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If you don't put lim->0 in differentiation from first principles do you still marks?

Hello, so we had a tiring exam on 6th as we never had sat an actual exam, it was a very stressing experience but we made it.

Now, the question on i think 13 or 14 cant remember on paper 1 edexcel a level on 2023 june 6.

it was to differentiate sin(x) from first principles, I used a different method where there was no mention of lim->0 source :https://youtu.be/cmBIyKQleWU

It was maths genie, he used a gradient method. What would be your opinion would this method be valid?
Reply 1
You need to take the limit somewhere, as that is the definition of a derivative.

That said, you don't need to write "lim" every line. In the video, the "lim" on every line is replaced by the sentence "as h tends to zero".
It is so key, that if you don't write it, your solution would be incorrect. (For instance, you'd end up claiming sin(x)/x = 1, which is nonsense)

The method is practically the same. It's just a different presentation.
(edited 11 months ago)
Reply 2
Original post by tonyiptony
You need to take the limit somewhere, as that is the definition of a derivative.

That said, you don't need to write "lim" every line. In the video, the "lim" on every line is replaced by the sentence "as h tends to zero".
It is so key, that if you don't write it, your solution would be incorrect. (For instance, you'd end up claiming sin(x)/x = 1, which is nonsense)

The method is practically the same. It's just a different presentation.


Yes,

My conclusion was at the end

"As h tends to 0, so does cosh -1/h tends to 0 and sinh/h tends to 1 and we get cosx as required "
Don't want to cause panic and got no idea how it's taught and how they mark but writing something like "d(sin(x))/dx = (sin(x+h) - sin(x)) / h" is not correct if we are being strict. (sin(x+h) - sin(x)) / h approaches the derivative of sin(x) as h->0 but it's not equal to the derivative while it is approaching..

The way it's done in the video is very sloppy.
Reply 4
Original post by hassassin04
Don't want to cause panic and got no idea how it's taught and how they mark but writing something like "d(sin(x))/dx = (sin(x+h) - sin(x)) / h" is not correct if we are being strict. (sin(x+h) - sin(x)) / h approaches the derivative of sin(x) as h->0 but it's not equal to the derivative while it is approaching..

The way it's done in the video is very sloppy.


here's what he did, i don't know if this method is correct or not but
(x, sinx) , (x+h, sinx + h)

d/dx = sin(x + h) - sin(x)/ x + h -x
you can simplify the bottom bit as just h because we have a positive and negative x

and then we use sin(a + b) formula then we can factorise cosx and sinx
we end up with sinh/h * cox + (cosh - 1)/h * (sin x) then we can use of the limits
that as h tends to 0 so does (cosh - 1)/h = 0 and sinh/h tends to 1 and we get cosx because cosx * 1 is cosx
(edited 11 months ago)
Reply 5
Original post by Zephyr0221
here's what he did, i don't know if this method is correct or not but
(x, sinx) , (x+h, sinx + h)

d/dx = sin(x + h) - sin(x)/ x + h -x


As hassassin04 said, this is incorrect. It should be

\displaystyle \frac{d}{dx} \sin x = \lim_{h\to 0} \frac{\sin(x+h)-\sin(x)}{h}

[To be clear: the important difference is that a limit is involved in the RHS, the rest is just cosmetic].
Original post by Zephyr0221
here's what he did, i don't know if this method is correct or not but
(x, sinx) , (x+h, sinx + h)

d/dx = sin(x + h) - sin(x)/ x + h -x
you can simplify the bottom bit as just h because we have a positive and negative x

and then we use sin(a + b) formula then we can factorise cosx and sinx
we end up with sinh/h * cox + (cosh - 1)/h * (sin x) then we can use of the limits
that as h tends to 0 so does (cosh - 1)/h = 0 and sinh/h tends to 1 and we get cosx because cosx * 1 is cosx



I understand, my issue is with writing "d/dx = sin(x + h) - sin(x)/ x + h -x". The derivative of sin(x) is the limit of sin(x + h) - sin(x)/ x + h -x.

If one wanted to be lazy and exclude the lim in each equality, then a better way would be to simply write
sin(x + h) - sin(x)/ x + h -x = ... = the final expression and then say that d(sin(x))/dx = lim the final expression = cos(x) while giving a justification for the final equality (follows from using well-known limit results).
(edited 11 months ago)
Reply 7
Original post by hassassin04
I understand, my issue is with writing "d/dx = sin(x + h) - sin(x)/ x + h -x". The derivative of sin(x) is the limit of sin(x + h) - sin(x)/ x + h -x.

If one wanted to be lazy and exclude the lim in each equality, then a better way would be to simply write
sin(x + h) - sin(x)/ x + h -x = ... = the final expression and then say that d(sin(x))/dx = lim the final expression = cos(x) while giving a justification for the final equality (follows from using well-known limit results).


thanks... it makes sense now
Reply 8
Original post by DFranklin
As hassassin04 said, this is incorrect. It should be

\displaystyle \frac{d}{dx} \sin x = \lim_{h\to 0} \frac{\sin(x+h)-\sin(x)}{h}


[To be clear: the important difference is that a limit is involved in the RHS, the rest is just cosmetic].
So, we can conclude the method shown on the video is incorrect? and the tutor should have added lim ->0 on RHS correct?
Reply 9
Original post by Zephyr0221
So, we can conclude the method shown on the video is incorrect? and the tutor should have added lim ->0 on RHS correct?

Yes. That said, it's really not a big deal. If he just hadn't written the ddθ=\dfrac{d}{d\theta} = bit, everything else is valid (the thing that's unequivocally wrong is to have the derivative on one side and not have the limit on the other). And then how much can you reasonably take off for putting the d/d theta = ...?

IMNSHO, leaving out the h->0 should only cost 1 mark - the "interesting bit" is manipulating the "(f(x+h) - f(x))/h" bit, not putting in lim h->0.
Reply 10
Original post by DFranklin
Yes. That said, it's really not a big deal. If he just hadn't written the ddθ=\dfrac{d}{d\theta} = bit, everything else is valid (the thing that's unequivocally wrong is to have the derivative on one side and not have the limit on the other). And then how much can you reasonably take off for putting the d/d theta = ...?

IMNSHO, leaving out the h->0 should only cost 1 mark - the "interesting bit" is manipulating the "(f(x+h) - f(x))/h" bit, not putting in lim h->0.


Alright, so I will leave an email to Maths Genie about it. As it could mislead or potentially student may lose marks. Thanks for the clarification.

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