Let g(t) denote the inner integral. By the fundamental theorem of calculus, the first derivative is
[tex]\displaystyle \frac{d}{dx} \int_0^x g(t) \, dt = g(x)[/tex]
Then using the FTC again, differentiating g gives
[tex]\displaystyle \frac{dg}{dx} = \frac{d}{dx} \int_1^{\sin(x)} \sqrt{1+u^4} \, du = \boxed{\cos(x) \sqrt{1+\sin^4(x)}}[/tex]
