Is this a valid representation of a derivative?

docnet · Tuesday, 6:56 PM

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!

Hill · Tuesday, 7:03 PM

My immediate observation is that in this "alternative" definition, the derivative in ##x=0## is identically ##0##. This is a problem.

docnet · Tuesday, 7:08 PM

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.

Hill · Tuesday, 7:20 PM

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)##.

docnet · Tuesday, 7:23 PM

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##.

FactChecker · Tuesday, 7:23 PM

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'##.

docnet · Tuesday, 7:33 PM

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.

Frabjous · Tuesday, 7:41 PM

Why do you think this will be useful? g(x,h) is always going to have to look like x+h for small h.

Hill · Tuesday, 7:43 PM

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##.

docnet · Tuesday, 7:45 PM

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.

WWGD · Tuesday, 8:20 PM

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.

FactChecker · Tuesday, 8:31 PM

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'##

Orodruin · 2024-03-20T01:12:46-0500

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.

Is this a valid representation of a derivative?

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