Question about polynomial functions and derivative

Let p,q be two polynomial functions. Show that the function:

1/x

cannot be the derivative

f(x)=p(x)/q(x)

Please, do not use integral operator, thanks :)

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I think that I can work on the polunomials degree but I don't know how.

Suppose that there exist (nonzero) polynomials p(x), q(x) such that
(d/dx) p(x)/q(x) = 1/x.

Expanding, we have
[p'(x) q(x) - p(x) q'(x)] / [q(x)]^2 = 1/x.
==> x [p'(x) q(x) - p(x) q'(x)] = [q(x)]^2.

Now, let deg p(x) = n and deg q(x) = m.

[We can assume that m, n > 1, since otherwise we can get quick contradictions.

Plainly, q(x) can not be constant, because the derivative of a polynomial is also a polynomial.
Moreover, p(x) can not be constant, because otherwise (say p(x) = c) we then have
-cx q'(x) = [q(x)]^2. Comparing degrees, we have m - 1 = 2m ==> m = -1, which is not possible for the degree of a polynomial.]
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Now, we compute degrees.
deg [p'(x) q(x) - p(x) q'(x)] = m + n - 1.
==> deg (x[p'(x) q(x) - p(x) q'(x)]) = m + n.

deg (q(x))^2 = 2m.

Hence, m + n = 2m ==> m = n.
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So, we can write (for some constants a(i) and b(i))
p(x) = a(n) x^n + a(n-1) x^(n-1) + ... + a(0) and p(x) = b(n) x^n + b(n-1) x^(n-1) + ... + b(0).

Substituting this into p'(x) q(x) - p(x) q'(x), we find that the highest degree terms (2n-1) cancel each other out (try writing these terms out).

Hence, deg[p'(x) q(x) - p(x) q'(x)] < 2n - 1.
==> deg(x [p'(x) q(x) - p(x) q'(x)]) < 2n, which is a contradiction.
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So, no such rational function exists.

I hope this helps!