Question
If $$y = \left| {\cos \,x} \right| + \left| {\sin \,x} \right|$$ then $$\frac{{dy}}{{dx}}$$ at $$x = \frac{{2\pi }}{3}$$ is :
A.
$$\frac{{1 - \sqrt 3 }}{2}$$
B.
0
C.
$$\frac{1}{2}\left( {\sqrt 3 - 1} \right)$$
D.
none of these
Answer :
$$\frac{1}{2}\left( {\sqrt 3 - 1} \right)$$
Solution :
In the neighborhood for $$x = \frac{{2\pi }}{3},$$ we have
$$\eqalign{
& \cos \,x < 0{\text{ and }}\sin \,x > 0 \cr
& \therefore \,y = - \cos \,x + \sin \,x \cr
& \Rightarrow \frac{{dy}}{{dx}} = \sin \,x + \cos \,x \cr
& \Rightarrow {\left[ {\frac{{dy}}{{dx}}} \right]_{x = \frac{{2\pi }}{3}}} = \sin \frac{{2\pi }}{3} + \cos \frac{{2\pi }}{3} \cr
& = \frac{{\sqrt 3 }}{2} - \frac{1}{2} \cr
& = \frac{{\sqrt 3 - 1}}{2} \cr} $$