If $y= |\sin x| + |\cos x|$, then $dy/dx$ at $x=2\pi/3$ is? If $$y= |\sin x| + |\cos x|,$$ 
then $dy/dx$ at $x = 2\pi/3$ is?          
Ans: $(\sqrt{3} - 1)/2$
How can we differentiate modulus functions? Can anyone explain? 

you only need to worry about the behaviour in the vicinity of $x=2 \pi /3$ where $\sin x $ is positive and $ \cos x $ is negative so
$$y=    \sin x - \cos x $$
$$y'=    \cos x + \sin x $$
$$y'(2 \pi /3)=    \cos (2 \pi /3) + \sin(2 \pi /3) = -\frac 12 + \frac{\sqrt 3}{2} =  \frac{\sqrt 3 -1}{2}$$
