constants. It's tempting to come to the conclusion that dE/dP = ds/dt which is a constant, since that can lead to P = c*ds/dt (or m) and all of the other mechanics laws. 
That last step seems like cheating, is it not possible to come to $F$ $\alpha$ $a$ using something else?
 A: My hint is too long for a comment, but maybe it is worth writing down. I'll not use symmetry arguments, but I'll try to get Newton's equations out of the non-degeneracy of a certain mathematical object.

Let $M$ be manifold and $\omega$ a 2-form on it. If $\omega$ is algebraically closed and non-degenerate, then the dimension of $M$ must be even (let's say $2n$) and, at least locally, $\omega = -d\alpha$. Now, since $\omega$ is non-degenerate, there's only one vector field $X_H$ associated to a differentiable function $H$ such that $\imath_{X_H} \omega = dH$, where $\imath$ stands for the inner product
\begin{align}
\imath: \Omega^n(M) &\to \Omega^{n-1}(M) \\
\imath_X \omega &\mapsto \omega(X) \, .
\end{align}
Consider now $n=1$ (generalization for any finite dimension is immediate), that is, the one degree of freedom case. Let's then express $X_H = a \partial_q + b \partial_p$, where $dq$ and $dp$ form a Darboux basis on $M$ (this means that, at least locally, $\omega = dq \wedge dp$). Then
\begin{align}
\imath_{X_H} \omega &= dH \\
\Leftrightarrow [dq \wedge dp] (a \partial_q + b \partial_p) &= dH \\
\Leftrightarrow a dp - bdq &= dH \\
\Leftrightarrow a dp - bdq &= \frac{\partial H}{\partial q} dq + \frac{\partial H}{\partial p} dp \\
\end{align}
which means that, 
\begin{align}
\begin{cases}
a = \partial_p H \\
b = -\partial_q H
\end{cases} \, .
\end{align}
If we consider $X_H$ to be the velocity field associated to $H$, that is, $a = \dot{q}$ and $b = \dot{p}$, then the non-degeneracy of $\omega$ leads unambiguously to Hamilton's equations. Starting from Hamilton's equations invariance under a certain group of transformations, called canonical, one can see that there is a certain integral that must be minimized along a classical allowed path (this can be done either in the Lagrangian or Hamiltonian formalisms). Using the inverse idea of d'Alemberts principle one can deduce Newton's second law and use it as a hint, together with some geometric considerations, to infere the first (if the acceleration, which is what changes momentum, is zero, then the momentum is constant in a certain class of reference frames, called inertial, which must exist if space is homogeneous). For a discussion about the third law, see P.S..
I believe what I've sketched is a hand-wavish way of getting to Newton's equations from pure mathematics. It is extremely artificial, since the area that gave birth to all Symplectic Geometry was Analytical Mechanics, which means each and every step I used in my "deduction" was historically done backwards.   
P.S.: In an earlier version of this answer I said that it might have been possible to infer the third law from the absence of a law that could explain the rest of Nature (as Newton did). Since this answer is about deducing all laws from mathematics, which is pretty much impossible, I've said that the second law hints at the other two, but even though the first law is sort of necessary to provide 0 acceleration a meaning, the third law may not be discerned from the other two (this is not my opinion). For a discussion on that read "the battle of the third law", which was fought in the comments to this answer with Timaeus, whom I thank dearly for the exposure of his views and many tries to clarify my misleading interpretation of the Laws. Readers have definitely enough material to choose a stance.
