Let $X$ be an exponential$(1)$ random variable defined on a probability space $(\Omega,F,P)$. That is, for any $a \geqslant 0$:
\begin{equation*} P \{ X \leqslant a \} = 1-e^{-a} \end{equation*}
Fix $t>0$. I would like to compute conditional expectation $\mathbb{E}[X \mid X \land t]$ in closed form in terms of $t$ ($X \land t$ means the minimum of $X$ and $t$). My attempt is to use the following machinery, which holds for every real random variable $X$ and $Y$ and every Borel function $h: \mathbb{R} \to \mathbb{R}$ such that $h(X) \in L^1(\Omega, F, P)$:
\begin{equation*} \mathbb{E}[h(X) \mid Y] = \int_{\mathbb{R}}h(x) f_{X \mid Y}(x \mid Y) dx \end{equation*}
Here $f_{X \mid Y}(x,y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}$, the quotient of joint probability density function and the probability density function of $Y$ (with $f_{X \mid Y}(x,y) = 0$ if $f_Y(y) = 0$).
It suffices to compute $ f_{X \land t}$, $f_{X, X \land t}$, and then the integral. Let $a>0$ be given.
First, we have $P\{ X \land t > a\} = P\{X > a\} = e^{-a}$ if $a<t$ and $P\{ X \land t > a\} = 0$ otherwise. Therefore, by taking complement and differentiation with respect to $a$, $f_{X \land t}(a) = e^{-a}$ if $a<t$ and $f_{X \land t}(a) = 0$ otherwise.
Next, let $a>0$ and $b>0$ be given. By using $X \land t = \frac{1}{2}(X+t-| X-t |)$ and discuss by cases whether $X>t$, I write:
\begin{equation*} P\{X > a, X \land t > b\} = P\{X>t, X>a, t > b \} + P\{X<t, X>a, X>b\} = P\{X \in (a \lor b,t)\} + P\{X > a \land t, t > b\} \end{equation*}
I conclude that $P\{X > a, X \land t > b\} $ has value $1$ if $t\leqslant b$, has value $1-e^{-t}$ if $t>b$ and $t \leqslant a$, and has value $1-e^{-a\lor b}$ if $t>b$ and $t>a$. If I again take complement, I get $1$ minus corresponding value on each of these cases. If I partial differentiate with respect to $a$ and then $b$, I get $0$ everywhere. What did I do wrong? Is there a more clever way to solve this problem?