If `f(x)=log{(u(x))/(v(x))},\ u(1)=v(1)` and `u^(prime)(1)=v^(prime)(1)=2` , then find the value of `f^(prime)(1)` .

Options

A. 0
B. 1
C. -1
D. none of these

Answer

To find the value of \( f'(1) \) for the function \( f(x) = \log\left(\frac{u(x)}{v(x)}\right) \), we will follow these steps:

### Step 1: Differentiate \( f(x) \)
Using the properties of logarithms, we can rewrite \( f(x) \):
\[ f(x) = \log(u(x)) - \log(v(x)) \]
Now, we differentiate \( f(x) \):
\[ f'(x) = \frac{u'(x)}{u(x)} - \frac{v'(x)}{v(x)} \]

### Step 2: Evaluate \( f'(1) \)
We need to evaluate \( f'(1) \):
\[ f'(1) = \frac{u'(1)}{u(1)} - \frac{v'(1)}{v(1)} \]

### Step 3: Substitute known values
From the problem, we know:
- \( u(1) = v(1) \)
- \( u'(1) = v'(1) = 2 \)

Substituting these values into the equation:
\[ f'(1) = \frac{2}{u(1)} - \frac{2}{v(1)} \]
Since \( u(1) = v(1) \), we can denote \( u(1) = v(1) = k \) (where \( k \) is some constant):
\[ f'(1) = \frac{2}{k} - \frac{2}{k} = 0 \]

### Final Answer
Thus, the value of \( f'(1) \) is:
\[ \boxed{0} \]