I know that the derivative of a constant is zero, but the only proof that I can find is:
given that $f(x) = {x}^{0}$,
$$ f'(x) = \lim_{h\to 0} \frac {f(x+h) - f(x)}{h} $$
$$ f'(x) = \lim_{h\to 0} \frac {{(x+h)}^{0} - {x}^{0}}{h} $$
and then because $ {(x+h)}^{0} - {x}^{0} = 1 - 1 = 0 $, then
$$ f'(x) = \lim_{h\to 0} \frac {0}{h} = 0 $$
This seems sort of fishy to me, however, as if you plug in 0 for h in the limit, you get an indeterminate. It might just be me, though, but it just doesn't seem entirely right to me. Is this proof actually fine, and there another proof that the derivative of a constant is zero, or is this the only one?
3 Answers
$\begingroup$The "$x^0$" really plays no role; besides, it is $1$ and any other constant works with the same proof.
In any case, as you write, if $f(x)=c$ for some number $c$, then $$ \frac{f(x+h)-f(x)}h=\frac{c-c}h=0. $$ So, the Newton quotient is already zero before taking the limit as $h\to0$.
$\endgroup$ 1 $\begingroup$Keep in mind that "limit" has a precise, formal meaning: $$\lim_{x\rightarrow c}f(x)=L \quad\iff\quad \forall \epsilon>0\exists \delta>0\forall x[0<\vert x-c\vert<\delta\implies \vert f(x)-L\vert<\epsilon].$$ In words, the limit as $x$ approaches $c$ of $f(x)$ is $L$ if, as we approach (but not reach - this is the "$0<$" clause) $c$, the value of $f$ approaches $L$.
For any $h$ other than $0$, we have ${0\over h}=0$; so indeed $\lim {0\over h}=0$. The fact that, at $h=0$, the expression $0\over h$ is undefined - and as a consequence, doesn't equal $0$ - doesn't effect the limit; it just means that the function isn't as nicely behaved at $h=0$ as it could be.
$\endgroup$ 2 $\begingroup$I had the same question, in an effort to understand the essence of the rule. I can't say this is rigidly correct, or even correct, but this is the most simple explanation I can come up with. It's nearly totally algebraic, and builds a calculus expression from an algebra expression using the "method of increments".
Note: I'm using the triangle here. 'Δy/Δx' refers simply to the change in y in relation to the change in x. I use this because dy/dx seems less intuitive and a little like black magic.
(1) y = C; which is equivalent to y = C + 0(x)
(2) Using the method of increments, a change in x corresponds to a certain change in y, so: y + Δy = C + 0(x + Δx)
(3) Since y equivalent to C + 0(x) ...
(4) C + 0(x) + Δy = C + 0(x) + 0(Δx)
(5) Algebraic manipulation: remove 'C + 0(x)' from both sides of the equation.
(6) Then divide both sides of the equation by Δx.
(7) You end up with Δy/Δx = 0(Δx/Δx).
(8) So, Δy/Δx = 0
(9) which is to say dy/dx = 0, meaning there is 0 change in y when there is a change in x. Normally in calculus, we're concerned with the dy/dx ratio in the case of dx (i.e., Δx) approaching 0, but that's irrelevant in this case.
