Question

5E%7B%20%5Clarge%20%5Cint%20%5Climits%20_%7B0%7D%5E%7B%20%7Bsin%7D%5E%7B%20-%201%7D%20%7D%20%5Csmall%7Bsect%20%5C%3A%20dt%7D%20%7D%20%5Cfrac%7B1%7D%7B1%20%2B%20%7Be%7D%5E%7Bt%7D%20%7D%20%5C%3A%20dt%5Cright%20%5C%7D%7D%20%5C%5C%20" id="TexFormula1" title=" \red{ \rm \frac{d}{dx} \left \{ \int \limits_{0}^{ \large \int \limits _{0}^{ {sin}^{ - 1} } \small{sect \: dt} } \frac{1}{1 + {e}^{t} } \: dt\right \}} \\ " alt=" \red{ \rm \frac{d}{dx} \left \{ \int \limits_{0}^{ \large \int \limits _{0}^{ {sin}^{ - 1} } \small{sect \: dt} } \frac{1}{1 + {e}^{t} } \: dt\right \}} \\ " align="absmiddle" class="latex-formula">​

Answer 1

I think you meant

$\displaystyle \frac{d}{dx} \int_0^{f(x)} \frac{dt}{1+e^t}$

where

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

By the fundamental theorem of calculus, we have

$\displaystyle \frac{d}{dx} \int_0^{f(x)} \frac{dt}{1+e^t} = \frac{\frac{df}{dx}}{1+e^{f(x)}}$

as well as

$\dfrac{df}{dx} = \sec\left(\sin^{-1}(x)\right) \times \dfrac{d\sin^{-1}(x)}{dx}$

The derivative of arcsine is

$\dfrac{d\sin^{-1}(x)}{dx} = \dfrac1{\sqrt{1-x^2}}$

and so the overall derivative we want is

$\displaystyle \frac{d}{dx} \int_0^{f(x)} \frac{dt}{1+e^t} = \frac{\sec\left(\sin^{-1}(x)\right)}{\left(1+e^{f(x)}\right)\sqrt{1-x^2}}$

We can further simplify $e^{f(x)}$, as

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

$\implies \displaystyle \int_0^{\sin^{-1}(x)} \sec(t) \, dt = \ln\left|\sec\left(\sin^{-1}(x)\right) + \tan\left(\sin^{-1}(x)\right)\right| = \ln\left|\frac{1+x}{\sqrt{1-x^2}}\right|$

$\implies e^{f(x)} = \dfrac{1+x}{\sqrt{1-x^2}}$

Then the fully simplified derivative would be

$\displaystyle \frac{d}{dx} \int_0^{f(x)} \frac{dt}{1+e^t} = \frac{\sec\left(\sin^{-1}(x)\right)}{\left(1+\dfrac{1+x}{\sqrt{1-x^2}}\right)\sqrt{1-x^2}} = \boxed{\frac{\sec\left(\sin^{-1}(x)\right)}{\sqrt{1-x^2}+1+x}}$