Constant Rule Proof?

1. Jun 28, 2014

TheKracken

1. The problem statement, all variables and given/known data
For the Constant function f(x)=c prove that f'(x)= 0

2. Relevant equations
Alright so the relevant equation would be the Difference quotient or definition of a derivative.

3. The attempt at a solution
This is what I said

I restated the equation as this
f(x)=c >>>> f(c)= c(x^0)

Then I imputed into definition of derivative

(c(x+h)^0 - c(x^0))/ h

Then I solved it to be

(c-c)/h = 0/h
and then set h to 0 and then you get 0/0 which is non determinant form. I am not sure if this "proves" the statements to be true or if since it is 0/0 that doesn't show anything.

2. Jun 28, 2014

LCKurtz

You don't need an artificial $x^0$ factor. It is just $c-c$.
Then I solved it to be

(c-c)/h = 0/h = 0
and then set h to 0 and then you get 0/0 which is non determinant form. I am not sure if this "proves" the statements to be true or if since it is 0/0 that doesn't show anything.[/QUOTE]

That comes out zero before you let $h\to 0$ so the limit is zero. It is not indeterminante 0/0. So you're done.

3. Jun 28, 2014

TheKracken

So by including the artificial x^0 I actually made the proof incorrect? I am curious what is the most proper way to represent this proof? As I am going to be a math major, I don't really know how they would prefer this to be represented.

4. Jun 28, 2014

LCKurtz

I would write it like this. Given $f(x)=c$ then$$f'(0) = \lim_{h\to 0}\frac {f(0+h)-f(0)}{h} = \lim_{h\to 0}\frac {c-c}{h}= \lim_{h\to 0}\frac {0}{h}=\lim_{h\to 0} 0=0$$to be explicit and correct.

5. Jun 29, 2014

TheKracken

did you mean f'(c) ?

6. Jun 29, 2014

LCKurtz

Woops, I was thinking you wanted f'(0). Just put $x$ instead of $0$ in that difference quotient in the first two expressions. It's the same arithmetic.

7. Jun 29, 2014

verty

Also you need to understand or be able to see that the limit is 0. I mean this expression: $lim_{h \to 0} {0 \over h}$ is 0 for every h, it is a horizontal line heading toward the origin, so of course the limit is 0.