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

[danne89]

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?

[arildno]

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

[Fredrik]

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

[danne89]

y=c, but \bigtriangleup{y}=c-c=0
Arghhh! :mad: 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!

[HallsofIvy]

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?)

[Muzza]

I think he said in another thread that he was reading about non-standard analysis.

[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.

[danne89]

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

[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)!

[HallsofIvy]

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.

And how do they actually DEFINE "infinitesmal"?

[Chrono]

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

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.

[danne89]

And how do they actually DEFINE "infinitesmal"?
As an infinite small number (short: infinitesimal).

[Hurkyl]

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:)

[Hurkyl]

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.

[danne89]

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:)
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.

[ChanDdoi]

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

[danne89]

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

[Hurkyl]

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


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