## Differentiation

- Abdul Muntakim Rafi
**Posts:**173**Joined:**Tue Mar 29, 2011 10:07 pm**Location:**bangladesh,the earth,milkyway,local group.

### Differentiation

I am sharing it because it seemed interesting to me...(And I haven't written any equation about calculus using the Equation editor )

Don't spoil the fun for others if u already know the answer...

\[\frac{\mathrm{d} x^2}{\mathrm{d} x}= 2x\]

Okay no confusion about that...

\[\frac{\mathrm{d} x^2}{\mathrm{d} x}= \frac{\mathrm{d}(x+x+x+x+.........x times) }{\mathrm{d} x}\]

\[\frac{\mathrm{d} x^2}{\mathrm{d} x}= \frac{\mathrm{d} x}{\mathrm{d} x}+\frac{\mathrm{d} x}{\mathrm{d} x}+..................................+\frac{\mathrm{d} x}{\mathrm{d} x}\]

\[\frac{\mathrm{d} x^2}{\mathrm{d} x}= 1+1+1+1+......................+1(x ones)\]

\[\frac{\mathrm{d} x^2}{\mathrm{d} x}= x\]

Where is the mistake?

Don't spoil the fun for others if u already know the answer...

\[\frac{\mathrm{d} x^2}{\mathrm{d} x}= 2x\]

Okay no confusion about that...

\[\frac{\mathrm{d} x^2}{\mathrm{d} x}= \frac{\mathrm{d}(x+x+x+x+.........x times) }{\mathrm{d} x}\]

\[\frac{\mathrm{d} x^2}{\mathrm{d} x}= \frac{\mathrm{d} x}{\mathrm{d} x}+\frac{\mathrm{d} x}{\mathrm{d} x}+..................................+\frac{\mathrm{d} x}{\mathrm{d} x}\]

\[\frac{\mathrm{d} x^2}{\mathrm{d} x}= 1+1+1+1+......................+1(x ones)\]

\[\frac{\mathrm{d} x^2}{\mathrm{d} x}= x\]

Where is the mistake?

**Man himself is the master of his fate...**

- Abdul Muntakim Rafi
**Posts:**173**Joined:**Tue Mar 29, 2011 10:07 pm**Location:**bangladesh,the earth,milkyway,local group.

### Re: Differentiation

it is for the beginners in Calculus(Like me)...

**Man himself is the master of his fate...**

### Re: Differentiation

No reply is coming.Will you please show the mistake?

### Re: Differentiation

The first question that tickles my mind: Can you actually define $x^2$ as a sum of $x$ number of $x$'s???

However, if yo keep insisting on relying on that definition, take it this way-

The derivative of a function $f(x)$ approaches $\frac {f(x+h) - f(x)} {h}$ as $h \rightarrow 0$.

When $f(x) = x^2$, then $f(x+h) = (x+h)^2$

This can be expressed as sum of $x$ number of $x+h$'s and a residual part of $h(x+h)$. You can consider this to be a nibble of $x+h$, only $h$ out of $1$ part is taken.

And then goes $x^2$, sum of $x$ number of $x$'s. Then the subtraction $f(x+h) - f(x)$ reduces to the sum of $x$ number of $(x+h) - x = h$'s and

Division by $h$ leaves us with the limiting value of $x$ number of $1$'s and

Disclaimer: I really don't think this explanation is technically correct. It just gives an intuitive idea why the addition rule for differentiation fails when the number of terms is also variable

However, if yo keep insisting on relying on that definition, take it this way-

The derivative of a function $f(x)$ approaches $\frac {f(x+h) - f(x)} {h}$ as $h \rightarrow 0$.

When $f(x) = x^2$, then $f(x+h) = (x+h)^2$

This can be expressed as sum of $x$ number of $x+h$'s and a residual part of $h(x+h)$. You can consider this to be a nibble of $x+h$, only $h$ out of $1$ part is taken.

And then goes $x^2$, sum of $x$ number of $x$'s. Then the subtraction $f(x+h) - f(x)$ reduces to the sum of $x$ number of $(x+h) - x = h$'s and

*an additional part*: $h(x+h)$Division by $h$ leaves us with the limiting value of $x$ number of $1$'s and

*one additional*$x$. So the derivative becomes $2x$Disclaimer: I really don't think this explanation is technically correct. It just gives an intuitive idea why the addition rule for differentiation fails when the number of terms is also variable

"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor

### Re: Differentiation

The addition rule fails because of countability. When we say "$x$ times", $x$ needs to be in a countable set. Then you can have a bijection with a sequence from that countable set and you stop when you reach $x$ in the sequence and call that being $x$ times. But the kind of calculus we are doing involves uncountable sets, because when you say differentiate $x^2$, you can not think $x$ to be one point, you actually need to be able to pick $x$ from an open interval around that point and that open interval is uncountable. So you can not say "$x$ times". You can only define the phrase "times" with respect to elements of a countable set.

P.S. of course you can try to define new things, that's how mathematics develops. There are theories of calculus where you don't need the notion of limit, and some of them are very algebraic. I have not read any of those yet. There are many examples of calculus on discrete and countable sets as well, and many of them are very useful in economics, physics etc. But you will need to define them in a intelligent way first.

