Chain Rule

It is very common in physics to have accelerations given as functions of variables other than time, like position or velocity. This is because accelerations are related to force by Newton's Law $\vec{F}=m\vec{a}$, and most forces are most naturally expressed as functions of position - gravity gets weaker the farther you get from a massive body, the tension in a spring increase as you pull it away from equilibrium. And sometimes we know forces as functions of velocity as in wind resistance - the faster you go the harder the wind pushes against you.

In each of these cases we do not have acceleration as a function of time, and thus we need to know how to handle these situations. And, as in most cases, we can write down the position dependence easily, but writing down the time-dependence will be very difficult or even impossible (in closed form - we can always make numerical approximations).

So, say we're given acceleration as a function of position $a=a(x)$ and we want to find the velocity $v$. We have the relationship

$$a = \frac{dv}{dt}$$

so we may attempt to rearrange this equation and solve:

$$\int dv=\int{a(x)dt}$$

but we cannot do this because we do not know $a$ as a function of $t$ - this expression cannot be integrated!

In order to make progress we have to deal with the dependence $a$ has on $x$. What we notice is that if we can express $a$ as a function of $x$, then naturally we can express $v$ as a function of $x$ as well. Thus we have:

$$a(x) = \frac{d}{dt}\left(v\left(x\right)\right)$$

and this is where the chain rule comes in. We need to take the time derivative, but we have to go through $x$ first. And this is perfectly acceptable as $x$ is definitely a function of time $t$.

So, by the definition of derivative for any function $f$ we have:

$$\frac{df}{dt} = \lim_{\Delta t \to 0}\frac{\Delta f}{\Delta t} = \lim_{\Delta t \to 0}\frac{f(x(t+\Delta t))-f(x(t))}{\Delta t}$$

But we have the relation

$$\Delta x = x(t+\Delta t)-x(t)$$

So that we can rewrite our derivative as

$$\frac{df}{dt}=\lim_{\Delta t \to 0}\frac{f(x+\Delta x))-f(x)}{\Delta t}$$

And if we multiply by 1= $\Delta x / \Delta x$ we get

$$\frac{df}{dt}=\lim_{\Delta t \to 0}\frac{f(x+\Delta x))-f(x)}{\Delta x}\frac{\Delta x}{\Delta t}$$

So that the first fraction looks like the limit definition of the derivative of $f$ with respect to $x$, while the second fraction is the derivative of $x$ with respect to $t$. And since $\Delta x \to 0$ as $\Delta t \to 0$, this is precisely what we have, and the final result is

$$\frac{df}{dt}=\frac{df}{dx}\frac{dx}{dt}$$

Now, in the current case we have $f=v$, and we can identify $dx/dt$ as the velocity $v$ of the particle so that we have

$$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$

and now we can certainly separate our variables and integrate for whatever particular $a(x)$ we have:

$$\int a(x)dx=\int vdv$$

Note, this formula is particularly easy to remember because you can just think of putting in a $dx/dx$ into the normal derivative formula. This can also be extended to functions of several variables in a straightforward way, but that would lead us too far astray.