Question

If $y=|\cos x|+|\sin x|$, then $\frac{d y}{d x}$ 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 $\sin x > 0$
$\therefore y=-\cos x+\sin x$
$\Rightarrow \frac{d y}{d x}=\sin x+\cos x$
$\Rightarrow \left(\frac{d y}{d x}\right)_{\left(x=\frac{2 \pi}{3}\right)}=\sin \frac{2 \pi}{3}+\cos \frac{2 \pi}{3}$
$=\frac{\sqrt{3}-1}{2}$