I think you meant
[tex]\displaystyle \frac{d}{dx} \int_0^{f(x)} \frac{dt}{1+e^t}[/tex]
where
[tex]f(x) = \displaystyle \int_0^{\sin^{-1}(x)} \sec(t) \, dt[/tex]
By the fundamental theorem of calculus, we have
[tex]\displaystyle \frac{d}{dx} \int_0^{f(x)} \frac{dt}{1+e^t} = \frac{\frac{df}{dx}}{1+e^{f(x)}}[/tex]
as well as
[tex]\dfrac{df}{dx} = \sec\left(\sin^{-1}(x)\right) \times \dfrac{d\sin^{-1}(x)}{dx}[/tex]
The derivative of arcsine is
[tex]\dfrac{d\sin^{-1}(x)}{dx} = \dfrac1{\sqrt{1-x^2}}[/tex]
and so the overall derivative we want is
[tex]\displaystyle \frac{d}{dx} \int_0^{f(x)} \frac{dt}{1+e^t} = \frac{\sec\left(\sin^{-1}(x)\right)}{\left(1+e^{f(x)}\right)\sqrt{1-x^2}}[/tex]
We can further simplify [tex]e^{f(x)}[/tex], as
[tex]\displaystyle \int \sec(t) \, dt = \ln|\sec(t) + \tan(t)| + C[/tex]
[tex]\implies \displaystyle \int_0^{\sin^{-1}(x)} \sec(t) \, dt = \ln\left|\sec\left(\sin^{-1}(x)\right) + \tan\left(\sin^{-1}(x)\right)\right| = \ln\left|\frac{1+x}{\sqrt{1-x^2}}\right|[/tex]
[tex]\implies e^{f(x)} = \dfrac{1+x}{\sqrt{1-x^2}}[/tex]
Then the fully simplified derivative would be
[tex]\displaystyle \frac{d}{dx} \int_0^{f(x)} \frac{dt}{1+e^t} = \frac{\sec\left(\sin^{-1}(x)\right)}{\left(1+\dfrac{1+x}{\sqrt{1-x^2}}\right)\sqrt{1-x^2}} = \boxed{\frac{\sec\left(\sin^{-1}(x)\right)}{\sqrt{1-x^2}+1+x}}[/tex]
