Cosine of the acute angle between the curve `y=3^(x-1)log_(e)x` and `y=x^(x)-1`, at the point of intersection `(1,0)` is

Options

A. 0
B. 1
C. `(sqrt3)/(2)`
D. `(1)/(2)`

Explanation

To find the cosine of the acute angle between the curves \( y = 3^{(x-1)} \log_e x \) and \( y = x^x - 1 \) at the point of intersection \( (1, 0) \), we will follow these steps:

### Step 1: Find the slopes of the curves at the point of intersection

1. **Differentiate the first curve**:
\[ y = 3^{(x-1)} \log_e x \]
Using the product rule:
\[ \frac{dy}{dx} = \frac{d}{dx}(3^{(x-1)}) \cdot \log_e x + 3^{(x-1)} \cdot \frac{d}{dx}(\log_e x) \]
The derivative of \( 3^{(x-1)} \) is:
\[ \frac{d}{dx}(3^{(x-1)}) = 3^{(x-1)} \ln(3) \]
The derivative of \( \log_e x \) is:
\[ \frac{d}{dx}(\log_e x) = \frac{1}{x} \]
Thus,
\[ \frac{dy}{dx} = 3^{(x-1)} \ln(3) \cdot \log_e x + 3^{(x-1)} \cdot \frac{1}{x} \]

2. **Evaluate at \( x = 1 \)**:
\[ \frac{dy}{dx} \bigg|_{x=1} = 3^{(1-1)} \ln(3) \cdot \log_e(1) + 3^{(1-1)} \cdot \frac{1}{1} \]
Since \( \log_e(1) = 0 \):
\[ \frac{dy}{dx} \bigg|_{x=1} = 1 \cdot 0 + 1 = 1 \]
Thus, the slope \( m_1 = 1 \).

3. **Differentiate the second curve**:
\[ y = x^x - 1 \]
Using implicit differentiation:
\[ \frac{dy}{dx} = \frac{d}{dx}(x^x) = x^x (\ln x + 1) \]

4. **Evaluate at \( x = 1 \)**:
\[ \frac{dy}{dx} \bigg|_{x=1} = 1^1 (\ln(1) + 1) = 1 \cdot (0 + 1) = 1 \]
Thus, the slope \( m_2 = 1 \).

### Step 2: Calculate the angle between the curves

Using the formula for the tangent of the angle \( \theta \) between two curves:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]
Substituting \( m_1 = 1 \) and \( m_2 = 1 \):
\[ \tan \theta = \frac{1 - 1}{1 + 1 \cdot 1} = \frac{0}{2} = 0 \]
Thus, \( \theta = 0 \).

### Step 3: Find the cosine of the angle

Since \( \theta = 0 \):
\[ \cos \theta = \cos(0) = 1 \]

### Final Answer
The cosine of the acute angle between the curves at the point of intersection \( (1, 0) \) is:
\[ \boxed{1} \]