Show that Laplace transform is differentiable

Let $g(x,t):=\exp(tx)f(x)$. We have to show that we can differentiate $g$ under the integral. To see that, we have to show that for all $t_0\in I$, we can find $\delta_t$ such that $|\partial_t g(x,u)|\leq G(x)$ for all $u\in (t-\delta_t,t+\delta_t)$ and $x\in\Bbb R$, where $G$ is integrable. We have $\partial_t g(x,u)=x\exp(xu)f(x)$. Let $r$ such that $(t-r;t+r)\subset I$. Then $$|\partial_t g(x,u)|=|x|\exp(x(u+r))e^{-rx}|f(x)|\\ \leq C_r \exp(x(u+r))|f(x)|\leq C_r\exp(x(t_0+2r))|f(x)|$$ for $x\geq 0$, and for $x<0$, $$|\partial_tg(x,u)|=|x|\exp(x(u-r))e^{rx}|f(x)|\leq C_r\exp(x(t_0-2r))|f(x)|.$$ As $x\mapsto \chi_{\Bbb R_+}(x)\exp(x(t_0+2r))|f(x)|$ and $x\mapsto \chi_{\Bbb R_-}\exp(x(t_0-2r))|f(x)|$ are integrable, we get the wanted result.