I have the following expression:
$$ \mathbb{E}\left[C\right] = \mathbb{E}\left[\alpha \max\{\tilde{\tau} - t , 0 \} + \gamma \tilde{\tau} \right]$$
where $\tilde{\tau} = \tau + \sigma \tilde{\varepsilon}$ and where $\tilde{\varepsilon}$ is a standard normal variable with mean zero and variance equal to one. $\alpha$, $\gamma$ and $\sigma$ are just parameters. I need to take the derivative of this expression with respect to $\tau$. Before doing that I re-write the expression above as:
$$ \mathbb{E}\left[C\right] = \mathbb{E}\left[ \alpha \max\{\tau +\sigma \tilde{\varepsilon} - t , 0 \} + \gamma \left(\tau + \sigma \tilde{\varepsilon} \right) \right]$$
$$ \mathbb{E}\left[C\right] = \mathbb{E}\left[ \alpha \max\{\tau +\sigma \tilde{\varepsilon} - t , 0 \} \right] + \gamma\tau + \sigma \mathbb{E} \left[\tilde{\varepsilon} \right]$$
$$ \mathbb{E}\left[C\right] = \alpha \mathbb{E}\left[\max\{\tau +\sigma \tilde{\varepsilon} - t , 0 \} \right] + \gamma\tau $$
The derivative with respect to $\tau$ is:
$$ \dfrac{\partial \mathbb{E}\left[C\right]}{\partial \tau} = \alpha \dfrac{\partial}{\partial \tau} \int \max\{\tau +\sigma \tilde{\varepsilon} - t , 0 \} f(\tilde{\tau}) d \tilde{\tau}+ \gamma $$
I assume the functions satisfy the conditions that allow us to interchange differentiation and integration, hence
$$ \dfrac{\partial \mathbb{E}\left[C\right]}{\partial \tau} = \alpha \int \dfrac{\partial}{\partial \tau} \max\{\tau +\sigma \tilde{\varepsilon} - t , 0 \} f(\tilde{\tau}) d \tilde{\tau}+ \gamma $$
Using the product rule:
$$ \dfrac{\partial \mathbb{E}\left[C\right]}{\partial \tau} = \alpha \int \max\{1,0\}f(\tilde{\tau}) d \tilde{\tau} + \int \max\{\tau +\sigma \tilde{\varepsilon} - t , 0 \} f'(\tilde{\tau}) d \tilde{\tau}+ \gamma $$
$$ \dfrac{\partial \mathbb{E}\left[C\right]}{\partial \tau} = \alpha \int f(\tilde{\tau}) d \tilde{\tau} + \int \max\{\tau +\sigma \tilde{\varepsilon} - t , 0 \} f'(\tilde{\tau}) d \tilde{\tau}+ \gamma $$
$$ \dfrac{\partial \mathbb{E}\left[C\right]}{\partial \tau} = \alpha F(\tilde{\tau}) + \int \max\{\tau +\sigma \tilde{\varepsilon} - t , 0 \} f'(\tilde{\tau}) d \tilde{\tau}+ \gamma $$
The next step is where I am a bit doubtful. Does the second term go to zero because I can redefine $f'(\tilde{\tau})$ with $\varepsilon$ so that computing the expectation of $\varepsilon$ gives me zero and makes the second term vanish... or can "I distribute" the integral inside the max operator and exploit the fact that $\varepsilon \sim \mathcal{N}(0,1)$, so that all of these terms in the max operator have mean zero?
Finally, last question: The official answer is:
$$ \dfrac{\partial \mathbb{E}\left[C\right]}{\partial \tau} = \alpha \mathbb{P}(\tau + \sigma \varepsilon - t > 0) + \gamma $$
I understand that $F(a)$ means $\mathbb{P}(\tilde{\tau}<a)$ but I am a bit confused with getting from $F(\tilde{\tau})$ to
$$ \mathbb{P}(\tau + \sigma \varepsilon - t > 0) $$