Question

If $y=|\cos x|+|\sin x|$, then $\frac{dy}{dx}$ at $x=\frac{2\pi }{3}$ is

• A. $0$
• B. $1$
• C. $\frac{1-\sqrt{3}}{2}$
• D. $\frac{\sqrt{3}-1}{2}$

Answer 1

Answer: (D)

In the neighborhood for $x=\frac{2\pi }{3}$, we have
$\cos x<0$ and $\sin x>0$
$\therefore y=-\cos x+\sin x$
$\Rightarrow \frac{dy}{dx}=\sin x+\cos x$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(x=\frac{2\pi }{3}\right)}=\sin \frac{2\pi }{3}+\cos \frac{2\pi }{3}$
$=\frac{\sqrt{3}-1}{2}$