Help bounding this function to apply dominated convergence theorem

Question

I'm trying to proof that if $f:\mathbb{R}\rightarrow\mathbb{R}$ is a Borel-measurable function such that the function $x\mapsto$$e^{tx}f(x)$ is Lebesgue-intergrable, the the function $h:I\rightarrow\mathbb{R}$ (where $I$ is an open interval) defined as $h(t)=\int e^{tx}f(x) d\lambda$ is differentiable and for every $t\in$$I$, $h'(t)=\int xe^{tx}f(x) d\lambda$. I know that if I find a function $g(x)$ such that $|\frac{e^{tx}f(x)-e^{{t_0}x}f(x)}{t-t_0}|\leq$$g(x)$ for any $t,t_0\in$$I$ and $x\in\mathbb{R}$ then using the dominated convergence theorem i can conclude that $h(t)$ is differentiable and that $h'(t)=\int xe^{tx}f(x) d\lambda$, but I don't know how to define a $g(x)$ that works. If anyone can help me to find that function I will be really thankfull.

Answer 1

Hint:

By the mean value theorem, with $I = (a,b)$ there is a point $\xi$ between $t$ and $t_0$ such that

$$\left|\frac{e^{tx}f(x)-e^{{t_0}x}f(x)}{t-t_0}\right| = |e^{\xi x}f(x)| \leqslant \max(e^{ax},e^{bx})|f(x)|, $$

and the RHS is integrable.