If `y=|cosx|+|sinx|`, then `(dy)/(dx)" at "x=(2pi)/(3)` is

Options

A. `(1-sqrt(3))/(2)`
B. 0
C. `(sqrt(3)-1)/(2)`
D. none of these

Explanation

To find the derivative \( \frac{dy}{dx} \) of the function \( y = |\cos x| + |\sin x| \) at \( x = \frac{2\pi}{3} \), we will follow these steps:

### Step 1: Determine the signs of \( \cos x \) and \( \sin x \) at \( x = \frac{2\pi}{3} \)

At \( x = \frac{2\pi}{3} \):
- \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \) (negative)
- \( \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \) (positive)

### Step 2: Rewrite the function \( y \)

Since \( \cos x \) is negative and \( \sin x \) is positive in this interval, we can rewrite \( y \) as:
\[ y = -\cos x + \sin x \]

### Step 3: Differentiate \( y \)

Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(-\cos x + \sin x) = \sin x + \cos x \]

### Step 4: Evaluate \( \frac{dy}{dx} \) at \( x = \frac{2\pi}{3} \)

Now we substitute \( x = \frac{2\pi}{3} \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x = \frac{2\pi}{3}} = \sin\left(\frac{2\pi}{3}\right) + \cos\left(\frac{2\pi}{3}\right) \]

Using the values we found earlier:
- \( \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \)
- \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \)

Thus,
\[ \frac{dy}{dx} \bigg|_{x = \frac{2\pi}{3}} = \frac{\sqrt{3}}{2} - \frac{1}{2} \]

### Step 5: Simplify the expression

Combine the terms:
\[ \frac{dy}{dx} \bigg|_{x = \frac{2\pi}{3}} = \frac{\sqrt{3} - 1}{2} \]

### Final Answer

The value of \( \frac{dy}{dx} \) at \( x = \frac{2\pi}{3} \) is:
\[ \frac{\sqrt{3} - 1}{2} \]

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