Rate of surface increase of a sphere related to its volume increase I'm trying to solve problems in the Thompson's classic "Calculus made easy" (not so easy by the way).

One of this problems is to find the rate increase of the surface of a sphere which volume is increasing at $4$ square feet/second rate.

I've found a solution which unfortunatly doesn't agree with author's one.

My solution:

$r$ = radius

$v$ (sphere volume) = $\frac{4}{3}\pi r^3$

$s$ (sphere surface) = $4\pi r^2$

I "link" the volume with the surface

$$v = \frac{4}{3}\pi r^3 = 4\pi r^2 \frac{r}{3}= s \frac{r}{3}\ \Longrightarrow s = \frac{3v}{r}.$$

Now I apply the derivatives chaine rule

$$d\left[\frac{s}{t}\right] = d\left[\frac{s}{v}\right]\cdot d\left[\frac{v}{t}\right]
=d\left[\frac{3v}{r}\right]\cdot d\left[\frac{v}{t}\right]
= \frac{3}{r}\cdot 4
= \frac{12}{r}.$$

The author finds $\frac{8}{r}$, but I suspect a typographical error as an other site has found also my result.

Thank you very much to say me if I've made a mistake somewhere.
Using the natural notation, we have that since $S=4\pi r^2$, by the Chain Rule we get
$$\frac{dS}{dt}=8\pi r\frac{dr}{dt}\tag{1}$$
and similarly that 
$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.$$
Since $\frac{dV}{dt}=4$ (cubic feet per second), we have $\frac{dr}{dt}=\frac{4}{4\pi r^2}$.
Substitute in (1). We get $\frac{dS}{dt}=\frac{8}{r}$. 

Remark: Since we have nice formulas for $S$ and $V$, it is best to differentiate immediately. That is often the case in "related rates" problems.
Note: The answer given in Calculus Made Easy-Third Edition is 8/r.
While the answer given for this problem in the 1998 edition (with Martin Gardner's additions) is 8*(pi^.5). I believe 8/r to be correct.