Question

`\red{ \rm (d)/(dx) \left \{ \int \limits_(0)^{ \large \int \limits _(0)^{ {sin}^( - 1) } \small{sect \: dt} } \frac{1}{1 + {e}^(t) } \: dt\right \}} \\`​

Answer 1

I think you meant

`\displaystyle (d)/(dx) \int_0^(f(x)) (dt)/(1+e^t)`

where

`f(x) = \displaystyle \int_0^{\sin^(-1)(x)} \sec(t) \, dt`

By the fundamental theorem of calculus, we have

`\displaystyle (d)/(dx) \int_0^(f(x)) (dt)/(1+e^t) = ((df)/(dx))/(1+e^(f(x)))`

as well as

`(df)/(dx) = \sec\left(\sin^(-1)(x)\right) * (d\sin^(-1)(x))/(dx)`

The derivative of arcsine is

`(d\sin^(-1)(x))/(dx) = \frac1{√(1-x^2)}`

and so the overall derivative we want is

`\displaystyle (d)/(dx) \int_0^(f(x)) (dt)/(1+e^t) = (\sec\left(\sin^(-1)(x)\right))/(\left(1+e^(f(x))\right)√(1-x^2))`

We can further simplify
`e^(f(x))`, as

`\displaystyle \int \sec(t) \, dt = \ln|\sec(t) + \tan(t)| + C`

`\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|(1+x)/(√(1-x^2))\right|`

`\implies e^(f(x)) = (1+x)/(√(1-x^2))`

Then the fully simplified derivative would be

`\displaystyle (d)/(dx) \int_0^(f(x)) (dt)/(1+e^t) = (\sec\left(\sin^(-1)(x)\right))/(\left(1+(1+x)/(√(1-x^2))\right)√(1-x^2)) = \boxed{(\sec\left(\sin^(-1)(x)\right))/(√(1-x^2)+1+x)}`