# Acknowledgement

I owe this to @Amrita Roychowdhury whom I thank for being one of the few people who could pay attention to my Kaboobly Doo.

#### For the Calculus Pro -

I understand that you do not like the way I explained this. But I have already mentioned in the comment that this is way too informal though it is actually a very good way to teach or to introduce the idea to a beginner.

#### For the Calculus Newbie -

Do not just read this like a story. View yourself as Amrita and when Agnishom asks a question to Amrita, try to work it out yourself. Hopefully, it will answer many questions and strengthen your foundations. If you really think that this seems to be more friendly than the usual Calculus text-book approaches, then I will recommend you read Calculus Made Easy by Silvanus Thompson.

# Day 1

### Agnishom:

Lets get started with some differential calculus. Are you ready Miss Roychowdhury?

### Amrita:

oh yes,Agnishom, let's start.

### Agnishom:

Okay;

First of all: d is an operator in calculus. it actually means a very small part of something. E.g, dx means a small part of x.

In differentiation, we are concerned with the ratio of change of one quantity at a particular instant with respect to another.

Consider the simple equation: y=2x

Now, we will consider how much y changes when x is increased by a little amount.

What can you say about the change of y with respect to x?

### Amrita:

i think if x increases,then y will also increase proportionally.

### Agnishom:

You are correct. Differentiating is the art of finding out this proportionality constant at a particular point on the graph.

y=2x

Suppose x=1, then y=2

Suppose x=1.0001, then y = 2.0002

Thus, dx=0.0001 and dy=0.0002 [dx is a very small quantity of x, very very small]

Can we say that dy/dx = 2 ?

### Amrita:

yes,we can say that.

### Agnishom:

Suppose, we are revolving a stone tied in a string. When the string breaks, the stone escapes tangentially? Why? Because that was the way the stone was looking at, at the instant of the breakage.

Similarly, at a particular point on the graph, the slope or the dy/dx we are considering is that of the tangent at that point. Will you see a figure?

### Amrita:

yes,show the fig.

### Agnishom:

derivatives

Do not look at the confusing notations yet.

The tangent on the curve is not the same everywhere. Somewhere, it has a steep slope, somewhere it hasn't. Do you agree?

# Day 2

### Amrita:

Hello Agnishom. Let's start Calculus.

### Agnishom:

Okay, a little recaptulation...

Do you remember what dx means?

If yes, did I tell you what dy/dx means?

### Amrita:

dx means a small part of x.

dy/dx refers to the change of y if x is changed by small amount.

am I right, Sir?

### Agnishom:

Ma'am, I'd like to change the second defintion a little. dy/dx refers to the rate of change of y if x is chnged by a small amount.

You can do this on a plane paper: Draw the parabola $y=x^2$

Now take two points on it and join them with a straight line. Again, take two points which are even closer nd join them by a line.

You will notice that the closer you get, the more the line looks like tangent at that region.

Yes?

### Amrita:

Yes Sir. I drew the parabola.

### Agnishom:

Call me Agnishom, or Sushil, if you like.

Do you agree that the closer the points, the more the line joining them looks like a tangent at that point?

### Amrita:

Yes. Btw is ur another name Sushil?

### Agnishom:

No, the guy in this picture is called Sushil:

sushil

Suppose the two points we chose were (x,y) and (x + dx, y + dy). Since, dx and dy are very very small this is the tangent exactly.

Suppose we, join (x,y) and (x + dx, y + dy). So, the slope of the line is $\frac{(y + dy) - y}{(x+dx) -x}=\frac{dy}{dx}$

In other words, when we are differentiating then we are essentially finding the slope of the tangent at that point on the function.

Do you agree?

If yes, lets get on to doing differentiation

### Amrita:

Yes Agnishom, I agree with it.

### Agnishom:

Lets do some differentiation then.

We do differentiation to find a derivative.

