Answer
\[F'(x)=-\sqrt{x+\sin x}\]
Work Step by Step
We will use the rule \[\frac{d}{dx}\int_{f(x)}^{g(x)}F(t)dt=F(g(x))\cdot g'(x)-F(f(x))\cdot f'(x)\;\;\; ...(1)\]
Given that \[F(x)=\int_{x}^{1}\sqrt{t+\sin t}\;dt\;\;\;...(2)\]
Differentiating (2) with respect to $x$ using (1)
\[F'(x)=\sqrt{1+\sin 1}\cdot (1)'-\sqrt{x+\sin x}\cdot (x)'\]
\[F'(x)=\sqrt{1+\sin 1}\cdot (0)-\sqrt{x+\sin x}\cdot (1)\]
\[\Rightarrow F'(x)=-\sqrt{x+\sin x}\]
Hence \[F'(x)=-\sqrt{x+\sin x}\]
