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Derivative means slope of a function. My question is why the slope of a function (when it's a polynomial specifically) is always one degree less than the polynomial itself? I know the derivation of $(x^n)' = (nx^{n-1})$ and it is the following: For $f(x) = x^n$, we have

\begin{align*} f'(x) &= \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h} \\ &= \lim_{h\to 0} \dfrac{(x+h)^n-x^n}{h} \\ &= \lim_{h\to 0} \dfrac{(\binom{n}{0}x^{n}+\binom{n}{1}x^{n-1}h+\binom{n}{0}x^{n-2}h^2+...+\binom{n}{n}x^{0}h^n)-(x^n)}{h} \\ &= \ldots \\ &= nx^{n-1} \end{align*}

Here in the next line, $x^n$ gets cancelled in the numerator since there are two terms and now we're only remaining with terms with degrees lesser than n. Of these only $\binom{n}{1}x^{n-1}$ remains and everything else becomes 0.

But why intuitively, why does the derivative become one degree lesser? We can see it gets cancelled in the proof but what does that signify on the graph?

stange
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Shreyansh
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  • Maybe starting with $n=1$ will help you to understand. And then look at an induction process. – mathcounterexamples.net Nov 28 '23 at 11:03
  • @Shreyansh,If it helps,You can notice that derivative is change in polynomial per small change in $x$,Indeed Change in polynomial will have same degree as polynomial itself,But when you divide by small change in x,it reduces each term's power by $1$,Just like area has units $l^2$ and length has units $l$,When you divide the two, You get the unit $l$ – Dheeraj Gujrathi Nov 28 '23 at 13:05

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