Let us consider a curve of the equation $y = f(x) = x^2$

We will differentiate f(x)

Suppose we have chosen the two points on the curve (x,x^2). If the x coordinate of the second point is $x+dx$, then what is the y coordinate of the second point?

Hint: Use the equation of the curve

### Amrita:

I think it's y+dy.

Am I right?

### Agnishom:

Can you express it in terms of x and dx?

### Amrita:

$y+dy=(x+dx)^2$

Is it? I am not sure. Please tell me.

### Agnishom:

Yes. It is.

Now, $y+dy=(x+dx)^2= x^2+2 \; x\; dx + (dx)^2 = x^2+2 \; x\; dx + d^2x$

We write (dx)^2 as d^2x just like (sin x)^2 is sin^2 x

Now, note that dx is a very very small part as compared to x. d^2x is even smaller.

Just for an example, if dx = 0.0001x, then d^2x = 0.0000001x which is even less small compared to x.

Since, we are only interested in dx and dy, we will neglect small quantities of the second order like d^2x.

So, $y+dy=(x+dx)^2= x^2+2 \; x\; dx + (dx)^2 = x^2+2 \; x\; dx + d^2x = x^2+2 \; x\; dx$

Do you understand why it is fair to do this? Is everything clear uptil now?

### Amrita:

And (dx)^2 should be d^2 x^2 not just d^2 x. Is it so?

### Agnishom:

Nope.

We write (dx)^2 as d^2x just like (sin x)^2 is sin^2 x

dx does not mean d multiplied with x. It is a special kind of a operator meaning a small part of x

### Amrita:

o. then we can proceed now or r u tired?

### Agnishom:

I am fine. I just got lost in a deep ocean, so I am late to reply...

We've got two equations: $y+dy= x^2+2 \; x\; dx$ $y = x^2$

1. Subtract the second equation from the first

2. Divide both sides by dx

Then, post the results.

### Amrita:

Then, Agnishom dy/dx=2x.

### Agnishom:

:D Great!

We say that the derivative of x^2 is 2x

Can you try and find out the derivative of x^3 in the same way?

### Amrita:

I think derivative of x^3 is 3x^2.

### Agnishom:

Very good.

There is a theorem that tells the derivative of $x^a$ is $a x^{a-1}$

You'll be taught the proof in school but just remember the rule for now.

O.K. then?

### Agnishom:

Let us study two more kind of cases.

1. Find the derivative of f(x) if f(x) = 7

2. Find the derivtive of 9x.

Try doing these

### Amrita:

I am feeling very tired now. I will try to do these and inform u later. Will that be o.k.?

And what will be y if f(x)=7? Similarly what about 9x?

### Agnishom:

y=f(x)=7 which means y is always 7.

These two cases are supposed to be solved by common sense or a graph rather than those type of calculations we did before.

Shall I explain it today or are you tired?

### Amrita:

First let me try, Agnishom.

### Agnishom:

Okay.

I've a request for you.

### Amrita:

Yes Agnishom, speak out.

### Agnishom:

Although trying to do these like the x^2 and x^3 will give correct results, I want you to do it by looking at their graphs or some simple reasoning.

There are two reasons why I want you to do it with the graphs:

1. You get a feeling of what you're doing.

2. Rakesh Babu would agree with it.

If you think plotting them by hand is tedious, I can plot the graphs with a computer and post here. Should I?

### Amrita:

Sure Agnishom, I must listen to you.

### Agnishom:

I mean do you want me to post the graphs or do you prefer drawing them yourself?

### Amrita:

Post it for me,plz.

### Agnishom:

1. Find the derivative of f(x) if f(x)= 7

Here is the plot of y = 7

y=7

1. Find the derivative of f(x) if f(x) = 9x

Here is the plot of y = 9x

f2

### Amrita:

Please give some hints to find out the Derivatives.

### Agnishom:

Hint 1: Derivative reffers to the slope of the tangent at a particular given point.

Hint 2: These two graphs have the same tangents and slope everywhere.

