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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?

nworb99
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    for $h\neq 0$, $\frac{0}{h}=0$. $lim_{h\to 0}0=0$ – user153582 Dec 09 '15 at 01:48
  • The point of a derivative limit is not to just plug in 0, it's the question of what value is being approached for smaller and smaller values of $h$. In this case your argument is perfectly fine since for smaller values the thing just remains as 0. Whoever told you to just plug in 0 did you a great disservice because when you think about it, the case for $h=0$ means the question doesn't exist at all! – Nεo Pλατo Mar 21 '20 at 10:54

3 Answers3

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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 $h\to0$.

Martin Argerami
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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.

Noah Schweber
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    There's a separate question, which is Why is this the right definition of "limit"? But that's a different issue. Part of the reason is that we want some language to talk about what expressions like "$f(x+h)-f(x)\over h$" ought to mean for $h=0$, and in such cases we clearly don't have good behavior of the function at the point of interest, so what we want is a definition which takes into account "local" behavior but isn't tethered to behavior at the "exact" value. – Noah Schweber Dec 09 '15 at 01:50
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    I guess I was overthinking it in that respect, thanks – nworb99 Dec 09 '15 at 01:51
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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.

Lymey
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