# Why the derivative of area is related to the graph of the function

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Frigus
the explanation about the question I got from internet is,
A very small change in area divided by the dx will give the function of graph so anti-derivative of function of graph should be equal to the area of the function.
It also seem quite obvious to me but I am not satisfied by it,
It seems to me that even for the tiniest of tiniest dx the derivative of area and function of graph should not be same.

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Mentor
Rather than attempt an explanation of why it is true, I would check out the 3blue1brown sequence of videos on the Essence of Calculus. The videos may give you new insight to how Calculus works:

member 587159
Frigus
Rather than attempt an explanation of why it is true, I would check out the 3blue1brown sequence of videos on the Essence of Calculus. The videos may give you new insight to how Calculus works:

The explanation I gave here is from 3b1b.😅
I watched his series 2-3 times but the problem is I can't really figure out why I am not satisfied.
I feel some void that I am not able to fill.

Mentor
2022 Award
The explanation I gave here is from 3b1b.😅
I watched his series 2-3 times but the problem is I can't really figure out why I am not satisfied.
I feel some void that I am not able to fill.
You have to forget that ##dx## and ##df## are tiny distances. They are abbreviations for a limit. E.g. ##\lim_{n \to \infty}\frac{1}{n}=0## although every single sequence member is strictly positive. The same limit processes define Riemann integration and differentiation. The imagination that the infinitesimals are tiny distances is a crutch, and it limps.

Frigus
Gold Member
I interpret your question as: "Let ##f## be a continuous function defined on an interval. Why is it true that ##\frac{d}{dx}\int_a^xf(t)dt=f(x)##?"

We can just check it directly: Consider the difference quotient ##\frac{1}{h}\left(\int_a^{x+h}f(t)dt-\int_a^xf(t)dt\right)=\frac{1}{h}\left(\int_x^{x+h}f(t)dt\right).## We wish to find the limit when ##h## goes to zero.

Let ##\varepsilon>0## be arbitrary, and then for sufficiently small ##h##, we have ##|f(x)-f(t)|<\varepsilon## when ##t\in [x,x+h].## Then ##h(f(x)-\varepsilon)\leq\int_x^{x+h}f(t)dt\leq h(f(x)+\varepsilon)##

Plug this into the difference quotient to find that ##f(x)-\varepsilon\leq\frac{1}{h}\int_x^{x+h}f(t)dt\leq f(x)+\varepsilon## for sufficiently small ##h##. Since ##\varepsilon>0## is arbitrary, this means that the limit is ##f(x)##.

Frigus
Frigus
Sorry for this but now I am not able to read and understand your post as it is already 2:20 am in my country,I am telling this because it feels little rude not to reply to people who are willing to help.

fresh_42
Frigus
You have to forget that ##dx## and ##df## are tiny distances. They are abbreviations for a limit. E.g. ##\lim_{h \to \infty}\frac{1}{n}=0## although every single sequence member is strictly positive. The same limit processes define Riemann integration and differentiation. The imagination that the infinitesimals are tiny distances is a crutch, and it limps.
The whole calculus series of 3b1b is based on tiny nudges and I studied calculus from his series so I can't imagine what does limits means except in the light of tiny nudges.
I think of limits like this,
For a function ##f(x)##(tells area of graph) if we move a tiny bit h then,
##lim_{h \to 0}\frac {f(x+h)-f(x)}{h}##
For smaller and smaller h this approaches the function of graph.
How can I imagine limits without small nudges?
There is also the notion of hyperreals that have been used as a foundation for Calculus:

https://en.wikipedia.org/wiki/Nonstandard_calculus

and this book on Calculus based upon it:

https://www.math.wisc.edu/~keisler/calc.html

Truthfully I think everyone trods this path in Calculus until the needs of graduating get in the way and we must buckle down to get the work done delaying our understanding until we retire.
It is now impossible for me to read this book as It can take too much time and devoting the time to it is not feasible for me as I have to spend time on other subjects too.

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Mentor
2022 Award
I think of limits like this,
For a function ##f(x)##(tells area of graph) if we move a tiny bit h then,
##\lim_{n \to 0}\frac{f(x+h)-f(x)}{h}##
This limit is ## \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}=\left.\dfrac{df}{dx}\right|_{x=x}.##
Now we have used the same letter ##x## in two roles, as variable and as point of evaluation.
Let us correct this.
$$\lim_{h \to 0}\frac{f(p+h)-f(p)}{h}=\left.\dfrac{df}{dx}\right|_{x=p}=f'(p) \in \mathbb{R}$$
This way the limit stands for the slope of the graph at point ##(p,f(p))##, not the function itself. However, the more we reduce ##h## the closer we get to ##f(p)##. But we still have a quotient and a difference in the numerator.

