Does f(x) = c Imply That f'(x) = 0?

[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?

 

[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! 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 don't 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. )

 

[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.

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

<br /> f&#039;(x) = \mathrm{Std} \frac{f^*(x^*+h) - f^*(x^*)}{h}<br />

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. )

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:

<br /> (\Delta_v p)(q) = p(q + v) - p(q)<br />