Consider the following equation for $X(t)$:

$$X(t)=e^{-bt}X(0)+\sigma\int_{0}^{b}e^{-b(t-s)}dW(t) \, ,$$

where $0 < b, \sigma\in\mathbb{R} $, $X(0)$ is the initial distribution of $X(t)$, independent of the Brownian motion $W(t)$. I want so show that $dX(t)= -bX(t)dt+\sigma dW(t)$, but I am getting stuck on computing the derivative of $$\sigma\int_{0}^{b}e^{-b(t-s)}dW(t) \, .$$

Could someone please give me some ideas? Thanks so much for your tim.

PS. the above equation is one of type of the Langevin's equation, more detail could be found here http://en.wikipedia.org/wiki/Langevin_equation