Find $dG_t$ where $G_t = \frac{1}{t} \int^t_0 e^{W_u} du$ and $W_t$ is a Weiner process.

Question

I am very new to stochastic calculus and I am trying to find the stochastic derivative $dG_t$ of the process $$G_t = \frac{1}{t} \int^t_0 e^{W_u}du$$ where $W_t$ is a standard Brownian motion. Here is what I have: we see \begin{align*} dG_t = G_{t+dt} - G_t &= \frac{1}{t+dt}\int^{t+dt}_0 e^{W_{u}}du -\frac{1}{t} \int^t_0 e^{W_u} du \\ &= \frac{t}{t(t+dt)} \int^{t+dt}_0 e^{W_u} du - \frac{t+dt}{t(t+dt)} \int^t_0 e^{W_u}du \\ &= \frac{1}{t+dt} \int^{t + dt}_t e^{W_u} du - \frac{dt}{t(t+dt)} \int^t_0 e^{W_u} du \\ &= \frac{1}{t+dt} \int^{t + dt}_t e^{W_u} du - \frac{G_t dt}{t+dt}. \end{align*} Now I believe that by a formal Taylor expansion, we have $\frac{1}{t+dt} = \frac{1}{t}(1 - \frac{dt}{t})$ and since $e^{W_u}$ is continuous (a.s.) we have $$\frac{1}{t+dt} \int^{t+dt}_t e^{W_u} du = \frac{1}{t} e^{W_t}dt.$$ Thus $$dG_t = \frac{1}{t} \left(e^{W_t} - G_t \right) dt.$$ Is this correct? If not, could you please point out my error and point me in the right direction?