Differentiation of first principles

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KingRich
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#1
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#1
I know this is probably going to seem extremely dumb. However, how do I prove using first principle that the derivative of y=8 is zero.

Does that mean my
f(x)=8

Or, am I trying to show that the end result from the function results in 8
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aaron2578
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#2
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#2
You already have y = 8.
You have to show that dy/dx = 0.

Alternatively, given that f(x) = 8 show that f'(x) = 0.

If you think about it, y = 8 is an equation of a straight line. Differentiating a function finds you the gradient of it. Seeing as you've been given a horizontal line, the derivative must be 0.
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davros
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#3
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(Original post by KingRich)
I know this is probably going to seem extremely dumb. However, how do I prove using first principle that the derivative of y=8 is zero.

Does that mean my
f(x)=8

Or, am I trying to show that the end result from the function results in 8
"First principles" usually means using the definition that involves limits and symbology like f(x+h) and f(x).

In this case you should be able to write down definitions for f(x) = ... and f(x+h) = ... .
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Anon???
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#4
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#4
(Original post by KingRich)
I know this is probably going to seem extremely dumb. However, how do I prove using first principle that the derivative of y=8 is zero.

Does that mean my
f(x)=8

Or, am I trying to show that the end result from the function results in 8
If you study edexcel use the formula booklet: https://qualifications.pearson.com/c...mulae_Book.pdf

It’s on page 7 in the PDF I think! (The one numbered page 3)

So it’s f(x) = 8 and f(x+h) = 8

f’(x) = lim f(x+h) - f(x)/h

F’(x) = lim 8-8/h

f’(x) = lim 0/h = 0 and your done!
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simon0
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#5
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#5
(Original post by KingRich)
I know this is probably going to seem extremely dumb. However, how do I prove using first principle that the derivative of y=8 is zero.

Does that mean my
f(x)=8

Or, am I trying to show that the end result from the function results in 8
Yes f(x) = 8.

As f(x) is continuous (for all  x \in \mathbb{R} ), use the formula Anon?? has mentioned where:

 \displaystyle f ^{\prime} (x) = \lim_{ h \rightarrow 0 } \frac{ f(x+h) - f(x) }{ h }.
Last edited by simon0; 6 months ago
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KingRich
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#6
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#6
I’m studying AQA. I did consider based on the formula
F(x)=f(x+h)-f(x)/h

If my f(x) =8 then i would sub the x for 8 but I don’t believe f(x+h) = 8 because h represents a small movement along, so wouldn’t it be f(x+h) =8+h??
And if that was the case I would end up with 1
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KingRich
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#7
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#7
(Original post by aaron2578)
You already have y = 8.
You have to show that dy/dx = 0.

Alternatively, given that f(x) = 8 show that f'(x) = 0.

If you think about it, y = 8 is an equation of a straight line. Differentiating a function finds you the gradient of it. Seeing as you've been given a horizontal line, the derivative must be 0.
How would I apply the principle to show this?
Would it be as someone mentioned
F(x)=8
F(x+h)=8
Doesn’t seem quite right to me
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simon0
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#8
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#8
(Original post by KingRich)
How would I apply the principle to show this?
Would it be as someone mentioned
F(x)=8
F(x+h)=8
Doesn’t seem quite right to me
You had it righ in your previous post:

- f(x+h) = 8,
as f(x) = 8 for all  x \in \mathbb{R} ,

- f(x) = 8 .
Last edited by simon0; 6 months ago
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mqb2766
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#9
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#9
(Original post by KingRich)
How would I apply the principle to show this?
Would it be as someone mentioned
F(x)=8
F(x+h)=8
Doesn’t seem quite right to me
If the function is constant,
F(whichever point you want to choose) = 8
So F(x+h) is indeed 8.
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KingRich
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#10
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#10
(Original post by simon0)
You had it righ in your previous post:

- f(x+h) = 8,
as f(x) = 8 for all  x \in \mathbb{R} ,

- f(x) = 8 .
Thank you. I will go apply with the formula
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KingRich
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#11
Report Thread starter 6 months ago
#11
Thank you for you help guys. Appreciate it
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