Derivation of the rocket equation I have read many different rocket equation derivations but I do not understand them. Here is my derivation:
At time $t$, assume we have an accelerating rocket with mass $m$ and velocity $v$. At time $t+\delta t$, the rocket would have ejected post-reaction propellant with mass $\delta m$ at exhaust velocity $v_e$. The rocket would have mass $m-\delta m$ and velocity $v+\delta v$. By the conservation of momentum from an observer's reference frame, we have
$$mv = (m-dm)(v+dv)+\delta m(v-v_e) \qquad\qquad \text{right is positive}$$
Note: Most of the derivations I have read use $(v+v_e)$ instead, but isn't the exhaust velocity in the opposite direction of the rocket? 
Expanding and simplifying leaves us with
$$m\delta v-v_e\delta m-\delta m\delta v=0$$
We can ignore higher order terms.
$$\delta v = v_e \frac{\delta m}{m}$$
Integrating gives
$$\Delta v = v_e ln{\frac{m_f}{m_0}}$$
I do not understand where I went wrong. I have read posts saying that $m -\delta m $ should be $m+\delta m$ but then the other $\delta m$ term would be affected too. 
 A: There is a subtle conceptual error in your derivation regarding the differantials; So let's review the problem: At the start the mass in our reference frame consists solely of the rocket and is described by a function $m(t)$. Now after the time $\mathrm{d}t$ has elapsed, a propellant particle with mass $\mathrm{d}m_e$ got ejected. So the total mass in our reference frame is $m(t + \mathrm{d}t) + \mathrm{d}m_e$. We may expand the function $m(t + \mathrm{d}t)$ and apply the principle of mass conservation:
$$m(t) = m(t + \mathrm{d}t) + \mathrm{d}m_e = m(t) + \mathrm{d}m + \mathrm{d}m_e.$$
From here it follows that the infinitesimal small mass $\mathrm{d}m$ has the opposite sign of $\mathrm{d}m_e$, otherwise the above equation couldn't be true. With this knowledge, you can correct the momentum equation (remember the principle of momentum tells us that the change of momentum equals the resulting force $R$):
\begin{align} R \, \mathrm{d}t & = I(t + \mathrm{d}t) - I(t) = (m(t) + \mathrm{d}m)(v(t) + \mathrm{d}v) - m(t)v(t) + \mathrm{d}m_e v_e \\
& = m(t) \, \mathrm{d}v + \mathrm{d}m \,v(t) + \mathrm{d}m_e v_e + \mathrm{d}m\,\mathrm{d}v \\
& = m(t) \, \mathrm{d}v + \mathrm{d}m \,v(t) - \mathrm{d}m v_e + \mathrm{d}m\,\mathrm{d}v.
\end{align}
Now dividing by $\mathrm{d}t$ and doing some algebraic manipulation gives you
$$m \frac{\mathrm{d}v}{\mathrm{d}t} = (v_e - v) \frac{\mathrm{d}m}{\mathrm{d}t} + R$$
Solving this differential equation for $R = 0$ and $v_e - v = \Delta v$ should result in a correct expression for $v(t)$. I hope I could clarify your issues (including why $\mathrm{d}m =\mathrm{d}m_{\text{rocket}} = -\mathrm{d}m_e$), the point is you should always be careful when it comes to expansion of functions.
A: Using $m+\delta m$ in the initial momentum will not change your result, if you change the right side of the equation accordingly. The result you got the wrong sign is  because when you integrate, you need to integrate the mass of the rocket, so for the rocket, so your $\delta m_{exhaust}=-\delta m_{rocket}$. See here.
