The values of $g'\left(\pm\frac\pi2\right)$ for $g(x)=\int_{\sin x}^{\sin 2x} \sin^{-1}(t) \,dt$ 
Question : If $g(x)=\displaystyle\int_{\sin x}^{\sin 2x} \sin^{-1}(t)  \,dt$, then :
$\text A) \, g'\left( -\dfrac{\pi}{2}\right)=2\pi$
$\text B) \, g'\left( -\dfrac{\pi}{2}\right)=-2\pi$
$\text C) \, g'\left( \dfrac{\pi}{2}\right)=2\pi$
$\text D) \, g'\left( \dfrac{\pi}{2}\right)=-2\pi$

Now, I used Newton-Leibniz rule and took derivative of $g(x)$ w.r.t $x$.
$$g(x)= \int_{\sin x}^{\sin 2x} \sin^{-1}(t)  \,dt$$
\begin{align}
\implies g'(x) &=\sin^{-1}(\sin 2x) \cdot (\sin 2x)'-\sin^{-1}(\sin x) \cdot (\sin x)'\\
&=\sin^{-1}(\sin 2x) \cdot 2 \cos 2x-\sin^{-1}(\sin x) \cdot \cos x
\end{align}
Hence, 
\begin{align}
g'\left( -\frac{\pi}{2}\right)&=\sin^{-1}(\sin (-\pi)) \cdot 2 \cos (-\pi)-\sin^{-1}\left(\sin \left(-\frac{\pi}{2}\right) \right) \cdot \cos \left(-\frac{\pi}{2}\right)\\
&=\sin^{-1}(0) \cdot 2 \cdot (-1)-\sin^{-1}(-1) \cdot 0\\
&=0
\end{align}
and
\begin{align}
g'\left( \frac{\pi}{2}\right)&=\sin^{-1}(\sin (\pi)) \cdot 2 \cos (\pi)-\sin^{-1}\left(\sin \left(\frac{\pi}{2}\right) \right) \cdot \cos \left(\frac{\pi}{2}\right)\\
&=\sin^{-1}(0) \cdot 2 \cdot (-1)-\sin^{-1}(1) \cdot 0\\
&=0
\end{align}
Where did I go wrong, because none of the options matches my answer.

Most probably this question is wrong. I am asking this question here because this is a question from a  very highly reputated engineering entrance examination.
All coaching institutes are saying that this question is bonus (i.e. since none of the options given is correct, its marks will be awarded to all the students, irrespective of what they marked). You can see it here (In case you are tired of searching, it's Q.49) But, I am doubtful for this because a similar scenerio once happened in year $2012$, all the coaching institutes uploaded their un-official answer key and answer given to a particular question was $0$ (in unofficial answer key), and when the official answer key was out, the answer for the same question was given to be $2$, which was the correct one. All the coaching institutes solved that question wrong.
Therefore, I request all mathematicians here to please verify my solution.
 A: Let $F(t)$ be the antiderivative of the integrand, $F'(t)=\sin^{-1}t$.
By the chain rule,
$$g'(x)=(F(\sin 2x)-F(\sin x))'=2\cos 2x\sin^{-1}(\sin 2x)-\cos x\sin^{-1}(\sin x).$$
This confirms your computation.
And indeed, $\sin 2x$ and $\cos x$ vanish at $\pm\pi/2$, making all expressions zero.

It is likely that the designer of the question erroneously simplified to
$$2\cos 2x\cdot2x-\cos x\cdot x,$$ giving
$$g'\left(\pm\frac\pi2\right)=\mp2\pi.$$
Unless this is a nasty trap.

By the way, notice that this antiderivative is 
$$t\sin^{-1}t+\sqrt{1-t^2}$$ which ranges in $[1,\frac\pi2]$ and can certainly not achieve a $2\pi$ difference.
A: $$g(x)=\int_{\sin x}^{\sin 2x} \sin^{-1}(t)dt=\int_{0}^{\sin 2x} \sin^{-1}(t)dt-\int_{0}^{\sin x} \sin^{-1}(t)dt$$
Next by denoting $h(u)=\int_{0}^{u}\sin^{-1}(t)dt$, one obtains
$$g(x)=h(\sin(2x))-h(\sin(x))$$
$$g'(x)=2\cos(2x)\arcsin((\sin(2x)))-\cos(x)\arcsin((\sin(x))).$$ 
Now placing $x= \frac{\pi}{2}$ (or $x=\pm\frac{\pi}{2}$) leads to
$$g'(\frac{\pi}{2})=2\cos(\pi)\arcsin((\sin(\pi)))-\cos(\frac{\pi}{2})\arcsin((\sin(\frac{\pi}{2})))=0,$$
confirming your calculations.
