# Functions that have the same derivative

Let’s say I have two continuous functions $$f(x)$$ and $$g(x)$$ , and both have the same derivative $$h(x)$$. How could I formally show that $$f(x)=g(x)+c$$ where $$c$$ is a constant. I know I have to show that $$f(x)-g(x)$$ is a constant function but not sure how? Thanks

• Welcome to MSE. In the future please learn to use MathJax to properly format math expressions. – Lee David Chung Lin Apr 21 at 16:34
• Yep thanks, its hard on my phone though but I’ll get there – Eden Hazard Apr 21 at 16:38

Beware that the statement is not true, unless you make some further assumptions.

If $$f$$ and $$g$$ are continuous functions over an interval $$I$$, differentiable in the interior of $$I$$, and $$f'(x)=g'(x)$$ for every $$x$$ in the interior of $$I$$, then there exists a constant $$k$$ such that $$f(x)=g(x)+k$$, for every $$x\in I$$.

This is nothing else than the mean value theorem: consider $$h(x)=f(x)-g(x)$$. Then, for $$a in $$I$$, $$\frac{h(b)-h(a)}{b-a}=h'(c)$$ for some $$c\in(a,b)$$. But then $$h'(c)=0$$, so $$h(b)=h(a)$$ and therefore $$h$$ is constant.

The usual counterexample is $$f(x)=\arctan x,\qquad g(x)=-\arctan\frac{1}{x}$$ defined over $$\mathbb{R}\setminus\{0\}$$. They have the same derivative, but the difference is not constant. Indeed $$f(x)-g(x)=\begin{cases} \pi/2 & x>0 \\[4px] -\pi/2 & x<0 \end{cases}$$

$$f'(x)-g'(x) \equiv 0$$ given that their derivatives are the same. Now we can apply FTC to this (integrate between some lower limit $$a$$ and $$x$$), and this will give you your answer!

• Careful. Some proofs of the FTC use the theorem the OP wants to prove. – Ethan Bolker Apr 21 at 16:16

It suffices to show that if $$A'=0$$ then $$A(x)=c$$ for some $$c$$. But this is clear by the mean value theorem. Indeed fix $$x\in \mathbb{R}$$ and apply the mean value theorem to the interval $$[ x, y]$$ to deduce that $$A(y)=A(x)$$ for all $$y>x$$. Similarly one can deduce that $$A(y)=A(x)$$ for all $$y whence $$A$$ is constant.

$$f^{'}(x) = g^{'}(x)$$ Integrating both sides with respect to x . $$\int f^{'}(x) = \int g^{'}(x)$$ $$f(x) +C_1= g(x) +C_2$$ $$f(x) - g(x) = C_2 - C_1$$ Since $$C_1$$ is a constant and also $$C_2$$ is a constant so $$C_2-C_1$$ is also a constant Thus $$f(x) - g(x) = C$$

• You don't multiply both sides by $\mathrm{d}x$ and integrate. You "integrate both sides with respect to $x$". Your phrasing makes it sound as if integration can occur independently of choice of variable of integration, i.e., that the left bracket, $\displaystyle \int$, and the right bracket, $\mathrm{d}\text{[some variable]}$, are separable. – Eric Towers Apr 21 at 16:17
• This begs he question. The "+C_2\$" in your integral uses what the OP wants to prove. – Ethan Bolker Apr 21 at 16:18