# 1st order derivative of x²

Gold Member

## Main Question or Discussion Point

d/dx (x²) = 2x ... eq 1
now we can write 2² = (2+2)
3² = (3+3+3)
4²=(4+4+4+4)
.
.
.
n² = (n+n+n+n+.........)n times
so here d/dx (x²) = d/dx (x+x+x+.......)x times
so ⇒ d/dx (x) +d/dx(x) +.......(x times) = 1+1+1+.....(x times) = x
⇒d/dx (x²) = x
where did I go wrong??

berkeman
Mentor
d/dx (x²) = 2x ... eq 1
now we can write 2² = (2+2)
3² = (3+3+3)
4²=(4+4+4+4)
.
.
.
n² = (n+n+n+n+.........)n times
so here d/dx (x²) = d/dx (x+x+x+.......)x times
so ⇒ d/dx (x) +d/dx(x) +.......(x times) = 1+1+1+.....(x times) = x
⇒d/dx (x²) = x
where did I go wrong??
When you wrote 2² = (2+2). 2² = (2*2)

Gold Member
When you wrote 2² = (2+2). 2² = (2*2)
i never said that! Did I?

berkeman
Mentor
You said this in your original post (OP):

now we can write 2² = (2+2)
Which I corrected to 2*2...

Mark44
Mentor
When you wrote 2² = (2+2). 2² = (2*2)
But it's also true that 22 = 2 + 2, and that 32 = 3 + 3 + 3

berkeman
Mentor
But it's also true that 22 = 2 + 2, and that 32 = 3 + 3 + 3
Oh, huh. Yeah, I guess it is. Okay, never mind...

Mark44
Mentor
It's true that 22 = 2 + 2, and 32 = 3 + 3 + 3, but what do you do with an expression such as 1.12? Can you have 1.1 terms being added together? That makes no sense.

The right way to do this is to recognize that x2 = x * x. Now d/dx(x2) = 2x, and d/dx(x * x) = x * 1 + 1 * x = 2x, using the product rule in differentiating.

mathman
d/dx (x²) = 2x ... eq 1
now we can write 2² = (2+2)
3² = (3+3+3)
4²=(4+4+4+4)
.
.
.
n² = (n+n+n+n+.........)n times
so here d/dx (x²) = d/dx (x+x+x+.......)x times
so ⇒ d/dx (x) +d/dx(x) +.......(x times) = 1+1+1+.....(x times) = x
⇒d/dx (x²) = x
where did I go wrong??
d/dx (x+x+x+.......)x times needs to take into account x times, which you did not.

Fredrik
Staff Emeritus
Gold Member
where did I go wrong??
By definition of derivative, we have
$$\frac{d}{dx}x^2 =\lim_{h\to 0}\frac{(x+h)^2-x^2}{h}.$$ That $x$ may be an integer, but the value of the expression on the right is completely determined by the values of $\frac{(x+h)^2-x^2}{h}$ for non-zero values of $h$ in the interval $(-1,1)$ (or any other open interval that contains $0$). So for all relevant values of $h$, the sum $x+h$ isn't an integer.

d/dx (x+x+x+.......)x times needs to take into account x times, which you did not.
My first thought was that he did, by writing $1+1+\cdots+1=x$. But you're right. We're differentiating with respect to x, so we can't treat any x in the expression as a constant.
\begin{align}
&\frac{d}{dx}x^2 =\lim_{h\to 0}\frac{ (x+h+x+h+\cdots)_{x+h\text{ times}}-(x+x+\cdots)_{x\text{ times}}}{h} =\lim_{h\to 0}\frac{(x+x+\cdots)_{h\text{ times}}+(h+h+\cdots)_{x+h\text{ times}}}{h}\\
&\lim_{h\to 0}\frac{hx+(x+h)h}{h} = x+\lim_{h\to 0}(x+h)= x+x=2x.
\end{align}
Edit: I should add that this calculation doesn't really make sense. Specifically, the notations "x+h times" and "h times" don't make sense, since h isn't an integer. This calculation is at best a reason to suspect that $\frac{d}{dx}x^2=2x$.

Last edited:
Gold Member
Thank you , all for those replies, thank you Fredrick for that first principle method, by my question still stands,I understand that by first principle and using x²=(x+x+x+...+x)x times we still get 2x but then why not by the common method of directly applying the operator. Is there some rule I'm violating?
PS. I'm sorry if you already mentioned it and i missed it!

