Rational function $(p/q)'=0$ implies $p'=0$ and $q'=0$ over an arbitrary field Lawrence Washington claims this in exercise 2.22 of his book Elliptic Curves - Number Theory and Cryptography (second edition). The precise formulation is:

Let p(x) and q(x) be polynomials with no common roots. Show that
  $$\frac{d}{dx}
\frac{p(x)}{
q(x)}=0$$
(that is, the identically $0$ rational function) if and only if both $p'
(x)=0$
  and $q'
(x) = 0$.
  (If p or q is nonconstant, then this can happen only in
  positive characteristic.)

His remark in the end remark makes it conceivable that this statement is meant to hold over an arbitrary field, the differentiation being formally the same as over the complex numbers.
Now setting $k=\mathbb{F}_3$ and taking $p(x)=0$ and $q(x)=x^2+1$ apparently gives a counterexample.
My question is: Is there an obvious way to change this statement making it correct?
 A: There are a few defects here.  As written,


*

*$q(x) = 0$ is permitted, for which $\frac{p(x)}{0}$ is undefined, so the formal process $$  \frac{\mathrm{d}}{\mathrm{d}x} \frac{p(x)}{0} = \frac{p'(x) 0 - p(x) 0}{0^2} = \frac{0}{0}  \text{,}  $$ undefined throughout, does not produce a rational function because its denominator is zero.

*With $p(x) = 0$ and any polynomial, $q(x)$, we have 
$$  \frac{\mathrm{d}}{\mathrm{d}x} \frac{0}{q(x)} = \frac{0 q(x) - 0 q'(x)}{q^2(x)} = 0  \text{.}  $$
For $\frac{p'(x) q(x) - p(x) q'(x)}{q^2(x)}$ to be a rational function, $q(x) \neq 0$.  Then, generically, for 
$$  \frac{p'(x) q(x) - p(x) q'(x)}{q^2(x)} = 0  \text{,}  $$
either 


*

*$p'(x) = q'(x) = 0$ or 

*$q(x) \neq 0$ and $p'(x) = \frac{q'(x)}{q(x)} p(x)$.


For the second case, since $p$ and $q$ are relatively prime, it must be that either $p = p' = 0$ or $q \mid q'$.  If $p \neq 0$ and $q$ is a nonconstant polynomial, $\deg q' < \deg q$, so $q \not \mid q'$.  Therefore, if $p \neq 0$, $q$ must be a constant polynomial and $\frac{q'}{q} = 0$.  This forces $p' = 0$, so $p$ is a constant polynomial.  Therefore, if $p \neq 0$, the second case forces $p$ and $q$ to be constants.
We can rescue the statement by changing the first sentence to "Let $p(x)$ and $q(x)$ be nonzero polynomials with no common roots."  (Here, we require $q(x)$ nonzero so that the result is actually a rational function and $p(x)$ nonzero to force $q$ and then $p$ constant in the second case above.)
