Derivative of $2\sqrt{1+\sin{t}}$ 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?
 A: What I think of as the standard method leads mechanically to $\dfrac{\cos t}{\sqrt{1+\sin t}}$. 
If we now replace $\cos t$ by $\sqrt{1-\sin^2 t}$, which is presumably how you obtained your expression, we are sometimes making a mistake. For when $\cos t$ is negative, we have $\cos t=-\sqrt{1-\sin^2 t}$. That would account for the trouble at $\pi$.  
A: $f(t)$ is, in fact, continuous since $1+\sin(t)\geq0$ for all $t$. The problem is the derivative you have. Here, you would have to use the chain rule with $g(t)=2\sqrt{t}$ and $h(t)=1+\sin(t)$. Then $f(t)=g\circ h(t)$, and when we take the derivative of $f$, we end up with $$f'(t)=g'(h(t))\cdot h'(t)=\frac{\cos(t)}{\sqrt{1+\sin(t)}}.$$ Hope this helps!
A: The derivative of this function is $f'(t) = \frac{\cos(t)}{\sqrt{1+\sin(t)}}$. Writing $\cos(t) = \sqrt{\cos^2(t)} = \sqrt{1 - \sin^2(t)}$ you get the simplification to your expression. However, this leads to a mistake, especially for $\frac{\pi}{2} < t < \pi$. Namely, for $t$ in this range, $\cos(t) <0$ which implies $f'(t)<0$, but the simplification absorbs the negative sign as a square inside the square root, hence the wrong answer. If you compute $f'(\pi) = \frac{\cos(\pi)}{\sqrt{1 + \sin(\pi)}} = -1$ you see that the function is decreasing, as expected. The function is not only continuous, it is differentiable for $t < \frac{3\pi}{2}$.
