Does $t^{p-1}$ have an antiderivative in $\Bbb F_q(t)$?

Question

The formal derivative on $\mathbb{F}_p[t]$ is not onto, since $t^{p-1}$ (for instance) has no antiderivative. Does it have one if we extend the formal derivative to $\mathbb{F}_p(t)$? By the quotient rule, this would be equivalent to finding out if there exist polynomials $f(t)$ and $g(t)$ with $f'(t)g(t)-f(t)g'(t)=t^{p-1}g(t)^2$. I'm guessing it's not possible but don't see how this could be shown.

Answer 1

$\Bbb{F}_p(t)$ is a $\Bbb{F}_p(t^p)$ vector space of dimension $p$, namely

$$\Bbb{F}_p(t)=\bigoplus_{n=0}^{p-1} t^n \Bbb{F}_p(t^p)$$ $( \Bbb{F}_p(t^p))'=0$ so $(t^n \Bbb{F}_p(t^p))' = nt^{n-1}\Bbb{F}_p(t^p)$ ie.

$$(\Bbb{F}_p(t))' = \bigoplus_{n=1}^{p-1} nt^{n-1} \Bbb{F}_p(t^p)= \bigoplus_{n=0}^{p-2} t^n \Bbb{F}_p(t^p)$$

Answer 2

HINT:

$D(t^p) = p t^{p-1}\cdot D t$, so in characteristic $p$, $t^p$ is a "constant". Therefore, if $D (\phi(t)) = t^{p-1}$, then $$D(\frac{\phi(t)}{t^p}) = \frac{1}{t}$$