Response function of derivative filter 
Problem: Consider a wide sense stationary stochastic process $X(t)$, with zero mean and auto correlation function $R_X(\tau)$. Consider its transformation by a linear time invariant derivative filter, which is the first derivative of $X(t)$, that is $Y(t)=X'(t)=\dfrac{\mathrm dX(t)}{\mathrm dt}$.
  Verify that the frequency response function of the associated linear invariant system in time is given by $H(w)=iw$($i$ is the imaginary unit).

Attempted solution: The inverse transform of $H(w)$ is given by
$$h(t)=\mathscr{F}^{-1}(H(w))=-\sqrt{2\pi}\delta(t),$$
so given the fact we have a linear system:
$$Y(t)=X(t)\ast h(t)=\int_{-\infty}^{\infty}X(\lambda)h(t-\lambda)\,\mathrm d\lambda=-\sqrt{2\pi}\int_{-\infty}^{\infty}X(\lambda)\delta(t-\lambda)\,\mathrm d\lambda.$$
However I have no idea on how this is going to yield me $\dfrac{\mathrm dX(t)}{\mathrm dt}$. I guess$$-\sqrt{2\pi}\int_{-\infty}^{\infty}X(\lambda)\delta(t-\lambda)\,\mathrm d\lambda=-\sqrt{2\pi}X(t-\lambda).$$
Question: Is this the way to go? How should I solve this exercise?
Thanks in advance!
 A: $\def\F{\mathscr{F}}\def\i{\mathrm{i}}\def\d{\mathrm{d}}\def\e{\mathrm{e}}$Lemma: If $X \in C^1(\mathbb{R})$ and $X, X' \in L^1(\mathbb{R})$, then $\F(X')(ω) = \i ω \F(X)(ω)$.
Proof: Since $\displaystyle \int_{-∞}^{+∞} |X'(t)| \,\d t < +∞$, then$$
\lim_{t → +∞} X(t) = X(0) + \lim_{M → +∞} \int_0^M X'(t) \,\d t = X(0) + \int_0^{+∞} X'(t) \,\d t,
$$
i.e. $X(+∞)$ exists, and $X \in L^1(\mathbb{R})$ implies that $X(+∞) = 0$. Analogously, $X(-∞) = 0$.
Now for any $ω \in \mathbb{R}$ and $M > 0$, integration by parts yields\begin{align*}
\int_{-M}^M X'(t) \e^{-\i ωt} \,\d t &= X(t) \e^{-\i ωt}\Biggr|_{-M}^M - \int_{-M}^M \frac{\partial}{\partial t}(\e^{-\i ωt}) X(t) \,\d t\\
&= X(M) \e^{-\i ωM} - X(-M) \e^{\i ωM} + \i ω \int_{-M}^M X(t) \e^{-\i ωt} \,\d t,
\end{align*}
letting $M → +∞$ yields$$
\int_{-∞}^{+∞} X'(t) \e^{-\i ωt} \,\d t = \i ω \int_{-∞}^{+∞} X(t) \e^{-\i ωt}.
$$
Thus $\F(X')(ω) = \i ω\F(X)(ω)$. $\qquad\square$
This exercise can be directly solved in the frequency domain. By the property of Fourier transform,$$
\F(Y)(ω) = \F(X')(ω) = \i ω\F(X)(ω),
$$
so$$
H(ω) = \frac{\F(Y)(ω)}{\F(X)(ω)} = \i ω.
$$
