# Using the chain rule backwards

I'm asking about a use of the chain rule that I've seen in a couple of derivations but that I don't understand, I hoping for it to be clarified.

Let's say we start with the gravitational acceleration,

$$\frac{dv}{dt} = -\frac{GM}{R^2}$$

The left-hand side of the equation can be turned easily into, $$\frac{dv}{dr} \frac{dr}{dt} = \frac{dv}{dr} v.$$ This is all fine but then the derivation that I'm looking through jumps to the following,

$$\frac{dv}{dr} v = \frac{d}{dr}\left(\frac12v^2\right).$$

I can work that step out backwards by recognizing that the derivative of $v^2/2$ gives back $v$ but I don't understand how you go from $\frac{dv}{dr} v$ to $\frac{d}{dr} (\frac12v^2)$.

• d/dr (v^2/2) = v * dv/dr. Where the first factor v is the outer derivative and the second factor dv/dr are due to the chain rules. Just like one have for d/dx f(x)^2 = 2*f(x)*df/dx. Another way to see this is d/dr v^2/2 = 1/2 d/dr v * v = 1/2(dv/dr * v+v * dv/dr) = v * dv/dr. Where we used the product rule. – Natanael Jan 9 '15 at 21:44
• Recognizing that (dv/dr)*v = d/dr(v^2/2) is common in math and physics. If we want to express (dv/dr)*v as just the derivative of some function, we can solve a differential equation (dv/dr)*v = df/dr. Thus df/dv = v and f = v^2/2. – nphirning Jan 9 '15 at 21:47
• Those could be answers. – jinawee Jan 9 '15 at 22:06

## 2 Answers

The answer to the question "given $g(x)$, what is the function $F(x)$ so that $\frac{dF}{dx} = g$" is commonly known as the integral of $g$.

If you look closely at your expression you will see that is exactly what is happening here. Unfortunately, while there are some tricks in integration, ultimately most of the come down to pattern recognition of one kind or another.

• Yep, or +1. There are some things you just have to know on sight to be able to see them for what they are. Compare with how wolfram alpha.com stupidly evaluates $\int_0^x \exp(-t)dt$. As of the writing of this comment, the result is $\sinh x - \cosh x + 1$. Seriously? This is technically correct, but seriously? What's wrong with $1-\exp (-x$)? – David Hammen Jan 10 '15 at 1:38
• @DavidHammen you can leave feedback and they do act on it - I did it three times, and they fixed all three. – Floris Jan 10 '15 at 1:41

Follow these steps for direct integration of the gravitational acceleration.

\begin{aligned} \frac{{\rm d}u}{{\rm d}r}v &= -\frac{G M}{r^2}\\ \int \frac{{\rm d}u}{{\rm d}r}v\,{\rm d}r=\int v\,{\rm d}v &= -\int \frac{G M}{r^2}\,{\rm d}r \\ \frac{1}{2} v^2 &= -\int \frac{G M}{r^2}\,{\rm d}r \end{aligned}