# Dy/dx = 0

1. Aug 16, 2008

### anantchowdhary

If dy/dx=0,
for all x in the function y= f(x)'s domain,then
how can we say... only by using limits, that the function is a constant one?

2. Aug 16, 2008

### morphism

Only by using limits? What's wrong with the way this is usually proved, i.e. via the mean value theorem?

3. Aug 16, 2008

### anantchowdhary

the mean value theorem here....im not sure...cuz we might get a limit of the function staying constant..but i dont see how the function is EXACTLY constant, being proved by only using the first principle...

thanks

4. Aug 16, 2008

### GoldPheonix

As far as a limit definition:

dy/dx = lim[ F(x+h) - F(x)]/h

In order for that to be true, then:

F(x+h) - F(x) = 0

=>

F(x+h) = F(x)

Which means that you have an equation which does not vary with respect to x, otherwise there would be something left over from that subtraction. The only equation which does not vary with respect to x is a function that is constant, relative to x.

5. Aug 16, 2008

### anantchowdhary

cant F(x+h)-F(x) be <<h ...I dont think it can as h is a dynamic variable,which is infinitesimally small...so am i correct
Please correct me if im wrong
thanks

6. Aug 17, 2008

### nicksauce

If (A-B)/C = 0, and C is not 0, then it has to follow that A=B.

7. Aug 17, 2008

### anantchowdhary

But this is a limit

i mean lim(a--->b)a-b/c=0

it duznt mean a=b!

8. Aug 17, 2008

### WarPhalange

No, it's lim(c -> 0) of (a-b)/c

9. Aug 17, 2008

### anantchowdhary

ok...i got my mistake...but cant $$a-b$$ be something like $$c^2$$

then the limit still is 0 but we cant say a=b...can we?

10. Aug 17, 2008

### HallsofIvy

Staff Emeritus
In order that this be true, f must be differentiable at every point in the interval. Your limit calculations are not using that.

11. Aug 17, 2008

### morphism

This is not true. Take F(x)=x^2 for instance, and apply the limit definition at x=0:

$$F'(0) = \lim_{h \to 0} \frac{F(0+h) - F(0)}{h} = \lim_{h \to 0} \frac{h^2}{h} = 0.$$

But F(0+h)-F(0)=h^2 is not zero for all h.

This is why you need to use a deeper result like the mean value theorem. It tells you that if x is not equal to y, then there is a c such that

$$\frac{F(x) - F(y)}{x - y} = F'(c) = 0 \implies F(x) = F(y).$$

[Note: I'm assuming that F is differentiable everywhere so that the hypotheses of the MVT are satisfied.]

12. Aug 17, 2008

### anantchowdhary

Is it correct to simply say ..that in a function we cant have an infinitesimal so the limit of the derivative is not zero unitl it IS constant
or like we have a function $$f(x)=a(x)-b(x)$$
if a(x)=b(x) for all x

then a(x)=b(x) necessarily using the above argument(infinitesimals)

Last edited: Aug 17, 2008
13. Aug 17, 2008

### morphism

That doesn't really make any sense.

14. Aug 17, 2008

### anantchowdhary

ok...

15. Aug 18, 2008

### anantchowdhary

So how would we exactly prove the required result using only limits and logic?

thanks

16. Aug 18, 2008

### HallsofIvy

Staff Emeritus
Use limits (and logic) to prove the mean value theorem! One point that has been made repeatedly here is that you can't use limits at a single point because saying that y= 0 identically (on some interval) is a property of that interval, not a single point.

17. Aug 18, 2008

### anantchowdhary

but what if we have a function like...
$$f(x+y)=f(x)+y^3$$
and it is provided that the function is continuous and differentiable
so this gives us
$$f(x+h)=f(x)+h^3$$

or
$$\frac {f(x+h)-f(x)}{h}=h^2$$

$$\lim_{h \to 0} \frac {f(x+h)-f(x)}{h}=0$$

but i dont think the function is constant....
is there anything wrong in the function?

18. Aug 18, 2008

### HallsofIvy

Staff Emeritus
What reason do you have to believe such a function exists? If so, taking x= 0, f(y)= y3. But then
$$f(x+y)= (x+y)^3= x^3+ 3x^2y+ 3xy^2+ y^3= f(x)+ 3x^2y+ 3xy^2+ y^3\ne f(x)+ y^3$$

19. Aug 18, 2008

### anantchowdhary

I agree.. that the function wont exist...

but,using the MVT arent we still just proving the tendency not the EXACT equality?
i know this is irritating

but

thanks anyways

20. Aug 18, 2008

### morphism

No. The MVT is an exact equality. See: http://planetmath.org/encyclopedia/MeanValueTheorem.html [Broken].

Last edited by a moderator: May 3, 2017