Let $F(s)$ be the Laplace transform of $f(t)$:
$$F\left(s\right)=\int_{0}^{\infty}e^{-st}f\left(t\right)dt$$
It then follows that $f(t)$ can be recovered from $F(s)$ by the inverse Laplace transform:
$$f\left(t\right)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}F\left(s\right)ds$$
From the Laplace transform formula one can prove by integration by parts that the Laplace transform of the derivative $f'(t)$ is given by $sF(s)-f(0)$; so that, applying the inverse Laplace formula again:
$$f'\left(t\right)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}\left[sF\left(s\right)-f\left(0\right)\right]ds$$
However, if one differentiates with respect to $t$ the inverse Laplace transform formula giving $f(t)$ from $F(s)$, one obtains:
$$f'\left(t\right)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}sF\left(s\right)ds$$
and from this it seems that the Laplace transform of $f'(t)$ is just $sF(s)$. Since these two results are inconsistent, I think I'm missing something here. Can someone help me?