Find the derivative of the function $f(x)=(P(x))^a$ where $a\in\Bbb{R}$ and $P(x)$ is a real polynomial with no real roots I need help to clarify this statement. I found it in a friend's old homework notebook and I think that the answer is
$$f'(x)=aP'(x)\cdot(P(x))^{a-1}$$
However, I just don't get why the fact of $P(x)$ having no real roots is relevant. Is there a way to state the answer using only $P(x)$ instead of $P'(x)$? Am I missing something?
I think his teacher was trying to make this problem look more complex than it is, but I just wanted to check if maybe I am the one who is mistaken.
 A: If you want to have defined $P(x)^a$ for all $a$, you need $P(x) > 0$ for all $x$.
Let us omit the case $a = 0$ which leads to $P(x)^0 \equiv 1$. Okay, it may be questionable what $0^0$ is (occurring if $P(x)$ has real zeros), but either we take $0^0 = 1$ or let it undefined which gives us a function not defined on all of $\mathbb R$.
So let $a \ne 0$. It is of course no problem to define $y^a$ for

*

*all $y \in \mathbb R$ and all $a \in \mathbb N$,

*all $y \in \mathbb R \setminus \{0\}$ and all $a \in -\mathbb N = \{-1,-2,-3,\ldots\}$.

For $y > 0$ one usually defines $y^a = e^{a \ln y}$. This definition does not make sense for $y \le 0$. Okay, if $a > 0$ and $y = 0$ we may take $0^a = 0$, but for $a < 0$ there is no $0^a$. For $y < 0$ there is no general method to give a reasonable interpretation of $y^a$ in case $a \notin \mathbb Z$. You might argue that it works for $a = p/q$ with $q$ odd because we could take $y^a = \sqrt[q]{y^p}$, but this is only a very special case and the function $y^a$ is not differentiable at $y  = 0$ (which transfers to $P(x)^a$ if $P(x)$ has real zeros).
Therefore, if $p(x)$ has a (real) zero $x_0$, then $P(x_0)^a$ is undefined for $a < 0$. This is the reason for excluding polynomials $P(x)$ with real zeros. Polynomials without zeros are either positive or negative, and the negative case gives problems as shown above.
A: Unless you are allowed to extract the coefficients and modify them, there is no way to obtain $P'(x)$ from $P(x)$. Indeed, there is no functional relation $P'(x)=f(P(x))$ between them, as $P(x)$ and $P(x)+C$ have the same derivative.
