Proof that $\frac{d}{dx}x^n = n x^{n-1}$ for positive integers

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Proof Builder

The proof has one decisive structural idea: after the binomial expansion, only the linear-in- $h$ term remains in the limit. Proof Builder keeps that sequence of moves visible instead of hiding it inside one compressed algebra block.

Where the binomial theorem enters

The derivative definition for $x^{n}$ produces the quotient $\frac{( x + h ) ^{n} - x ^{n}}{h}$ . The main obstacle is expanding $(x + h)^{n}$ in a useful way.

The binomial theorem does exactly what the proof needs: it separates one linear term $n x^{n - 1} h$ from the higher-order terms that contain $h^{2}, h^{3},$ and beyond.

Why only one term survives

After subtracting $x^{n}$ and dividing by $h$ , the linear term becomes $n x^{n - 1}$ and every other term still contains a positive power of $h$ .

That is the whole proof. As $h \to 0$ , every term with a remaining factor of $h$ vanishes, leaving only $n x^{n - 1}$ .

Why this theorem matters later

Students often meet the power rule as a memorized shortcut. This proof shows that the shortcut is justified by the derivative definition itself.

Once this argument is clear, later derivative rules feel less arbitrary because they can be read as compressed versions of first-principles reasoning.

It also explains why first-principles examples for $x^{2}$ , $x^{3}$ , and $x^{4}$ all keep showing the same cancellation pattern. The binomial theorem is the general statement behind that repetition.

Common Pitfall

The higher-order terms do not disappear by magic. They disappear because after dividing by $h$ , each of them still contains a positive power of $h$ , so each one tends to $0$ as $h \to 0$ .

Try a Variation

Try the same first-principles method on $f (x) = x^{4}$ . Can you see the pattern without writing the full general summation first?

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