Is this a valid representation of a derivative?

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Homework Statement
The standard definition of the derivative is ##f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}##.

What would happen we replace the term ##f(x+h)## inside the limit with a function composition ##f(g(x))##, with ##\lim_{h\to 0} g(x) = x##? Then ##f(g(x))## converges to ##f(x)## as ##h\to 0##.
Relevant Equations
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If ##f(g(x)) =f(x+h)##, then ## \lim_{h\to 0} \frac{f(g(x))-f(x)}{h}= \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=f'(x) ##.

What if we let ##g(x) = x+\sin(h)##?

Then ## \lim_{h\to 0} \frac{f(g(x))-f(x)}{h}= \lim_{h\to 0} \frac{f(x+\sin(h))-f(x)}{h}##

This is not equivalent to that the standard definition of ##f'(x)##. It seems very similar because ##\lim_{h\to 0} x+\sin(h)=x## makes the numerator of the limit converges to ##0## just differently than with ##f(x+h)##. Can we also call this ##f'(x)##?

Thanks for commenting!
 
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  • #2
My immediate observation is that in this "alternative" definition, the derivative in ##x=0## is identically ##0##. This is a problem.
 
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  • #3
Hill said:
My immediate observation is that in this "alternative" definition, the derivative in ##x=0## is identically ##0##. This is a problem.
Sorry! Do you mean that it wouldn't work because ##\lim_{h\to 0} x(1-\cos(h))=0##? Thank you for pointing out the example was a bad one. I was thinking more like ##g(x) = x + \sin(h)##. Would this work for a derivative? Thanks.
 
  • #4
docnet said:
Sorry! Do you mean that it wouldn't work because ##\lim_{h\to 0} x(1-\cos(h))=0##? Thank you for pointing out the example was a bad one. I was thinking more like ##g(x) = x + \sin(h)##. Would this work for a derivative? Thanks.
Just for the record, the example I referred to, was ##g(x)=x(1-\cos(h))##.
With the new example, I don't see how it differs from the standard definition. You could define there, ##h = \sin(p)##.
 
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  • #5
I want to add some extra 'conclusions' related to the 'alternative definition of derivative'. I am aware that anything I cooked up this quickly has a high probability of being wrong, and I'm not attaching anything of value in these statements. If this is correct, then,

If ##\lim_{h\to 0} g(x)=x##, then ##\lim_{h\to 0}\frac{f(g(x))-f(x)}{h}=f'(x).##

If ##\lim_{h\to 0^+} g(x)=x## but ##\lim_{h\to 0^-} g(x)= \text{undefined},## then ##g(x)## can be used to exactly describe the right derivative of ##f## at ##x##.

If ##\lim_{h\to 0^-} g(x)=x## but ##\lim_{h\to 0^+} g(x)= \text{undefined},## then ##g(x)## can be used to exactly describe the left derivative of ##f## at ##x##.
 
  • #6
docnet said:
Relevant Equations: .

If ##f(g(x)) =f(x+h)##,
##f(g(x))## is a constant. It is not a function of ##h##. If you want it to change as ##h \to 0##, you will have to specify how that happens. You might be interested in the chain rule: ##(f\circ g)' = (f' \circ g) \cdot g'##. So the result depends a lot on ##g'##.
 
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  • #7
FactChecker said:
##f(g(x))## is a constant. It is not a function of ##h##. If you want it to change as ##h \to 0##, you will have to specify how that happens. You might be interested in the chain rule: ##(f\circ g)' = (f' \circ g) \cdot g'##. So the result depends a lot on ##g'##.
Okay. I'm having a difficult time understanding. Can you explain more? Thank you.
Also, Is it fair to say ##f(g(x))## is a function of ##x## with a parameter ##h##, which is also the parameter that the limit is with respect to?

I'm a bit confused by why we have to differentiate ##g##. Are you implying that ##g## needs to be continuous and differentiable with respect to ##h##, as well as converging to ##x## as ##h\to 0##?Thank you.
 
  • #8
Why do you think this will be useful? g(x,h) is always going to have to look like x+h for small h.
 
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  • #9
docnet said:
I want to add some extra 'conclusions' related to the 'alternative definition of derivative'. I am aware that anything I cooked up this quickly has a high probability of being wrong, and I'm not attaching anything of value in these statements. If this is correct, then,

If ##\lim_{h\to 0} g(x)=x##, then ##\lim_{h\to 0}\frac{f(g(x))-f(x)}{h}=f'(x).##

If ##\lim_{h\to 0^+} g(x)=x## but ##\lim_{h\to 0^-} g(x)= \text{undefined},## then ##g(x)## can be used to exactly describe the right derivative of ##f## at ##x##.

If ##\lim_{h\to 0^-} g(x)=x## but ##\lim_{h\to 0^+} g(x)= \text{undefined},## then ##g(x)## can be used to exactly describe the left derivative of ##f## at ##x##.
Still a problem.
For example, let ##g(x)=x##.
Then ##\lim_{h\to 0} g(x)=x##.
But ##\lim_{h\to 0}\frac{f(g(x))-f(x)}{h}=\lim_{h\to 0}\frac{f(x)-f(x)}{h}=0##.
 
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  • #10
Frabjous said:
Why do you think this will be useful? g(x,h) is always going to have to look like x+h for small h.
I'm not sure if it's useful on its own. The reason I asked is because there was a similar formulation in a stochastic analysis homework at school, and I just wanted to understand it correctly.
 
  • #11
An informal comment. There's this integral called the Stieljes Integral. Yours is a sort of Stieljes derivative. Or maybe a sort of directional derivative.
 
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  • #12
You are replacing ##x+h##, a function of ##x## and ##h##, with ##g(x)##, a function only of ##x##. So ##f(g(x))## has nothing to do with ##h## and never changes as ##h \to 0##.

UPDATE: If you really meant ##\lim_{h \to 0} \frac {f(g(x+h))-f(g(x))} {h}## then you have the chain rule:
##(f\circ g)' = (f'\circ g)\cdot g'##
 
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  • #13
Finding counter examples is trivial. If you take ##g(x) = x + 2h## you get twice the usual derivative. If you take ##g(x) = x + h^2## you get zero.
 
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