# 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.

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

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

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 $g$.
• 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
\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}