satsun
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I keep seeing questions where after dy/dx people put lim as H tends to 0. I have never done this, I just differentiate and at the very end say as H tends to 0 xyz. Is this incorrect?
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hajima
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You must include the limit in every step when differentiating from first principles, otherwise every step until the last is incorrect as the definition of the differential relies on there being a limit.
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satsun
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(Original post by hajima)
You must include the limit in every step when differentiating from first principles, otherwise every step until the last is incorrect as the definition of the differential relies on there being a limit.
I’ll give you an example of my working:

Prove from first principles that the derivative of 10x² is 20x

1. let y = 10x²
2. consider points P (𝑥, 10x²) and Q (𝑥+h, 10(𝑥+h)²)
3. find dy/dx = 10𝑥² + 20xh + 10h² - 10x² / 𝑥+h-𝑥
4. simplify = 20xh + 10h² / h so = 20x + 10h
5. as h tends to 0, 20x + 10h tends to 20x.

Is this correct? This is how I always do it. I never mention limits, would I lose marks?
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davros
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(Original post by hajima)
You must include the limit in every step when differentiating from first principles, otherwise every step until the last is incorrect as the definition of the differential relies on there being a limit.
Depends how you lay it out surely?

If you have y = f(x) and you know f(x) explicitly then you can do all the working for f(x + h), f(x+h) - f(x) all the way up to [f(x+h) -f(x)]/h and then at that point write dy/dx = lim(h->0) whatever
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DFranklin
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(Original post by satsun)
I’ll give you an example of my working:

Prove from first principles that the derivative of 10x² is 20x

1. let y = 10x²
2. consider points P (𝑥, 10x²) and Q (𝑥+h, 10(𝑥+h)²)
3. find dy/dx = 10𝑥² + 20xh + 10h² - 10x² / 𝑥+h-𝑥
4. simplify = 20xh + 10h² / h so = 20x + 10h
5. as h tends to 0, 20x + 10h tends to 20x.

Is this correct? This is how I always do it. I never mention limits, would I lose marks?
It's not correct. The correct version of line 3 would be:

\dfrac{dy}{dx} = \lim_{h \to 0} \dfrac{10x^2+20xh+10h^2 - 10x^2}{x+h-x}

and then you'd need to maintain that \displaystyle \lim_{h \to 0} on line 4 and then line 5 would lose the limit and simply replace 20x+10h by 20x.

But the whole thing with introducing P and Q doesn't look right to me (but I am not 100% sure how they teach this at A-level) - it's something you'd do as an illustration of why the formal definition of dy/dx (*) makes sense, but not something you'd do during evaluation (certainly without a bit of explanation of how it connects to dy/dx).

(*): \displaystyle \dfrac{dy}{dx} = \lim_{h\to 0} \dfrac{y(x+h) - y(x)}{h}

Also, as davros suggests, you may want to largely consider (y(x+h)-y(x))/h for h non-zero without taking limits (as purely algebraic manipulation) and then take limits at the end.

Something like:

Let y(x) = 10x^2.
Then for h \neq 0, \dfrac{y(x+h) - y(x)}{h} = \dfrac{10x^2+20xh+10h^2 - 10x^2}{x+h-x} = 20x + 10h.
\displaystyle \dfrac{dy}{dx} = \lim_{h\to 0} \dfrac{y(x+h) - y(x)}{h} = \lim_{h\to 0} 20x + 10h = 20x
Last edited by DFranklin; 1 month ago
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satsun
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(Original post by DFranklin)
It's not correct. The correct version of line 3 would be:

\dfrac{dy}{dx} = \lim_{h \to 0} \dfrac{10x^2+20xh+10h^2 - 10x^2}{x+h-x}

and then you'd need to maintain that \displaystyle \lim_{h \to 0} on line 4 and then line 5 would lose the limit and simply replace 20x+10h by 20x.

But the whole thing with introducing P and Q doesn't look right to me (but I am not 100% sure how they teach this at A-level) - it's something you'd do as an illustration of why the formal definition of dy/dx (*) makes sense, but not something you'd do during evaluation (certainly without a bit of explanation of how it connects to dy/dx).

(*): \displaystyle \dfrac{dy}{dx} = \lim_{h\to 0} \dfrac{y(x+h) - y(x)}{h}

Also, as davros suggests, you may want to largely consider (y(x+h)-y(x))/h for h non-zero without taking limits (as purely algebraic manipulation) and then take limits at the end.

Something like:

Let y(x) = 10x^2.
Then for h \neq 0, \dfrac{y(x+h) - y(x)}{h} = \dfrac{10x^2+20xh+10h^2 - 10x^2}{x+h-x} = 20x + 10h.
\displaystyle \dfrac{dy}{dx} = \lim_{h\to 0} \dfrac{y(x+h) - y(x)}{h} = \lim_{h\to 0} 20x + 10h = 20x
Hi. Thank you for this. I’ve just checked a similar question within the Edexcel papers.

