Consider `f (x) = x ^(ln x), and g (x) = e ^(2) x.` Let `alpha and beta` be two values of x satisfying `f (x) = g(x)( alpha lt beta)`
If `h (x) = (f (x))/(g (x))` then `h'(alpha)` equals to:

Options

A. (a) e
B. (b)`-e`
C. (c) `3e`
D. (d)`-3e`

Answer

To solve the problem, we need to find the derivative \( h'(\alpha) \) where \( h(x) = \frac{f(x)}{g(x)} \) and \( f(x) = x^{\ln x} \) and \( g(x) = e^{2x} \). We know that \( \alpha \) and \( \beta \) are values of \( x \) such that \( f(x) = g(x) \).

### Step 1: Set up the equation
Given:
\[ f(x) = g(x) \implies x^{\ln x} = e^{2x} \]

### Step 2: Rewrite the equation
Rearranging gives:
\[ \frac{x^{\ln x}}{e^{2x}} = 1 \]
This can be rewritten as:
\[ x^{\ln x - 2} = e^0 \]

### Step 3: Take the natural logarithm of both sides
Taking the natural logarithm:
\[ \ln(x^{\ln x - 2}) = \ln(1) \implies (\ln x - 2) \ln x = 0 \]

### Step 4: Solve the equation
This gives us two cases:
1. \( \ln x = 0 \) which implies \( x = e^0 = 1 \)
2. \( \ln x - 2 = 0 \) which implies \( \ln x = 2 \) or \( x = e^2 \)

Thus, the solutions are \( x = 1 \) and \( x = e^2 \). Since \( \alpha < \beta \), we have \( \alpha = 1 \) and \( \beta = e^2 \).

### Step 5: Define \( h(x) \)
Now we define:
\[ h(x) = \frac{f(x)}{g(x)} = \frac{x^{\ln x}}{e^{2x}} \]

### Step 6: Differentiate \( h(x) \)
Using the quotient rule:
\[ h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2} \]

### Step 7: Find \( f'(x) \) and \( g'(x) \)
1. For \( f(x) = x^{\ln x} \):
- Using logarithmic differentiation:
\[ \ln f(x) = \ln x \cdot \ln x \implies f'(x) = f(x) \left( \frac{\ln x + 1}{x} \right) = x^{\ln x} \left( \frac{\ln x + 1}{x} \right) \]

2. For \( g(x) = e^{2x} \):
\[ g'(x) = 2e^{2x} \]

### Step 8: Substitute into \( h'(x) \)
Now substituting back into the derivative:
\[ h'(x) = \frac{e^{2x} \cdot x^{\ln x} \left( \frac{\ln x + 1}{x} \right) - x^{\ln x} \cdot 2e^{2x}}{(e^{2x})^2} \]
Simplifying:
\[ h'(x) = \frac{x^{\ln x} e^{2x} \left( \frac{\ln x + 1 - 2}{x} \right)}{e^{4x}} = \frac{x^{\ln x} \left( \ln x - 1 \right)}{e^{2x} x} \]

### Step 9: Evaluate \( h'(\alpha) \)
Substituting \( \alpha = 1 \):
\[ h'(1) = \frac{1^{\ln 1} \left( \ln 1 - 1 \right)}{e^{2 \cdot 1} \cdot 1} = \frac{1^0 \cdot (0 - 1)}{e^2} = \frac{-1}{e^2} \]

### Final Answer
Thus, \( h'(\alpha) = -\frac{1}{e^2} \).