Let f(x) and g(x) be defined and differentiable for all $x \geq x_0$ and $f\left(x_{0)}=g\left(x_0\right), f'(x)>g'(x)\right.$ for $x>x_0$, then

$f(x)>g(x), x>x_0$

Let $h(x)=f(x)-g(x)$ for all $x \geq x_0$.

Since, f(x) and g(x) are differentiable for all $x \geq x_0$. Therefore, so is h(x).

Now,

$h(x)=f(x)-g(x)$ for all $x \geq x_0$

$\Rightarrow h'(x)=f'(x)-g'(x)$

$\Rightarrow h'(x)>0 $ for all $x>x_0$             [∵ f'(x) > g'(x) for all $x>x_0$]

⇒ h(x) is an increasing function for all $x>x_0$

$\Rightarrow h(x)>h\left(x_0\right) $ for all $x>x_0 $

$\Rightarrow h(x)>0 $ for all $x>x_0$          $\left[∵ h\left(x_0\right)=f\left(x_0\right)-g\left(x_0\right)=0\right]$

$\Rightarrow f(x)>g(x) $ for all $x>x_0$