### Agnishom:

Look at the first problem

We say that dy/dx is the rate of change of y with respect to small changes of x.

Suppose, x=1 => y=7

If, x = 1.0001, y = 7

If x = 2, y = 7

So, y does not change with x

Fill up the blank: Thus, the change of y is _

### Amrita:

Not possible as it is constant for any value of x.

### Agnishom:

Can we say that the change in y is 0?

### Amrita:

Yes,we can surely say that, Agnishom.

### Agnishom:

So, for a small change of x, there is no change in y at all

Thus derivative of f(x) = 7 is 0

Look at the first graph again. Doesn't it have slope 0 everywhere?

Yes as dy=0.

### Agnishom:

Also, the derivative of any constant is 0.

Any queries? May I proceed to the 9x problem?

### Amrita:

I am liking ur teaching very much, Agnishom. Thnx 4 taking the trouble of teaching me.

### Agnishom:

Thanks

Let us proceed to the next question now.

What is the slope of the straight line y=9x?

### Amrita:

Agnishom,I am really feeling very tired now. I will understand it later from u. Bye,gdn8,ttyl.

# Day 3

### Agnishom:

Let's deal with the other question

### Amrita:

Then we know that y=9x.

### Agnishom:

What can you say about the slope of the straight line y= 9x?

### Amrita:

Slope is same everywhere.

Btw,can u plz tell me the def of slope once again?

### Agnishom:

The slope of a curve at any point reffers to the rate of change of y with a very small change in the value of x.

In case, of straight lines which have the equation y = mx + c. m is the slope.

Can you figure out its value?

Is m=9-c/x?

### Agnishom:

y = 9x

Or, y = 9x + 0 [Compare this with y = mx + c]

m is just 9

In other words, if we increase the value of x by some amount, the value of y will change by 9 times that amount.

Numerical Example:

If x changes from 1 to 1.000001

y will become 9.000009 from 9

Since the slope is the same everywhere, we can say that the derivative of 9x is 9.

Okay?

### Amrita:

And another question,in this graph y is changing with a small change of x. Then the slope should'nt be 0.

### Agnishom:

The slope isn't 0, the slope is 9.

### Amrita:

So the derivative of f(x)=9x is 9, right?

Plz tell me the def of derivative.

### Agnishom:

Derivative of a function f(x) reffers to the rate of change of f(x) with respect to a infinitesimal change in x.

It is the same thing as dy/dx. Since, usually we call y = f(x), $\frac{d\;f(x)}{dx}=\frac{dy}{dx}$

Do you get it?

Yes,Then?

### Agnishom:

We have seen atleast three things till now:

1. The derivative of $x^a$ is $ax^{a-1}$E.g, The derivative of x^3 is 3x^2

2. The derivative of any constant is 0. E.g, The derivative of 9 is 0.

3. The derivative of $ax$ is $a$. E.g, The derivative of 7x is 7

Have we seen them?

### Amrita:

Yes,we have seen them.Then?

### Agnishom:

Notation: The derivative of f(x) is written as f'(x)

Another Rule: The derivative of af(x) is af'(x)

Proof: Let y = af(x)

With a small change in y, y + dy = af(x+dx)

Subtracting, the first equation from the second: dy = a f(x+dx)-f(x)

Dividing both sides by dx, dy/dx = a (f(x+dx) - f(x))/dx

But, (f(x+dx) - f(x))/dx is the derivative of f(x)

So, dy/dx = a f'(x)

Do you understand the proof? Please remember this rule too. Let me show an example usage of this rule, okay?

### Amrita:

Agnishom, I will understand the rest of the Calculus from u later.Now going 2 watch Mahabharat.Bye,gdn8,ttyl.

# Day 3

### Amrita:

Now plz start teaching me Calculus,if u have time.

### Agnishom:

(Please look at the last thing I posted to you)

### Amrita:

I was just recapitulating it.Now,u can show this with an example.

