For convenience I use $x$ as the variable of integration. Let $ F(t)=I(u(t),v(t),t)=\int_{u(t)}^{v(t)}f(x,t)dx $
You have variable limits and need to differentiate under the integral sign. By the Leibniz integral rule, the derivative is
$\begin{eqnarray*} F^{\prime }(t) &=&\frac{\partial I}{\partial t}\frac{dt}{dt}+\frac{\partial I }{\partial v}\frac{dv}{dt}+\frac{\partial I}{\partial u}\frac{du}{dt} \\ &=&\int_{u(t)}^{v(t)}\frac{\partial f(x,t)}{\partial t}dx+f(v(t),t)v^{\prime }(t)-f(u(t),t)u^{\prime }(t). \end{eqnarray*}$
Hence $ F^{\prime }(0)=\int_{u(0)}^{v(0)}\left. \frac{\partial f(x,t)}{\partial t} \right\vert _{t=0}dx+f(v(0),0)v^{\prime }(0)-f(u(0),0)u^{\prime }(0). $
In the present case $ \begin{eqnarray*} u(t) &=&\sin t,\qquad u(0)=0 \\ u^{\prime }(t) &=&\cos t,\qquad u^{\prime }(0)=1 \\ v(t) &=&\cos t,\qquad v(0)=1 \\ v^{\prime }(t) &=&-\sin t,\qquad v^{\prime }(0)=0. \end{eqnarray*}$
and $ \begin{eqnarray*} f(x,t) =e^{x^{2}+xt}, \qquad f(1,0) =e, \qquad f(0,0) =1. \end{eqnarray*}$
So $ F^{\prime }(0)=\int_{0}^{1}\left. \frac{\partial f(x,t)}{\partial t} \right\vert _{t=0}dx+0-1 $
Since $ \left. \frac{\partial f(x,t)}{\partial t}\right\vert _{t=0}=\left. \frac{ \partial }{\partial t}e^{x^{2}+xt}\right\vert _{t=0}=xe^{x\left( x+t\right) }\vert _{t=0}=xe^{x^{2}} $
and $ \int_{0}^{1}xe^{x^{2}}dx=\frac{1}{2}e-\frac{1}{2} $
we finally get $ F^{\prime }(0)=\frac{1}{2}e-\frac{1}{2}-1=\frac{1}{2}e-\frac{3}{2}. $
You find another example here.