Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Good evening,

I need to prove that $f'(x) = g'(x)$ for all $x \in (a,b)$ if and only if there exist a $c \in \mathbb{R}$ with $f = g+c$.
We know that $f:[a,b] \rightarrow \mathbb{R}$ and $g:[a,b] \rightarrow \mathbb{R}$ are continuous and differentiable on $(a,b)$.

My attempt:

Define $t(x):=c$
Let $t:[a,b] \rightarrow \mathbb{R}$ be continuous and differentiable on $(a,b)$.
$t'(x)=0$ for all $x \in (a,b)$ $\Rightarrow t(x)=c$ , $c \in \mathbb{R}$ for all $x \in [a,b]$.

Proof (with contraposition):

assume that $t(x)$ isn't $const$ on $[a,b]$
$\Rightarrow \exists x_0 \in [a,b]$ that $t'(x_0) \neq 0$
Under the assumption there exists a $x,y \in [a,b] : f(x) \neq f(y)$
$w.l.o.g.$ assume that $x<y \Rightarrow (x,y) \leq (a,b)$
$\Rightarrow t(x)$ is differentiable on $(x,y)$ and continuous on $(x,y) \leq (a,b)$
With the mean value theorem $\Rightarrow \exists \eta_0 \in (x,y): t(y)-t(x) = t'(\eta_0) \cdot (y-x)$
Under the assumption we know that $(y-x) > 0$ and $t(y)-t(x) \neq 0$.
$\Leftrightarrow \frac {t(y)-t(x)}{y-x} = t'(\eta_0) \Rightarrow f'(\eta_0) \neq 0 \Rightarrow Assumption.$

Now since we know that the differentiation of a constant function is zero we can define our $c:= f-g \Rightarrow f'(x)-g'(x)= 0 \Rightarrow f'(x) = g'(x)$.

Am I on the right track?

share|cite|improve this question
Well, it would be much easier to start with $(f-g)' = 0 \iff (f-g) = c$. – dtldarek Dec 2 '12 at 22:54
I think it is intended to show it that way since I came to a conclusion but not to a equivalence. Thank you. – Just a Student Dec 2 '12 at 23:09

Suppose $f'=g'$. $h=f-g$ must be differentiable and its derivative is $h'=f'-g'$.Since it identically vanishes, $h$ must be a constant from Mean Value theorem.

share|cite|improve this answer
I will prove it your way, thanks. – Just a Student Dec 2 '12 at 23:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.