### Agnishom:

Example: Find the derivative of 9x^2

We can split this into two parts 9 and x^2

Since 9 is a constant we can apply the rule we just discussed.

Thus derivative of 9x^2 = 9(x^2)' = 92x = 18x

Can you find the derivative of 3x^3?

### Amrita:

Derivative of 3x^3 is 9x^2.Right?

### Agnishom:

Yes, very good.

Let us look at a problem: What is the slope of the tangent to the curve y=x^3 at x=4? Can you find out the equation for the tangent?

### Amrita:

I can understand that y=64.Then plz give me a hint.

### Agnishom:

Had we discussed that the derivative of a function is the function that desrcibes the slopes of the tangents on that function?

### Amrita:

I think we haven't. Or maybe I have forgotten it. Anyway,what's it?

### Agnishom:

(Oh, come on! That is the very first idea I started my lectures with)

Anyway, I will explain it once again.

Please find the derivative to x^3 for now.

It's 3x^2.

### Agnishom:

The value of 3x^2 at x=4 is ....?

### Amrita:

I am confused. What will be the answer?

### Agnishom:

Value of 3x^2 with x=4 is 3(4^2)=48

Since 3x^2 is the derivative of x^3, and the value of 3x^2 at x=4 is 48, we can say that the slope of the tangent of x^3 at x=4 is 48.

Do you get what I am saying? If yes, then let's verify it.

### Amrita:

Yes,now I got it.

### Agnishom:

Now, that was theory. We must check out if that is the real tangent. Lets do an experiment

Please go to http://www.mathsisfun.com/data/function-grapher.php and do the following settings:

1. Set y = x^3 in one function

2. Set y = 48*x-128 [Note that this 48 is the slope we got]

3. Click the Out 10x button to zoom out

You should get something like this:

experiment

Getting that?

### Amrita:

I am getting that.

### Agnishom:

If you say so....

# Some Other Day

### Amrita:

Agnishom,u told me that u will show me whether it's a real tangent. Well I have plotted the graph. Then?

### Agnishom:

Can you see that we have actually found the tangent?

Yes.

### Agnishom:

Do you believe that differential calculus works, now? :)

By the way, here is another rule:

(f+g)' = f' + g'

Example Usage:

Find the derivative of x^2 + 9x

Answer = (x^2)' + (9x)' = 2x + 9

### Amrita:

Yes,I understood what u did. Then?

### Agnishom:

I was talking about the graph that you plotted. It is a visual proof that calculus works! Do you believe it?

### Amrita:

Yes, I believe it completely.

### Agnishom:

Thank you.

Situation: the relation between the distance travelled, denoted by x, and the time taken by a body, denoted by t, is given below:

$x=7t^2+5t+8$

Please find out the equation for the velocity of the body

### Agnishom:

Instantenous velocity is the derivative of displacement with respect to time.

$v=\frac{dx}{dt}$

Can you differentiate the function in the previous post?

# Day after Some Other Day

### Agnishom:

By the way, velocity is the rate of increase in displacement with respect to time.

That means it is the derivative of displacement with respect to time. Yes?

Yes, Agnishom.

### Agnishom:

This was my equation: x=7t^2+5t+8

Can you differentiate the right side with respect to t?

### Amrita:

It's 14t+5,as u told me just now.

### Agnishom:

Velocity = dx/dt = 14t+5

If acceleration is defined as the change of velocity with respect to time, please find the acceleration of the body.

Acceleration = dv/dt [where v is velocity]

### Agnishom:

I suppose it is 14.

dv=14?

No, dv/dt = 14

### Agnishom:

Acceleration is change of velocity/change of time = dv/dt.

We must compute the derivative of v with respect to t. We have got v = 14t+5

Differentiating 14t+5 with respect to t, we get (14t)'+(5)'=14+0

Do you get this?

Then, dt = ?

### Agnishom:

dt is a small part of t. When we are differentiating v with respect to t, it means that we want the ratio of change of v with respect to a very small change of t.

