# How could I prove this? Would this be using a theorem from real analysis?

May 17, 2018

$\text{Use the definition of derivative :}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

$\text{Here we have}$

$f ' \left({x}_{0}\right) = {\lim}_{h \to 0} \frac{f \left({x}_{0} + h\right) - f \left({x}_{0}\right)}{h}$
$g ' \left({x}_{0}\right) = {\lim}_{h \to 0} \frac{g \left({x}_{0} + h\right) - g \left({x}_{0}\right)}{h}$

$\text{We need to prove that}$

$f ' \left({x}_{0}\right) = g ' \left({x}_{0}\right)$
$\text{or}$
$f ' \left({x}_{0}\right) - g ' \left({x}_{0}\right) = 0$
$\text{or}$
$h ' \left({x}_{0}\right) = 0$
$\text{with } h \left(x\right) = f \left(x\right) - g \left(x\right)$
$\text{or}$
${\lim}_{h \to 0} \frac{f \left({x}_{0} + h\right) - g \left({x}_{0} + h\right) - f \left({x}_{0}\right) + g \left({x}_{0}\right)}{h} = 0$
$\text{or}$
${\lim}_{h \to 0} \frac{f \left({x}_{0} + h\right) - g \left({x}_{0} + h\right)}{h} = 0$
$\text{(due to "f(x_0) = g(x_0)")}$

$\text{Now}$

$f \left({x}_{0} + h\right) \le g \left({x}_{0} + h\right)$
$\implies \lim \le 0 \text{ if "h>0" and "lim >= 0" if } h < 0$

$\text{We made the assumption that f and g are differentiable}$
$\text{so "h(x) = f(x) - g(x)" is also differentiable,}$
$\text{so the left limit must be equal to the right limit, so}$

$\implies \lim = 0$
$\implies h ' \left({x}_{0}\right) = 0$
$\implies f ' \left({x}_{0}\right) = g ' \left({x}_{0}\right)$

May 18, 2018

I will provide a quicker solution than the one in https://socratic.org/s/aQZyW77G. For this we will have to rely on some familiar results from calculus.

#### Explanation:

Define $h \left(x\right) = f \left(x\right) - g \left(x\right)$

Since $f \left(x\right) \setminus \le g \left(x\right)$, we have $h \left(x\right) \le 0$

At $x = {x}_{0}$ , we have $f \left({x}_{0}\right) = g \left({x}_{0}\right)$, so that $h \left({x}_{0}\right) = 0$

Thus $x = {x}_{0}$ is a maximum of the differentiable function $h \left(x\right)$ inside the open interval $\left(a , b\right)$. Thus

${h}^{'} \left({x}_{0}\right) = 0 \implies$

${f}^{'} \left({x}_{0}\right) - {g}^{'} \left({x}_{0}\right) \implies$

${f}^{'} \left({x}_{0}\right) = {g}^{'} \left({x}_{0}\right)$