Answer
$\cos x\sqrt{1+(\sin x)^{4}}$
Work Step by Step
$\frac{d^{2}}{dx^{2}}\int_{0}^{x}\left(\int_{1}^{\sin t}\sqrt{1+u^{4}}du \right)dt$
$\frac{d}{dx}\left(\frac{d}{dx}\left(\int_{0}^{x}\left(\int_{1}^{\sin t}\sqrt{1+u^{4}}du \right) dt\right)\right)$
Suppose that $g(t)=\int_{1}^{\sin t}\sqrt{1+u^{4}}du$ it follows:
$\frac{d}{dx}\left(\frac{d}{dx}\left(\int_{0}^{x}\left(\int_{1}^{\sin t}\sqrt{1+u^{4}}du \right) dt\right)\right)=\frac{d}{dx}\left(\frac{d}{dx}\left(\int_{0}^{x}g(t) dt\right)\right)$
Using the fundamental theorem of calculus it follows:
$\frac{d}{dx}\left(\frac{d}{dx}\left(\int_{0}^{x}g(t) dt\right)\right)=\frac{d}{dx}\left(g(x)\right)$
We have $g(t)=\int_{1}^{\sin t}\sqrt{1+u^{4}}du \to g(x)=\int_{1}^{\sin x}\sqrt{1+u^{4}}du $
$\frac{d}{dx}\left(g(x)\right)=\frac{d}{dx}\left(\int_{1}^{\sin x}\sqrt{1+u^{4}}du\right)$
Use $\frac{d}{dx}\left(\int_{a}^{f(x)}h(u)du\right)=f'(x)h(f(x))$ it follows:
$\frac{d}{dx}\left(\int_{1}^{\sin x}\sqrt{1+u^{4}}du\right)=(\sin x)'\sqrt{1+(\sin x)^{4}}$
$=\cos x\sqrt{1+(\sin x)^{4}}$