I think you are confused.

### Amrita:

No no,it's O.K. Please proceed on.

### Agnishom:

Ok, notice something.

We were given that x = 7t^2+5t+8

Differentiating: We get $\frac{dx}{dt}=14t+5$

But again, we call dx/dt as v

We redifferentiate v with respect to t

$\frac{dv}{dt}=14$

But wasn't v = dx/dt?

yeah, So We have actually done this

$\frac{dv}{dt}=\frac{\frac{dx}{dt}}{dt}$

We have differentiated 7t^2+5t+8 two times consecutively.

The result after differentiating twice is called a second derivative. For example, the second derivative of 7t^2+5t+8 is 14

We can also write $\frac{\frac{dx}{dt}}{dt}$ as$\frac{d^2x}{dt^2}$.

### Amrita:

I couldn't understand the last step. Can u plz xplain it 2 me?

### Agnishom:

It is just a notation. (Though there is a good reason for it)

One more notation: f'(x) is the first derivative of f(x) f''(x) is the second derivative of f(x) f'''(x) is the third derivative of f(x) and so on

O.k. Then?

## ---

The conversation continued to the chain rule, partial derivatives, total derivatives, the exponential function, extrema, taylor series, integration and more but as for the introduction to derivatives this is it.

4 years, 8 months ago

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I think my vocabulary is weak enough to convey you an infinitely large Thanks!

- 4 years ago

Mention not :)

Staff - 4 years ago

I can't find words to thank you! I would love to learn things from you in person, or like Ms Amrita did...

- 4 years, 6 months ago

By the way, you already know a lot of Calculus.

img

Staff - 4 years, 6 months ago

Yeah, my knowledge of Calculus got me to a rating of 229. What's better, it doesn't get any lower. I want to learn from where you left in this note.

- 4 years, 6 months ago

Staff - 4 years, 6 months ago

That was luck.

- 4 years, 6 months ago

You mean coincidence?

Staff - 4 years, 6 months ago

We would start studying Calculus this month now that we had done Trigonometry and Statistics. I learned so much here, very nice note and very detailed! I also like how you made it into a conversation as it was unique among others!(Or was it a real one?) Keep up the good work :)

- 4 years, 6 months ago

This conversation, indeed, is real. This is Miss Amrita: @Amrita Roychowdhury

I updated a few things on the note about $e$. It is now ready to read. Let me know what you think about it.

Hope you enjoy the calculus class. Feel free to ask me and the community and share your thoughts :)

Staff - 4 years, 6 months ago

So we started with limits (technically we started with a seatwork D:) and one of the questions was: $\displaystyle\lim_{x\rightarrow 1}\dfrac{\sqrt{x}-1}{x-1}$ I'm not sure about my answer:

$\displaystyle\lim_{x\rightarrow 1}\dfrac{\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+1)} = \dfrac{1}{\sqrt{x}+1}=\dfrac{1}{2}$

@Agnishom Chattopadhyay Sorry if this is not related to derivatives.

- 4 years, 6 months ago

That is correct.

Staff - 4 years, 6 months ago

Thnx Agnishom for dedicating this to me......it's a great honour for me.

- 4 years, 6 months ago

The honor is mine!

Remember what Cantor said?

In mathematics the art of proposing a question must be held of higher value than solving it.

You and @megh choksi are the only ones

Staff - 4 years, 6 months ago

wow! its amazing.GOD bless u all and keep it up please.

- 4 years, 8 months ago

Whee, thanks

Staff - 4 years, 6 months ago

Thanks!

Staff - 4 years, 8 months ago

If you can explain the true idea behind the limit definition of the derivative, I'll give you props. What I mean, is that if you can prove, without rigorous math, that the limit-definition of the instantaneous rate of change is the EXACT instantaneous rate of change, and not some approximation, unlike some people are made to believe (primarily because they're taught to evaluate limits by plugging in numbers closer and closer to the desired number, hence "approaching" but never actually "touching" the desired quantity).

