# F(x) = c => f'(x) = 0 ?

1. Nov 2, 2004

### danne89

f(x) = c => f'(x) = 0 ??

Hi again! Now I can't understand $f(x) = c\Rightarrow f'(x) = 0$, where c is a constant. I think I should be undefined.
y = c
$\Delta x$= c
$$\frac{\Delta y}{\Delta x} = \frac{c}{\Delta x}$$
$$st(\frac{c}{\Delta x}) = Undefined$$
What am I doing wrong this time?

Last edited: Nov 2, 2004
2. Nov 2, 2004

### arildno

y=c, but $$\bigtriangleup{y}=c-c=0$$

3. Nov 2, 2004

### Fredrik

Staff Emeritus
If f(x)=c for all x, then

$$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(h)}{h}=\lim_{h\rightarrow 0}\frac{c-c}{h}=\lim_{h\rightarrow 0}0=0$$

4. Nov 2, 2004

### danne89

Arghhh! That demostrates the importance of written every stage out.

y = c
y + $\Delta y$ = c + c
$\Delta y$ = c - c = 0
$$\frac{\Delta y}{\Delta x} = \frac{0}{\Delta x} = 0$$
Thus f'(x) = 0

Sorry, Fredrik. Haven't learned limits yet. Thanks for trying to explain.

Thanks!

5. Nov 3, 2004

### HallsofIvy

Staff Emeritus
Haven't learned limits yet???

Then what are you doing with the derivative? That's a lot like working with fractions before you have learned to multiply!

(Do you know the "slope" of a straight line? What's the slope of a horizontal straight line?)

6. Nov 3, 2004

7. Nov 3, 2004

### danne89

Sure I know that a constant function has zero slope, but I want to solve it the algebra-way.

The book I'm reading is "Elementary Calculus: An Approach Using Infinitesimals", which, as the titel states, uses infinitesimals to introduce both derivatives and integrals - before limits.

8. Nov 3, 2004

### danne89

Hmm. When I think about it: $\Delta y + y$ can't be c + c. Then
$$\Delta y = c + c - c = c$$
Argh!

9. Nov 3, 2004

### ReyChiquito

again, $y+\Delta y=c$, $\Delta y =0$

You have to be more carefull with algebra if you dont want to go insane (trust me)!

10. Nov 3, 2004

### HallsofIvy

Staff Emeritus
And how do they actually DEFINE "infinitesmal"?

11. Nov 3, 2004

### Chrono

It's a lot easier to understand derivatives if you know limits. I suggest you hit that before trying to understand the derivative of a constant.

12. Nov 4, 2004

### danne89

As an infinite small number (short: infinitesimal).

13. Nov 4, 2004

### Hurkyl

Staff Emeritus
Why should $\Delta x = c$?!?!

I think you've totally forgotten what $\Delta$ usually means:

$$(\Delta p)(q) = p(q + v) - p(q)$$

For some nonzero v. (For your purposes, v should be a nonzero infinitessimal)
(And yes, I know I used unusual variable names. :tongue:)

14. Nov 4, 2004

### Hurkyl

Staff Emeritus
I believe I've seen the text danne is using, it's not bad. It doesn't rigorously define infinitessimals, but normal calc textbooks don't rigorously define the real numbers either. :tongue:

In nonstandard analysis, the derivative of a standard function f is defined by:

$$f'(x) = \mathrm{Std} \frac{f^*(x^*+h) - f^*(x^*)}{h}$$

whenever the right hand side is independent of your choice of nonzero infinitessimal h. "Std" means "round to the nearest real number", and the * denotes the nonstandard version of that symbol.

Note that the notion of limit, here, has been replaced with the notion of nearest standard number.

Last edited: Nov 4, 2004
15. Nov 4, 2004

### danne89

Hmm. Isn't it $(\Delta p)(q) = p(q) + p(v) - p(q)$.
$$y = f(x)$$
$$y + \Delta y = f(x) + f(\Delta x)$$
$$\Delta y = f(x) + f(\Delta x) - f(x) = f(\Delta x)$$
And then, because f(x) = c
$$\Delta y = c + c - c = c$$

I think you denote your std() with st() and it's called the "Standard Part" of the hyperreal (reals + infinitesimals) number.

16. Nov 4, 2004

### ChanDdoi

hmmm ...isn't it (Delta p)(q) = q(p+v) - q(p) .

17. Nov 4, 2004

### danne89

Oops. My mistake. I should have checked that up before bother you. :(

18. Nov 4, 2004

### Hurkyl

Staff Emeritus
I think you're thinking of the more explicit notation:

$$(\Delta_v p)(q) = p(q + v) - p(q)$$