Prove that: $\frac{d}{dt} \int\_{-\infty}^{\infty} e^{-x^2} \cos(2tx) dx=\int\_{-\infty}^{\infty} -2x e^{-x^2} \sin(2tx) dx$

This is my proof:

$\forall \ t \in \mathbb{R}$ (the improper integral coverge absolutely $\forall \ t \in \mathbb{R}$) I consider: $g(t)=\int\_{-\infty}^{\infty} e^{-x^2} \cos(2tx) dx$. 

Let $h \ne 0$ 

$\left| \frac{g(t+h)-g(t)}{h}-\int\_{-\infty}^{\infty} -2x e^{-x^2} \sin(2tx) dx \right|\le\int\_{-\infty}^{\infty} \left|\frac{\cos(2(t+h)x)-\cos(2tx)}{h}-(-2x)\sin(2tx)\right| e^{-x^2}dx$

For the main value theorem and since $\int\_{-\infty}^{\infty} e^{-x^2} dx=\sqrt{\pi}$

$\int\_{-\infty}^{\infty} \left|\frac{\cos(2(t+h)x)-\cos(2tx)}{h}-(-2x)\sin(2tx)\right| e^{-x^2}dx=\left|\frac{\cos(2(t+h)\bar{x})-\cos(2t\bar{x})}{h}-(-2\bar{x})\sin(2t\bar{x})\right| \sqrt{\pi}$

$\cos(2tx)$ is derivable in $\bar{x}$ then fixed a $\epsilon>0 \ $ if $\ 0<|h|<\delta$:

$\left|\frac{\cos(2(t+h)\bar{x})-\cos(2t\bar{x})}{h}-(-2\bar{x})\sin(2t\bar{x})\right| \sqrt{\pi}<\sqrt{\pi} \ \epsilon$

It is correct?

There are other ways?

**UPDATE**

probably the proof is incorrect, because when I use the mean value theorem $x$ depends also from $h$ and hence the continuity of $x(h)$ is not obvious, then I can't guarantee the derivability of $\cos(2 t x(h))$ in $x$ for $h \rightarrow 0$. Am I right?