Oh, and in this, there'll be the need to explain what allows you to cancel 0/0 in limits and then plug in the "forbidden" value at the end; to an algebra person, this seems like a hogwash - just illegal mathematics. But that's not true, if you know the exact concepts behind the limit.

And as far as Calculus is concerned, the #1 spot for the "Heart of Calculus" ranking, for me, goes to

$\boxed{f'(x)=\displaystyle \lim_{\Delta x \rightarrow 0}{\frac{f(x+\Delta x)-f(x)}{(x+\Delta x) - x}}}.$

P.S. - if you're planning on replying, no need to make it extensive. A bullet-point summary is fine; I have it figured out, so your explanation shouldn't be too different from mine. And this is my choice for the #1 equation in calculus because, without it, you cannot understand the integral, at all. Why does the anti-derivative give the area under the curve? If you say Riemann-sums, I'll slap you. This equation, if understood, explains the entire concept of calculus, including integral calculus when you apply to anti-differentiation.

Cheers

P.S. 2 - what are "total derivatives"? :O

- 4 years, 8 months ago

can you prove the fact that the limit-definition of the instantaneous rate of change is the EXACT instantaneous rate of change, and not some approximation???

- 3 years, 10 months ago

- 3 years, 10 months ago

that's cool man

I just won a bet, thanks to you.

- 3 years, 10 months ago

Np ;)

- 3 years, 10 months ago

Hm, here is my explanation:

1. To define a line we need atleast two points.

2. A tangent is a line that touches a curve at only one point.

Definition 2 seems incompatible with 1, so we modify it to

A tangent is a line that passes through the given point and another point on the curve which is infinitesimally far apart from the first point. It is the limiting case of any other line that passes through the first point and any other point on the curve.

Same thing for rate of change - To measure the rate of change, you cannot just keep time freezed, for that would not allow the change at all. So, we must choose two points in time which are very close to each other.

The lower the dx, the more closer you are to the theoretical INSTANTENOUS rate of change.

That is where we come in with the idea of limits. By saying dx -> 0, what we basically mean is that our dx is approaching zero, i.e, there is no other real number between 0 and dx.

By now, I think I should have established that this idea that of 'just the real number after 0' is similar to saying just the moment in time immediately after this moment.

[That is what the weird epsilon and delta idea means as far as I got it]

Staff - 4 years, 8 months ago

Haha! NOOOOOB!

Nope!

The truth is, there is absolutely no distance between the two points of the tangent line. The tangent line is 100% the instantaneous rate of change, and in no shape or form is it an infinitely-accurate approximation. It's EXACT.

I'll show you later how and why. Stay tuned (and awesome)!

[P.S. - I replied to your comment on my troll physics problem but then I deleted it; I forgot I'm not supposed to "ruin" the fun ;) Here's my reply, if you're interested:

Um yeah actually there is - I doubt your winnings are going to cover the medical costs. So your "infinite money" will all be used up to cover your infinite hospital bills. [and hence, cyborgs. no need for expensive cures - just tweak a few parts and we good :)]

- 4 years, 8 months ago

Ofcourse the tangent line is 100% the rate of change. When did I say otherwise?

There is a need to define the tangent with two infinitesimally close points.

That is why f(x) = |x| is not differentiable at x=0. Because there is no way you can tell what the next point for the tangent is.

Based on the tangent idea, one could say that the derivative there is 0 considering x=0 the tangent at that point which is certainly not true.

Saying that the two points on the curve are same is close to saying that the distance between them tends to 0, except that the former renders you clueless on how to draw the tangent.

What is a cyborg?

Ofcourse, that is a fallacy. However, it is not a mathematical fallacy. Another fallacy could be that no one will bet on me after a few iterations - because they do not want to loose their money.

Staff - 4 years, 8 months ago

cyborg

s

robot

ss

cyborg = robot + human (no, not science fiction. this is real.)

- 4 years, 8 months ago

Are you a cyborg?

