If $gcd(f(x),`f(x))=1$ then f(x) is square free. But what is the reason behind taking derivative of f(x)? How one came to this conclusion?
2026-04-01 21:59:23.1775080763
Significance of derivative in finding square free decomposition
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If $f(x)=g(x)^2h(x)$ then by the product rule $$f'(x)=2g(x)g'(x)h(x)+g(x)^2h'(x)=g(x)\cdot(2g'(x)h(x)+g(x)h'(x))$$ so that $g(x)$ is a common factor of $f$ and $f'$