The slope of tangent to the curve `y=2 cos^2(3x)` at `x=pi/6`

Options

A. 0
B. 6
C. `-6`
D. `pi`

Answer

To find the slope of the tangent to the curve \( y = 2 \cos^2(3x) \) at \( x = \frac{\pi}{6} \), we will follow these steps:

### Step 1: Differentiate the function
We begin by differentiating the function \( y = 2 \cos^2(3x) \) with respect to \( x \).

Using the chain rule, the derivative of \( \cos^2(u) \) is \( -2 \cos(u) \sin(u) \cdot \frac{du}{dx} \). Here, \( u = 3x \).

So, we have:
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\cos^2(3x)) = 2 \cdot (-2 \cos(3x) \sin(3x) \cdot 3) \]
This simplifies to:
\[ \frac{dy}{dx} = -12 \cos(3x) \sin(3x) \]

### Step 2: Substitute \( x = \frac{\pi}{6} \)
Now, we need to evaluate the derivative at \( x = \frac{\pi}{6} \):
\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{6}} = -12 \cos(3 \cdot \frac{\pi}{6}) \sin(3 \cdot \frac{\pi}{6}) \]

Calculating \( 3 \cdot \frac{\pi}{6} \):
\[ 3 \cdot \frac{\pi}{6} = \frac{\pi}{2} \]

### Step 3: Evaluate \( \cos(\frac{\pi}{2}) \) and \( \sin(\frac{\pi}{2}) \)
Now we find:
\[ \cos\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad \sin\left(\frac{\pi}{2}\right) = 1 \]

### Step 4: Substitute back into the derivative
Substituting these values back into the derivative:
\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{6}} = -12 \cdot 0 \cdot 1 = 0 \]

### Conclusion
Thus, the slope of the tangent to the curve \( y = 2 \cos^2(3x) \) at \( x = \frac{\pi}{6} \) is:
\[ \boxed{0} \]

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