Here is the mark scheme for the question:Name:  7EB92F22-EAF9-45F0-9F52-9005812F0E7B.jpeg
Views: 10
Size:  12.2 KBName:  6219F032-2158-426B-ABAD-A1F526116BC4.jpeg
Views: 8
Size:  70.5 KB

I cannot see the (lim as h tends to 0) sign. Is this an error in the mark scheme then?
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DFranklin
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(Original post by satsun)
Hi. Thank you for this. I’ve just checked a similar question within the Edexcel papers.

Here is the mark scheme for the question:

I cannot see the (lim as h tends to 0) sign. Is this an error in the mark scheme then?
"as h \to 0, gradient \to 6x" is equivalent to \lim_{h\to 0} \text{ gradient } = 6x.

In case it's not clear, the error is in equating the derivative to "the gradient of the line joining (x, f(x)) to (x+h, f(x+h))". You can equate the derivative (i.e. dy/dx) to the limit of the gradient as h goes to 0, but you can't directly equate them. You'll note the mark scheme doesn't even mention the derivative until after it's taken the limit.

[Of course, I personally think that mark scheme looks like a steaming mess, but that's why I think it's stupid to be teaching this at A-level as a topic to be examined on. You're just going to end up learning it all over again properly at university].
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hajima
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(Original post by davros)
Depends how you lay it out surely?

If you have y = f(x) and you know f(x) explicitly then you can do all the working for f(x + h), f(x+h) - f(x) all the way up to [f(x+h) -f(x)]/h and then at that point write dy/dx = lim(h->0) whatever
I suppose so? I find it easier to do it in one go by writing dy/dx = ..., and then follow it down but as long as you don’t write dy/dx = ..., without a limit at each step then it’s mathematically correct.
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DFranklin
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(Original post by hajima)
I suppose so? I find it easier to do it in one go by writing dy/dx = ..., and then follow it down but as long as you don’t write dy/dx = ..., without a limit at each step then it’s mathematically correct.
davros's point is you said "you must include the limit in every step " (emphasis mine). It's fine to do it your way, but it's also fine to not talk about limits until near the end, as long as you're not ever saying dy/dx equals (the expression without a limit).
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hajima
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(Original post by DFranklin)
davros's point is you said "you must include the limit in every step " (emphasis mine). It's fine to do it your way, but it's also fine to not talk about limits until near the end, as long as you're not ever saying dy/dx equals (the expression without a limit).
Yes I know, I did say that:
"...but as long as you don’t write dy/dx = ..., without a limit at each step then it’s mathematically correct."
I also did talk specifically about the definition of a differential so I was originally talking about starting from dy/dx = ... anyway, but their method would be accepted in any mark scheme.
Last edited by hajima; 1 month ago
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satsun
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(Original post by hajima)
Yes I know, I did say that:
"...but as long as you don’t write dy/dx = ..., without a limit at each step then it’s mathematically correct."
I also did talk specifically about the definition of a differential so I was originally talking about starting from dy/dx = ... anyway, but their method would be accepted in any mark scheme.
So I am correct and I don’t need to include the limit sign?
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DFranklin
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(Original post by satsun)
So I am correct and I don’t need to include the limit sign?
Your post #3 is incorrect for the reasons I previously stated.
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hajima
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(Original post by satsun)
So I am correct and I don’t need to include the limit sign?
If you evaluate each part separately and then say "as h --> 0" at the end, you can avoid using "lim", yes.
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satsun
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(Original post by DFranklin)
"as h \to 0, gradient \to 6x" is equivalent to \lim_{h\to 0} \text{ gradient } = 6x.

In case it's not clear, the error is in equating the derivative to "the gradient of the line joining (x, f(x)) to (x+h, f(x+h))". You can equate the derivative (i.e. dy/dx) to the limit of the gradient as h goes to 0, but you can't directly equate them. You'll note the mark scheme doesn't even mention the derivative until after it's taken the limit.

[Of course, I personally think that mark scheme looks like a steaming mess, but that's why I think it's stupid to be teaching this at A-level as a topic to be examined on. You're just going to end up learning it all over again properly at university].
I said as H tends to 0 which is just the written form of that sign?
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DFranklin
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(Original post by satsun)
I said as H tends to 0 which is just the written form of that sign?
In your step 3 you wrote:

3. find dy/dx = 10𝑥² + 20xh + 10h² - 10x² / 𝑥+h-𝑥
this is unambiguously wrong.
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satsun
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(Original post by DFranklin)
In your step 3 you wrote:



this is unambiguously wrong.
If I changed dy/dx to gradient of PQ it would be correct?
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DFranklin
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(Original post by satsun)
If I changed dy/dx to gradient of PQ it would be correct?
Yes, except if you make that change you never mention dy/dx or the derivative at all in the entire proof. So after you take the limit you'd need to at least remark that you now had the derivative required.
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