Staff - 4 years, 8 months ago

Aren't we all... ;)

But eh... If you checked out my profile, you'd know what I am.

- 4 years, 8 months ago

Hmm.... Done

Staff - 4 years, 8 months ago

cyb

cc

org

s

- 4 years, 8 months ago

How did you inserted an image in this comment box?

- 4 years, 8 months ago

@U Z lol donwvotes after downvotes, huh? I'm not even gonna comment on that... some people need to...

! [ anything ] ( URL link of the image ) - with no spacing

- 4 years, 8 months ago

"What is a cyborg"

s

ss

sss

yo

• cyborg:

a person whose physiological functioning is aided by or dependent upon a mechanical or electronic device - often perceived as a person whose physical abilities are extended beyond normal human limitations by mechanical elements built into the body.

sssssss

A quick question - do they ever show sci-fi or American movies in India?

- 4 years, 8 months ago

A lot of them.. we have more than a dozen of Hollywood movie Channels.. and yes we enjoy them too..( atleast I)!!

- 4 years, 8 months ago

@megh choksi Please read this. Reading the full discussion should refresh your calculus concepts and questions like 'Why is d/dx of x 1?' will be clearer. :) Thanks for the patient reading in advance.

Staff - 4 years, 8 months ago

Thank you for such effortless writing and a good response . can you tell me what f" (x) tells - rate of change of tangent at any point, ( if true) then what does f "' (x) tells us ?

you asked me in this note that what type of doubts?

These type of silly doubts engage on my mind whenever i learn a new topic

Thank you , you are great

- 4 years, 8 months ago

I updated the description of the second derivative a little in the other reply, please check.

A fool no questioner is. By asking, wiser he becomes. -Yoda

Your doubts are not silly. They are legitimate questions. The way calculus (or any math) is taught in school or atleast introduced in the elevneth standard is horrible and it would be a miracle if anyone got it.

By the way, effortless writing?

Staff - 4 years, 8 months ago

Writing on brilliant sometimes becomes tedious and especially with the signs with \ ( { etc for displaying the math content . you wrote such a long note that's why effortless

- 4 years, 8 months ago

@U Z I used stackedit.io

Is there anything else you'd like to ask me?

Staff - 4 years, 8 months ago

Oh yes poor at english

- 4 years, 8 months ago

@U Z LOL! Thanks for reading this, btw

Staff - 4 years, 8 months ago

f''(x) [second derivative] is the rate of change of the rate of change!

For example, two mountaineers are talking to each other:

A: Isn't the mountain too steep here? B: yes but do not worry, the higher we go, the less steep it is.

Mountaineer A is talking about the first derivative of the slope, i.e, how high you have to climb when you go forward by a small amount.

Mountaineer B is talking about the second derivative of the mountain. It is the rate of change of the rate of change. He claims that the second derivative is negative, i.e, the tangent is becoming less and less steep as we go ahead.

I agree that it is harder to get than the first derivative since it is more abstract.

Update:

Consider another example.

Suppose you are driving at an insane speed realisng which you press the breaks. At the instant when you press the breaks, the acceleration (second derivative) is negative, i.e, trying to slow down the car while the speed( first derivative) is still very high

Staff - 4 years, 8 months ago

How do you get such good examples @Agnishom Chattopadhyay

- 4 years, 8 months ago

@U Z Mathwmatics is a good deal about imagination. Fortunately, we can apply some of the imaginations to reality. That is where good examples come.

I am grateful to my father who started teaching me Calculus when I was 12 or so.

I willexplain the tan(x) issue tomorrow

Staff - 4 years, 8 months ago

A beautiful note , reading again and again

- 4 years, 7 months ago

@U Z You know, I'm going to have to explain elementary calculus to someone again, tomorrow...

Staff - 4 years, 7 months ago

@U Z Glad that you like it.

Staff - 4 years, 7 months ago

You deserve it

- 4 years, 7 months ago