# Thread: equivalence of functions and their derivatives

1. ## equivalence of functions and their derivatives

Hello!

I hope this is the correct place to ask this.
So, I need to find out this :
If f~g then f'~g'? Where x seeks to 0, and f,g - functions, that can be diferentiated.
Checked some examples with functions and they all gives positive answer to the question. But cannot find how to proove it in general, or maybe it is impossible?

2. What exactly do you mean by f~g in this context? ( what kind of equivalence)

3. Originally Posted by Dinkydoe
What exactly do you mean by f~g in this context? ( what kind of equivalence)
well, i hadn't thought about it. but in way i look at this task, f~g if dom(f)=dom(g), ran(f)=ran(g) and f(x)=g(x).
example: f(x)=sin(arccos(x))
g(x)=sqrt(1-x^2), both of them in my opinion are equivalent.

f'(x)=(sin(arccos(x)))'=-x/sqrt(1-x^2)
g'(x)=(sqrt(1-x^2))'=-x/sqrt(1-x^2), those obviously are equivalent.

4. Well yes, if that's what you mean, those two functions are identical in the litteral sense. What you said is in fact nothing more then stating: $x= \frac{x^2}{x}$

Ofcourse, you can define a equivalence relation on a set of functions: Something like f~g iff f(x)=g(x) almost everywhere (that is, these functions may disagree on a finite (or countable set of points).

5. Originally Posted by Dinkydoe
Well yes, if that's what you mean, those two functions are identical in the litteral sense. What you said is in fact nothing more then stating: $x= \frac{x^2}{x}$

Ofcourse, you can define a equivalence relation on a set of functions: Something like f~g iff f(x)=g(x) almost everywhere (that is, these functions may disagree on a finite (or countable set of points).
I don't see how this is an equivalence relation? What is to guarantee transitivity?

6. $f,g,h:[a,b]\to\mathbb{R}$. Given f~g, g~h. Then f,g disagree on a countable set of points, g,h disagree on a countable set of points. Then f,h disagree on a countable set of points as well. Hence f~h. I don't exactly see the problem there.

7. Now, do you have any ideas about how to proove that if f~g, then f'~g' ? I was thinking about L'Hopital rule, maybe I can use that?

8. Originally Posted by Dinkydoe
Well yes, if that's what you mean, those two functions are identical in the litteral sense. What you said is in fact nothing more then stating: $x= \frac{x^2}{x}$
Except that, as functions, x is NOT equal to $\frac{x^2}{x}$.
The function, x, is defined for all x, the function, $\frac{x^2}{x}$ is defined for all x except 0. They are not the same function because the have different domains.

Ofcourse, you can define a equivalence relation on a set of functions: Something like f~g iff f(x)=g(x) almost everywhere (that is, these functions may disagree on a finite (or countable set of points).

9. Originally Posted by waytogo
Now, do you have any ideas about how to proove that if f~g, then f'~g' ? I was thinking about L'Hopital rule, maybe I can use that?
You have still not made clear what you mean by "~". You said before "well, i hadn't thought about it. but in way i look at this task, f~g if dom(f)=dom(g), ran(f)=ran(g) and f(x)=g(x).
example: f(x)=sin(arccos(x))
g(x)=sqrt(1-x^2), both of them in my opinion are equivalent."
In that case, f and g are the same function, regardless of what "formula" you use to represent them.
sin(arccos(x)) and $\sqrt{1- x^2}$ are just different ways of writing the same function.

What is true is that if f(x)= g(x) on some interval around $x_0$, then $f'(x_0)= g'(x_0)$ because the derivative is a local property.

10. Except that, as functions, x is NOT equal to .
Yeah, good point. Bad example, meant to get the same point across ;p

(I shouldve apprached them as functions: $\mathbb{R}\setminus\left\{0\right\} \to \mathbb{R}$))

11. ok, good lesson for me. i agree that the task is not given clearly.
now - i precised this detail of equivalence with my professor.
f~g in this way means that (limit f)/(limit g)=1. i guess x seeks to the same for both functions.