Can some one please (in simple terms) put reasoning behind why:

A: the derivative of the area of a circle=the circumference \(\frac{d}{dx} \pi r^2=2\pi*r\)

B: the derivative of the volume of a sphere=the surface area \(\frac{d}{dx} 4/3\pi r^3=4\pi*r^2\)

Yes I know what derivatives are. But I'm guessing that the two above statements aren't just coincidences, there has to be a reason for this.

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## Comments

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TopNewestWell, take a circle of radius r. And a circle of radius (r+dr)

What is the difference between this two circles? It's only a bit of boundary. I mean, you can get the larger circle just by adding a very thin hollow ring onto the smaller circle. But the area of the ring is 2

pir dr , since it's thickness is negligible.Consider, the circle,

It's just the same thing here. Much like an onion, when you increase the volume of the sphere by adding a infinitesimally thin peel with surface area equal to that of the smaller sphere, we are doing the same thing by adding a hollow layer of volume 4

pir^2 drLog in to reply

That was a question worth asking. I will volunteer to explain again if you do not get it.

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yes i too

i am not able to understand how \(\frac{d}{dx}\) tells the rate of change

here also in a) how is the rate of change of area of circle \(2\pi r\) . can you tell me an practical example

here to

i was amazed when i saw the application part

the question was

which is greater \(tanx or x\)

solution

let

\[f(x) = tanx - x\]

\[ f ' (x) = sec^{2} x - 1 = tan^{2}x\]

since tanx is an increasing function and \[ tan^{2}x\] will remain postive for negative values of x too

therefore tanx is greater than x

now how the rate of change of \[\frac{perpendicular}{base}\] is \[\frac{hypotenuse}{base}\]

and too how the rate of change of x is 1

please explain

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Who said tan(x) is greater than x? \(\tan (7 \frac{\pi}{4} ) = -1\) and \( -1 < 7 \frac{\pi}{4}\) That interpretation only holds good as long as tan(x) is increasing. (It is not strictly increasing all the time)

Okay, so what are derivatives? They point out the rate of change of a small amount of f(x) with respect to a small amount of change in x. How small? Infinitesimally small.

In fact it is easier to perceive it as the slope of the tangent of the function at a particular point.

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@megh choksi

I have some more detailed explanations on this topic. I'll upload them here as a note as soon as I can. I believe reading through them will clarify your doubts. Please let me know if you want me to upload themLog in to reply

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this note

Kindly readIt is a transcript from a person I taught Calculus.

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Instead of looking at the circle's circumference as the derivative (lit. rate of change) of the area of a circle, you could view the area of the circle as the definite integral of the circumference as the infinitesimal radial element varies from 0 to r, or mathematically : \(\displaystyle\int_{0}^{r} 2\pi r dr = \pi r^{2}\)

This seems more intuitively sensible since if you multiply the circumference by infinitely smaller radial measures from the origin(0) to the very end of the circle(r), and add them all up, you will get the 'thickness' of the circle(area).

Now, we see why the rate of change of the area of the circle should be it's circumference (because the rate of change is the sum of the circumferences of infinitesimal thicknesses \(dr\))

Along the same lines, the rate of change of volume of a sphere is it's surface area because the change in volume can really be represented as the sum of the surface areas of infinitesimal thicknesses \(dr\))

Hope that helped.

Warning : This might be more confusing than helpful since I've never written/explained things like this in calculus before.

P.S: in A. it should be the circumference, and not the surface area (guess you repeated that) and in both A. and B., it should be \(dr\), not \(dx\).

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