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Why does diffrentiation work the way it does?

When you diffrentiate 2x2 Why do we get 4x as the diffrential?

Why does the method work? I was just curious.

NOTE: I am not asking how we diffrentiate. I want to know why it works.
Original post by SirRaza97
When you diffrentiate 2x2 Why do we get 4x as the diffrential?

Why does the method work? I was just curious.

NOTE: I am not asking how we diffrentiate. I want to know why it works.


https://en.wikipedia.org/wiki/Derivative
Reply 2
Avatar for Zacken
Zacken
22
Original post by SirRaza97
When you diffrentiate 2x2 Why do we get 4x as the diffrential?

Why does the method work? I was just curious.

NOTE: I am not asking how we diffrentiate. I want to know why it works.


lim⁡h→02(x+h)2−2x2h\displaystyle \lim_{h \to 0} \frac{2(x+h)^2 - 2x^2}{h}
Reply 3
Avatar for SirRaza97
SirRaza97
OP
7
Original post by zetamcfc


Anything but Wikipedia?
Original post by SirRaza97
Anything but Wikipedia?


It's explained clearly.
Reply 5
Avatar for lerjj
lerjj
13
Original post by SirRaza97
When you diffrentiate 2x2 Why do we get 4x as the diffrential?

Why does the method work? I was just curious.

NOTE: I am not asking how we diffrentiate. I want to know why it works.


Broadly, because of the binomial expansion theorem:

ddxxn=lim⁡h→0(x+h)n−xnh \displaystyle \frac{d}{dx}x^n = \lim_{h\rightarrow 0} \frac{(x+h)^n-x^n}{h}

When you expand out with the binomial theorem, the only term that survives is the term hnxn−1h=nxn−1 \frac{hnx^{n-1}}{h} = nx^{n-1}

Spoiler

(edited 7 years ago)
Reply 6
Avatar for Zacken
Zacken
22
Original post by SirRaza97
Anything but Wikipedia?


https://www.khanacademy.org/math/differential-calculus/taking-derivatives/derivative-intro/v/calculus-derivatives-1-new-hd-version
Reply 7
Avatar for Zacken
Zacken
22
Original post by lerjj
Broadly, because of the binomial expansion theorem:

ddxxn=lim⁡h→0(x+h)n−xnh \displaystyle \frac{d}{dx}x^n = \lim_{h\rightarrow 0} \frac{(x+h)^n-x^n}{h}

When you expand out with the binomial theorem, the only term that survives is the term hnxn−1h=nxn−1 \frac{hnx^{n-1}}{h} = nx^{n-1}


What do you mean by survives?

We have (x+h)n−xn=hnxn−1+o(h2)(x+h)^n - x^n = hnx^{n-1} + o(h^2).
Reply 8
Avatar for Notnek
Notnek
21
Original post by SirRaza97
When you diffrentiate 2x2 Why do we get 4x as the diffrential?

Why does the method work? I was just curious.

NOTE: I am not asking how we diffrentiate. I want to know why it works.

I recommend you watch this video:

https://www.youtube.com/watch?v=Ayf9gKwjXlY
Reply 9
Avatar for lerjj
lerjj
13
Original post by Zacken
What do you mean by survives?

We have (x+h)n−xn=hnxn−1+o(h2)(x+h)^n - x^n = hnx^{n-1} + o(h^2).


Sorry, by "survive" I mean that in the limit h→0h \rightarrow 0 that's the only term that doesn't vanish (after dividing by h in every term, of course).
Reply 10
Avatar for Zacken
Zacken
22
Original post by lerjj
Sorry, by "survive" I mean that in the limit h→0h \rightarrow 0 that's the only term that doesn't vanish (after dividing by h in every term, of course).


Ah, okay - I wasn't sure because you left an hh\frac{h}{h} in your answer so was wondering whether you'd taken the limit or not yet. S'alright now, thanks. :smile:

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