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Mainly I am thinking about second degree polynomials.
For instance
$f(x)=2x^2$
$f'(x)=4x$
or
$f(x)=ax^n$
$f'(x)=a*n*x^{n-1}$
What I want to know is why we multiply n in front.

From my understanding the derivative is what x is increasing with so I would think that in
$x^2$
x is increasing with x per x because
$x^2=x*x$
and the derivative becomes just x. Like how in
$2x$
x is increasing with 2 per x because
$2x=2*x$
and the derivative becomes just 2

I am also deeply sorry if this way of thinking is making someone cringe, for instance like my maths teacher, but it makes sense to me.

I already know some proofs like how the area of a second degree polynomial derivative is a triangle and that's why you have to divide by two if you want to find the integral, but it just doesn't make much sense the other way around to me.
I can also probably think of some complex algebraic way of proving this with the definition of the derivative, but I just want a logical explanation of the correlation. Like how I showed with my stupid way of thinking.

Tuskies
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    This might be a symptom of the way calculus is taught. We multiply by $n$ because it’s precisely correct, from the definition of derivative. A proof of correctness comes from a simple study of the derivative limit. $(x+h)^n-x^n=nhx^{n-1}+O(h^2)$. – FShrike Sep 03 '22 at 10:15
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    Do you know the product rule? $(x^2)'=(xx)'=(x')x+x(x')=x+x$. You are dropping one of those terms, "breaking a symmetry" by giving a different interpretation to each factor $x$ in your $x^2=xx$. – Wouter Sep 03 '22 at 10:22
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    I don't think I have seen the product rule before, but I think I kind of understand what you mean by "giving a different interpretation to each factor x in your $x^2=x∗x$". I still don't see how that makes things different though. I don't understand algebra very well. – Tuskies Sep 03 '22 at 10:26
  • See Why do differentiation rules work? What's the intuition behind them? (Not asking for proofs). It specifically asks why the derivative of $x^2$ is $2x.$ There are several answers explaining why. Do any of them answer your question? – David K Sep 04 '22 at 15:17
  • Have you seen the idea that derivative is, in a sense, the ratio of two small changes? The change in the value of the function divided by the change in the value of the variable. When differentiating $x^2$ the formula encapsulates the idea that when $x$ changes a little, the value of $x^2=x\cdot x$ change for two reasons. The first factor changes, and the second factor changes. When we take both of those into account, we need to add the two ratios together. Similar thinking appears (further in your studies) in the product rule as well as the chain rule of several variable calculus. – Jyrki Lahtonen Sep 04 '22 at 16:23
  • @Davidk looks very interesting, will try to read it – Tuskies Sep 05 '22 at 19:08

2 Answers2

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Clearly, $$f(x)*g(x) = g(x)*f(x)$$ Thus, $$(f(x)*g(x))' = (g(x)*f(x))'$$ Thus, it cannot be the case that $$(f(x)*g(x))' = f(x)'*g(x)$$ Replace both $f(x)$ and $g(x)$ by $x$, and we see that it cannot be that $$(x*x)' = x'*x = x$$ It we don't break the symmetry between $f$ and $g$, and use the correct product rule, $$(f(x)*g(x))' = f(x)'*g(x) + f(x)*g(x)'$$ we see that $$(x*x)' = x'*x + x*x' = x + x = 2x$$

(ah, but why isn't the product rule $(f(x)*g(x))' = \frac{1}{2}f(x)'*g(x) + \frac{1}{2}f(x)*g(x)'$? That would also have the right symmetry!)

Wouter
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  • Sorry for being late to answer and I see that my question has also been marked as a duplicate now. Even though I don't quite understand why $(xx)'=x'x+x*x'$, this answer made the most sense to me and also kind of explains why for example $x^3=3x$. – Tuskies Sep 05 '22 at 18:49
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The answer is that the derivative of a function is defined as the gradient of the tangent to the function, which you can write as $$f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$$

When $f(x) = a x^n$, we can calculate this directly as:

$$\begin{eqnarray} f'(x) & = & \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} \\ & = & \lim_{h \rightarrow 0} \frac{a(x + h)^n - a x^n}{h} \\ & = & \lim_{h \rightarrow 0} a \frac{(x^n + n x^{n - 1} h + O(h^2)) - (x^n)}{h} \\ & = & \lim_{h \rightarrow 0} a \left(n x^{n - 1} + O(h) \right) \\ & = & a n x^{n - 1} \end{eqnarray}$$

where the $O(h)$ terms are multiples of $h$ and hence vanish in the limit.

In particular, with $f(x) = x^2$, notice that $(x + h)^2 - x^2 = (x + h - x)(x + h + x) = h(2x + h)$, so if you increase $x$ slightly, then $x^2$ increases proportional to $2x$ (plus a little bit more).

ConMan
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