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

• B
• Frigus
In summary, I think that the derivative of the area and the function of the graph should not be the same for the tiniest of dx. Rather, you should watch the 3blue1brown series on the Essence of Calculus to gain a deeper understanding of how Calculus works.
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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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
jedishrfu said:
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.

Hemant said:
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
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
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
fresh_42 said:
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?
jedishrfu said:
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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Hemant said:
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
fresh_42 said:
This way the limit stands for the slope of the graph at point (p,f(p)), not the function itself.
I have a doubt about this point,
I was trying to take derivative of area and I denoted area function by ##f(x)##.
I mentioned it here too,
Hemant said:
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.

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.

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
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.

## 1. Why is the derivative of area related to the graph of the function?

The derivative of area is related to the graph of the function because it represents the rate of change of the area under the curve. This means that as the function changes, the area under the curve also changes, and the derivative tells us how quickly that change is happening.

## 2. How is the derivative of area calculated?

The derivative of area is calculated using the fundamental theorem of calculus, which states that the derivative of the area under a curve is equal to the height of the curve at a specific point. This can be represented mathematically as the integral of the function over a given interval.

## 3. What is the significance of the derivative of area in real-world applications?

The derivative of area has many real-world applications, such as in physics, where it is used to calculate the velocity of an object at a specific point in time. It is also used in economics to determine the rate of change of a company's profits over time.

## 4. How does the graph of the function affect the derivative of area?

The shape and behavior of the graph of the function directly affect the derivative of area. For example, if the function is increasing at a steady rate, the derivative of area will also be constant. However, if the function is constantly changing, the derivative of area will also vary.

## 5. Can the derivative of area ever be negative?

Yes, the derivative of area can be negative. This occurs when the function is decreasing at a specific point, meaning that the area under the curve is decreasing as well. This can be seen as a downward slope on the graph of the function.

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