Using the standard method to find the derivative of $f(t)=2\sqrt{1+\sin{t}}$ yields $f'(t)=\sqrt{1-\sin{t}}$ but plotting both of the equations into a graph shows that this is not true.
The plot can be seen here: WolframAlpha
The thing to note is that at around $\pi$, $f(\pi)$ is decreasing, meaning that its derivative $f'(\pi)$ should be negative; but instead, $\sqrt{1-\sin{\pi}} = 1$. And the definite integral of $f'(t)$ from $0$ to $2\pi$ should not be $0$, but $f(2\pi)-f(0)=0$.
Is this because $f(t)$ is not continuous?