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We know that a function which is continuous on intreval $[a,b]\subset\mathbb{R}$ then it has certainly has IVP. But how can we guarantee that if a function $f$ is differentiable on $(a,b) $ then $f'$ has IVP?

if $f$ is diffentiable then $f$ must be continuous but this doesn't tell anything about the the continuity or monotonicity of $f'$. So how these two concepts are related to each other? HELP PLEASE. Thank You!

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The proof is different than for continuous case.

Take $x_1

Take $y$ with $f(x_1)

  1. Take the antiderivative of $f(t)-y$, which is $h(t)=f(t)-yt$
  2. $h$ is continuous on $[x_1,x_2]$ so it has a min and a max
  3. $h'(x_1)<0$ and $h'(x_2)>0$
  4. $h'(x_1)<0$ implies that the min of $h$ is not attained in $x_1$
  5. $h'(x_2)>0$ implies that the min of $h$ is not attained in $x_2$

So the min of $h$ is attained somewhere in $x \in (x_1,x_2)$. Now apply Fermat's theorem to obtain $h'(x)=f(x)-y=0$

You may also take a look here for the proof.