in the context of Brownian motions I have to solve this expression here: $\frac{\partial}{\partial{t}}\int^{\infty}_{x}{g_t(y)dy}$
where $g_t(y) = \frac{1}{\sqrt{2\pi t}}e^{\frac{-y^2}{2t}}$. I tried to swap derivative and integral and I came up with
$$\int^{\infty}_{x}{\frac{-1}{2t}g_t(y)+\frac{y^2}{2t^2}g_t(y)dy}$$ I know that the result should be $$\frac{x}{t}g_t(x)$$ but I don't know why. Can anyone help me?