Answer
$$s'(t)=\frac{\sin t+\cos t+1}{2\sqrt{1+\sin t}(1+\cos t)^{3/2}}$$
Work Step by Step
$$s'(t)=\left(\sqrt{\frac{1+\sin t}{1+\cos t}}\right)'=\frac{1}{2\sqrt{\frac{1+\sin t}{1+\cos t}}}\left(\frac{1+\sin t}{1+\cos t}\right)'=
\frac{\sqrt{1+\cos t}}{2\sqrt{1+\sin t}}\frac{(1+\sin t)'(1+\cos t)-(1+\sin t)(1+\cos t)'}{(1+\cos t)^2}=
\frac{\sqrt{1+\cos t}}{2\sqrt{1+\sin t}}\frac{\cos t(1+\cos t)-(1+\sin t)(-\sin t)}{(1+\cos t)^2}=
\frac{\sqrt{1+\cos t}}{2\sqrt{1+\sin t}}\frac{\cos t+\cos^2t+\sin t+\sin^2t}{(1+\cos t)^2}=
\frac{\sqrt{1+\cos t}(\sin t+\cos t+1)}{2\sqrt{1+\sin t}(1+\cos t)^2}=
\frac{\sin t+\cos t+1}{2\sqrt{1+\sin t}(1+\cos t)^{3/2}}$$
