Is this a valid representation of a derivative?

[docnet]

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!

 

[Hill]

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

 

[docnet]

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]

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]

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]

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]

$f(g(x))$ 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]

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]

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]

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]

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]

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]

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.