If `y = sqrt((cos x - sin x)(sec 2x + 1)(cos x + sin x))` ,
Then `(dy)/(dx)` at `x = (5 pi)/(6)` is :

Options

A. `(1)/(sqrt(2))`
B. `-(1)/(sqrt(2))`
C. `(sqrt(3))/(2)`
D. `-(sqrt(3))/(2)`

Explanation

To find \(\frac{dy}{dx}\) for the function

\[ y = \sqrt{(\cos x - \sin x)(\sec 2x + 1)(\cos x + \sin x)} \]

at \(x = \frac{5\pi}{6}\), we will follow these steps:

### Step 1: Simplify the expression for \(y\)

We start with:

\[ y = \sqrt{(\cos x - \sin x)(\sec 2x + 1)(\cos x + \sin x)} \]

We can simplify \(\sec 2x + 1\):

\[ \sec 2x = \frac{1}{\cos 2x} \implies \sec 2x + 1 = \frac{1 + \cos 2x}{\cos 2x} \]

Thus, we can rewrite \(y\) as:

\[ y = \sqrt{(\cos x - \sin x)(\frac{1 + \cos 2x}{\cos 2x})(\cos x + \sin x)} \]

Next, we know that:

\[ \cos x - \sin x = \sqrt{2} \cos\left(x + \frac{\pi}{4}\right) \]
\[ \cos x + \sin x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]

Using the identity \(\cos^2 x - \sin^2 x = \cos 2x\), we can express \(y\) as:

\[ y = \sqrt{(\cos^2 x - \sin^2 x)(\sec 2x + 1)} = \sqrt{\cos 2x \cdot \frac{1 + \cos 2x}{\cos 2x}} = \sqrt{1 + \cos 2x} \]

### Step 2: Further simplify \(y\)

Using the identity \(1 + \cos 2x = 2 \cos^2 x\):

\[ y = \sqrt{2 \cos^2 x} = \sqrt{2} |\cos x| \]

### Step 3: Determine the sign of \(\cos x\) at \(x = \frac{5\pi}{6}\)

At \(x = \frac{5\pi}{6}\), we know that:

\[ \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} \]

Thus, since \(\cos x\) is negative in the second quadrant, we have:

\[ y = \sqrt{2} \left(-\cos\left(\frac{5\pi}{6}\right)\right) = \sqrt{2} \left(-\left(-\frac{\sqrt{3}}{2}\right)\right) = \frac{\sqrt{6}}{2} \]

### Step 4: Differentiate \(y\)

Now, we differentiate \(y\):

\[ y = \sqrt{2} |\cos x| \]

Since \(\cos x\) is negative in the second quadrant, we can express this as:

\[ y = -\sqrt{2} \cos x \]

Differentiating \(y\):

\[ \frac{dy}{dx} = -\sqrt{2} \frac{d}{dx}(\cos x) = -\sqrt{2} (-\sin x) = \sqrt{2} \sin x \]

### Step 5: Evaluate \(\frac{dy}{dx}\) at \(x = \frac{5\pi}{6}\)

Now we evaluate \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \sqrt{2} \sin\left(\frac{5\pi}{6}\right) \]

We know:

\[ \sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

Thus:

\[ \frac{dy}{dx} = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2} \]

### Conclusion

The value of \(\frac{dy}{dx}\) at \(x = \frac{5\pi}{6}\) is:

\[ \frac{dy}{dx} = \frac{1}{\sqrt{2}} \]