Fredrik
Staff Emeritus
Gold Member
Thank you , all for those replies, thank you Fredrick for that first principle method, by my question still stands,I understand that by first principle and using x²=(x+x+x+...+x)x times we still get 2x but then why not by the common method of directly applying the operator. Is there some rule I'm violating?
PS. I'm sorry if you already mentioned it and i missed it!
The main problem was identified by mathman. You didn't correctly account for the last x in $(x+x+\cdots)_{x\text{ times}}$. In other words, what you did isn't equivalent to defining f by $f(x)=(x+x+\cdots)_{x\text{ times}}$ and then computing $f'(x)$. It's equivalent to defining $f(x,y)=(x+x+\cdots)_{y\text{ times}}$ and then computing the partial derivative $D_1f(x,x)$.

Gold Member
oh ok i understood.! Thank you all.

Intrastellar
Gold Member
By definition of derivative, we have
$$\frac{d}{dx}x^2 =\lim_{h\to 0}\frac{(x+h)^2-x^2}{h}.$$ That $x$ may be an integer, but the value of the expression on the right is completely determined by the values of $\frac{(x+h)^2-x^2}{h}$ for non-zero values of $h$ in the interval $(-1,1)$ (or any other open interval that contains $0$). So for all relevant values of $h$, the sum $x+h$ isn't an integer.

My first thought was that he did, by writing $1+1+\cdots+1=x$. But you're right. We're differentiating with respect to x, so we can't treat any x in the expression as a constant.
\begin{align}
&\frac{d}{dx}x^2 =\lim_{h\to 0}\frac{ (x+h+x+h+\cdots)_{x+h\text{ times}}-(x+x+\cdots)_{x\text{ times}}}{h} =\lim_{h\to 0}\frac{(x+x+\cdots)_{h\text{ times}}+(h+h+\cdots)_{x+h\text{ times}}}{h}\\
&\lim_{h\to 0}\frac{hx+(x+h)h}{h} = x+\lim_{h\to 0}(x+h)= x+x=2x.
\end{align}
Is there a way to find
\begin{align}
\frac{d}{dx} \sum_{1}^{x} f(x)
\end{align}
without calculating the sum explicitly ?

Last edited:
Fredrik
Staff Emeritus
Gold Member
Is there a way to find
\begin{align}
\frac{d}{dx} \sum_{1}^{x} f(x)
\end{align}
without calculating the sum explicitly ?
You'd have to define the notation before you can start thinking about how to find the number it represents.

Intrastellar
Gold Member
You'd have to define the notation before you can start thinking about how to find the number it represents.
I realised that shortly after I made my post.

Surely one cannot differentiate factorials ? They have values only when $x$ is an integer, continuous does not even make sense there when you try to plot it.
However, one can extend the definition of factorials to the gamma function, which one can actually differentiate.

I suppose for my question to make sense, I have to first ask whether there is a way to extend the definition of summation to non-integer values of $x$ ?
Then my question becomes whether one can apply some sort of a chain rule so that one can evaluate the derivative of the extended summation without needing to evaluate the summation first.

If I understood the OP's question correctly, he has the following:

\begin{align}
\sum_{k=1}^{x} x = x^2
\end{align}
But when he tries to calculate the derivatives of both sides, he doesn't get matching answers. I presume that is because there is some hidden chain rule which he is not applying ?

My guess would be that when he differentiates $(x+x+x+....)_x$, he has to apply the chain rule this way:
\begin{align}
\frac{d}{dx} (x+x+x+....)_x = (\frac{d}{dx}[x+x+x+....] )_x + (x+x+x+....)_\frac{d(x)}{dx}
\end{align}
\begin{align}
&= (1+1+1+...)_x + (x+x+x+...)_1
\end{align}
\begin{align}
&= x + x
\end{align}
\begin{align}
&=2x
\end{align}
Is that correct ?

Last edited:
Gold Member
My guess would be that when he differentiates $(x+x+x+....)_x$, he has to apply the chain rule this way:
\begin{align}
\frac{d}{dx} (x+x+x+....)_x = (\frac{d}{dx}[x+x+x+....] )_x + (x+x+x+....)_\frac{d(x)}{dx}
\end{align}
\begin{align}
What did you do here Montadhar?
I didn't understand the way you applied the operator (d/dx)..
Is this some other rule, that i'm not aware of??

Homework Helper
in "so here d/dx (x²) = d/dx (x+x+x+.......)x times" there is no guarantee that x is an integer, so "x times" is ill defined.

Intrastellar
Gold Member
What did you do here Montadhar?
I didn't understand the way you applied the operator (d/dx)..
Is this some other rule, that i'm not aware of??
Do you mean why did I differentiate the 'x-times' in this expression ?
\begin{align}
(x+x+x+....)_\frac{d(x)}{dx}
\end{align}
mathman and Fredrik said earlier that you differentiated as if the 'x-times' was a constant, which is why you got the wrong result.
So I was trying to come up with a way that does not use 'x-times' as a constant, but takes it into account too when you differentiate. So I used some kind of a product rule as a guess, and it turns out that it gave the right answer in this case.