What is commonly and sloppily written ##\lim_{n \to 0}\frac {f(x+h)-f(x)}{h}## means a function:
$$p \longmapsto f'(p) = \lim_{h \to 0}\frac{f(p+h)-f(p)}{h}=\left.\dfrac{df}{dx}\right|_{x=p} = f'(p)$$
where we can use ##p=x## if we do not need the ##x## in the limits anymore.

All these expressions are the slope of the function / graph, not the function itself.
That would be ##\lim_{h \to 0}f(p+h)##.
For smaller and smaller h this approaches the function of graph.
No. see above.
How can I imagine limits without small nudges?
You can see it that way, but the crucial point in your sentence is missing. It should read:
'I imagine limits by arbitrary small nudges.'

It is not one tiny difference as physicists like to take it, it is an ever smaller difference: the secant that becomes a tangent. If it is a tangent, then there are no tiny bits anymore.

etotheipi
Frigus
This way the limit stands for the slope of the graph at point (p,f(p)), not the function itself.
I was trying to take derivative of area and I denoted area function by ##f(x)##.
I mentioned it here too,
For a function f(x)(tells area of graph)
Sorry if I have not used correct notations or their is something I am still missing.
I have not taken any maths courses so my maths notations and concepts are very flawed.

Mentor
2022 Award
The topic itself is confusing. I once listed the different views of what a derivative is and ended up with 10. And the word slope wasn't even among them. Now if you also put integration in the bucket, things are even more confusing. I only answered to the limit which you wrote, and that was a differentiation, not an integration.

As tangent (one possible view of differentiation), it is of course an approximation of the graph at a certain point. That's why we use it. But again no integration.

The integration part is completely explained in @Infrared's post #5.

If you need a heuristic, then you can see it this way (attention, this is only a heuristic and as such inaccurate). Say we have ##F'(x)=f(x)## with ##F(0)=0.## Then
\begin{align*}
f(x)&=F'(x)=\dfrac{dF}{dx}\approx \dfrac{\Delta F}{\Delta x} \quad\text{ a quotient of difference, hence an incline }\\
F(x) &=F(x)-F(0)=\int_0^xf(t)\,dt \approx \Delta f\cdot\Delta x=\Delta F\quad \text{ a product: height times length, hence an area }
\end{align*}
Thus we have approximately an area ##\Delta F## which is height ##\Delta f## times length ##\Delta x##, or the other way around, a height ##\Delta f## which we get as quotient ##\Delta F / \Delta x,## area divided by length.

If we make these ideas rigorous and suited for curved boundaries, we end up with differentiation and integration, where limits play the central role, because we approach curved lines by straights, and areas by rectangular boxes.

Homework Helper
area is base times height, and when you take the derivative you divide by the base, as an approximation, so when you take derivative of area, you get the height (of the graph), as an approximation. Then as the approximation is made better, you get exactly the height.

In more detail, assume the graph of f is increasing for simplicity. when you look at the graph and draw a vertical line at x=a, then move over to a nearby point x = a+h, draw another vertical line, and ask what is the change in area, ie what is the area between the two vertical lines, the answer is the base times the average height, i.e. h times some height between f(a) and f(a+h). If h is very small, this average height is close to f(a).

Thus when you ask for the derivative of area wrt x, you are asking for the limit of the change in area, which is close to h times f(a), divided by h, which is close to f(a). As h is taken smaller, this gets closer and closer to f(a).

I.e. since area is base times average height, the derivative of area, wrt the base length, i.e. area divided by base, is the average height, but when you take the base smaller and smaller, i.e. as h goes to zero, in the limit the average height taken between a and a+h, is just the height at a.

All this assumes we are in the case of a graph where the height at a point near a approximates the height at a, i.e. the graph of a "continuous" function.

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Frigus
Frigus
I think I got the concept,
If we have a equation ##y=x²## then if we plot it and if I forgot the function that gives the area for a moment but I know that the area function derivative will give us the height (y) of the graph even if it is not exact but only an approximation and the other point i know is that their will be only one function whose derivative is function of graph and if I combine these two points then my problem is solved.
thanks to all for helping me.