P.S. of course you can try to define new things, that's how mathematics develops. There are theories of calculus where you don't need the notion of limit, and some of them are very algebraic. I have not read any of those yet. There are many examples of calculus on discrete and countable sets as well, and many of them are very useful in economics, physics etc. But you will need to define them in a intelligent way first.

### Re: Differentiation

আমার মনে হয় এখানে সমস্যা টা হয়েছে এই জন্য যে এখানে আমরা ধ্রুবক আর চলকের মধ্যে ঝামেলা পাকিয়ে ফেলেছি। সমস্যাটা যেভাবে solve করেছি সেটার সাধারন সুত্র মনে হয় এই রকমঃ \[\frac{\mathrm{d} (nx)}{\mathrm{d} x}\]

\[\Rightarrow n\frac{\mathrm{d} x}{\mathrm{d} x}\]

দেখুন এখানে n=x হলে \[x^2\] এর derivative \[x\]

হয়। কিন্তু ভুলটা হল n অবশ্যই ধ্রুবক হতে হবে। কিন্তু আমরা যখন n=x লিখেছি। তখন এটা আর ধ্রুবক থাকে নি। তাই সমস্যা হয়েছে।

আমরা আসলে যেটা করেছি সেটা হল\[\frac{\mathrm{d} (x+x+x+....+x)(x times)}{\mathrm{d} x}\]

এর মানে হল\[x.\frac{\mathrm{d} (x+x+x+....+x)}{\mathrm{d} x}\]

যেটা আমার মনে হয় ঠিক না । কারন x পরিবর্তনের সাথে সাথে (x+x+x......+x) এবং (x times) দুটোই পরিবর্তিত হচ্ছে।

এই জন্যই ঝামেলা হয়েছে।

\[\Rightarrow n\frac{\mathrm{d} x}{\mathrm{d} x}\]

দেখুন এখানে n=x হলে \[x^2\] এর derivative \[x\]

হয়। কিন্তু ভুলটা হল n অবশ্যই ধ্রুবক হতে হবে। কিন্তু আমরা যখন n=x লিখেছি। তখন এটা আর ধ্রুবক থাকে নি। তাই সমস্যা হয়েছে।

আমরা আসলে যেটা করেছি সেটা হল\[\frac{\mathrm{d} (x+x+x+....+x)(x times)}{\mathrm{d} x}\]

এর মানে হল\[x.\frac{\mathrm{d} (x+x+x+....+x)}{\mathrm{d} x}\]

যেটা আমার মনে হয় ঠিক না । কারন x পরিবর্তনের সাথে সাথে (x+x+x......+x) এবং (x times) দুটোই পরিবর্তিত হচ্ছে।

এই জন্যই ঝামেলা হয়েছে।

*হার জিত চিরদিন থাকবেই*

তবুও এগিয়ে যেতে হবে.........

বাধা-বিঘ্ন না পেরিয়ে

বড় হয়েছে কে কবে.........তবুও এগিয়ে যেতে হবে.........

বাধা-বিঘ্ন না পেরিয়ে

বড় হয়েছে কে কবে.........

### Re: Differentiation

Good point

I think my argument with countability was wrong. Need to think about it more.

I think my argument with countability was wrong. Need to think about it more.

- Abdul Muntakim Rafi
**Posts:**173**Joined:**Tue Mar 29, 2011 10:07 pm**Location:**bangladesh,the earth,milkyway,local group.

### Re: Differentiation

Avik bhaiya wrote,

s.m. joty, Yeah... I think you got the point... Bravo! Here we are taking $x$ as a variable and constant... That's created the confusion...

And I think u made a typing mistake in the last line...

Please, explain why we can't define $x^2$ like this.The first question that tickles my mind: Can you actually define x2 as a sum of x number of x 's???

s.m. joty, Yeah... I think you got the point... Bravo! Here we are taking $x$ as a variable and constant... That's created the confusion...

And I think u made a typing mistake in the last line...

**Man himself is the master of his fate...**

### Re: Differentiation

Answer to that is my post above about countabilityAbdul Muntakim Rafi wrote:Avik bhaiya wrote,

Please, explain why we can't define $x^2$ like this.The first question that tickles my mind: Can you actually define x2 as a sum of x number of x 's???

When we say "$x$ times", $x$ needs to be from a countable set.

### Re: Differentiation

I think Tanvir vai is right.Answer to that is my post above about countability

When we say "x times", x needs to be from a countable set.

'Rafi' can you say where I made a mistake. Post a quote please .s.m. joty, Yeah... I think you got the point... Bravo! Here we are taking x as a variable and constant... That's created the confusion...

And I think u made a typing mistake in the last line...

*হার জিত চিরদিন থাকবেই*

তবুও এগিয়ে যেতে হবে.........

বাধা-বিঘ্ন না পেরিয়ে

বড় হয়েছে কে কবে.........তবুও এগিয়ে যেতে হবে.........

বাধা-বিঘ্ন না পেরিয়ে

বড় হয়েছে কে কবে.........