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I know that the derivative of $x^2$ is $2x$ when derived classically, but I was thinking about the function $f(x)=x^2$ and I realized that you could potentially interpret it as $f(x)=x*x$, by which I mean classify the first occurrence of $x$ as a sort of slope variable, just like one would in $y=mx+n$.

At first glance, this seems reasonable. In it's essence, a quadratic equation is just a line with a linearly increasing slope after all. What's bothering me is that when I approach it in this way, I get $x$ as the derivative (since the derivative of $y=mx+n$ is simply $m$) instead of $2x$.

I know that this confusion almost definitely stems from a naive application of logic, but I'm having trouble formalizing what exactly prevents it from behaving this way. Can someone explain to me why one cannot think about the situation in this way?

Edit: Maybe I should have been more clear about what I want: Like I said above, I understand that the derivative is $2x$ when classically derived, be it through the power rule or through the full differential quotient. What I want to know is an intuitive reason why one cannot look at the problem through the lense of $y=mx+n$. I'll ask differently: Someone who doesn't know anything about differentiating would probably assume the slope to be x, because of the similarites with $y=mx+n$, as mentioned above. How would one explain to him intuitively, without actually teaching him differentiation, why his thought process doesn't work?

Rodrigo de Azevedo
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Nik Tedig
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    In $y=mx+n$, $m$ is constant. In $y=x\cdot x+n$, $m=x$ is not constant. This makes talking about the gradient as $m$ impossible. How do you even define gradient? Well, rise/run of course... which leads to $2x$ as the derivative – FShrike Jun 04 '22 at 09:33
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    There’s a theorem that says “if there is a real number $m$ such that $f(x) = mx$ for all $x \in \mathbb R$ then $f’(x) = m$.” The function $f(x) = x^2$ does not satisfy the hypothesis of that theorem. – littleO Jun 04 '22 at 09:37
  • I recommend this video (actually the entire playlist) to help with intuition about the derivative. It clearly shows why $\frac{d}{dx} \bigl( x^2 \bigr) = 2x$. – Sammy Black Jun 04 '22 at 09:40
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    I don't think anyone is correctly addressing the question that is asked. – JP McCarthy Jun 04 '22 at 09:41
  • @JPMcCarthy I think FShrike adresses to the question and gives a perfect answer where the flaw is. – Peter Jun 04 '22 at 09:57

4 Answers4

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If you think about $x^2$ as $mx+n$, your $m$ is actually a function of $x$: you have $f(x) = m(x)x+n$ with $m(x)=x$ and $n=0$. Hence the derivative is given by $f’(x)= m’(x)x+m(x)x’ = x’x+xx’ = x+x = 2x$.

Michaël
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Following your line of argument, write $y=x^2$ in the form $$y=\color{red}{x}\color{green}{x}$$

Then $\color{red}{x}$ is the ‘gradient” of $\color {green}{x}$ but also $\color{green}{x}$ is the ‘gradient’ of $\color{red}{x}$.

So for every unit increase in the value of $\color{green}{x}$, $y$ increases by amount $\color{red}{x}$, and for every unit increase in the value of $\color{red}{x}$, $y$ increases by amount $\color{green}{x}$.

This makes the total increase in $y$ for every unit of $x$ equal to $\color{red}{x}+\color{green}{x}=2x$ as expected.

David Quinn
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I think that you are thinking of the wrong rule as "primary". The "constant multiplier" rule is only taught first because it is easier to learn, not because it is more fundamental. If you think of the product rule instead, then the situation becomes a lot more understandable.

In the product rule, let's say you had two variables, $a$ and $b$. The differential of their product is like this:

$$ d(a b) = a\, db + b \, da$$

(NOTE - If you aren't familiar with differentials, a differential is a derivative before dividing by $dx$ or whatever variable the derivative is being taken "with respect to")

In the case where they are both $x$, this becomes:

$$ d(x x) = x, dx + x \, dx = 2x\,dx $$

In the case where one is a constant (we'll say 5 to be concrete) and the other is $x$, this becomes:

$$ d(5x) = 5\,dx + x\,d5 $$

However, the differential of a constant is always zero (because a constant isn't changing at all). Therefore, we can further expand this to say:

$$ d(5x) = 5\,dx + x\,d5 = 5\,dx + x\cdot0 = 5\,dx + 0 = 5\,dx$$

So, the constant multiplier rule is just a special case of the more general product rule when one of the terms is a constant, because the differential of a constant wipes out half of the result.

johnnyb
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I'm very late, but I also had this question, and I finally had an epiphany yesterday, so I figured that I should share it.

Rate of change is the amount of change in y occuring after a change in x. Hence, let's look at how it compares to when multiplier is a constant. For the sake of convinience, let change be +k.

y=mx

y=m(x+k)

y=mx + mk

x is essentially amount of m in y, and value changed in x adds that value of m into y. While if m=x, then it itself changes alongside it

y=mx

y=(m+k)(x+k)

It doesn't simply turn m into current x and then add it by changed amount, which a derivative x would suggest, it warps the m of previous product as well.

y=(m+k)x + k(m+k)

So what is the other value that's been warped in? If previous value of product was mx, then:

(m+k)x=mx+xk

A previous x added by amount of change. So, in other words, the changed amount added isn't simply of current x, but also of previous x. Let's rephrase x as being current

k(x+x-k)

Let k be dx. Since derivative is amount of change in y per change in x, we get

y/dx=2x

TL;DR: the difference made in y with change of x is actually sum of current x and x before change, because if x changes m it doesn't just add a new amount, but warps one before change as well. (in fact, with k(x+x-k) you can calculate difference between squared values of any 2 values differing by k amount, simply by plugging in bigger one in place of x)

Null
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