Consider the non-constant differentiable function f of one variable which obeys the relation `(f (x))/( f (y)) =f (x-y). If f '(0) =p and f '(5) =q,` then `f '(-5) ` is

Options

A. `(p ^(2))/(q)`
B. `(q)/(p)`
C. `(p)/(q)`
D. `q`

Answer

To solve the problem, we start with the given functional equation:

\[ \frac{f(x)}{f(y)} = f(x - y) \]

### Step 1: Assume a form for the function
We can assume that \( f(x) \) is of the form \( f(x) = a^{kx} \), where \( a \) is a constant and \( k \) is a non-zero constant (since \( f \) is non-constant).

### Step 2: Verify the functional equation
Substituting \( f(x) = a^{kx} \) into the functional equation:

\[ \frac{f(x)}{f(y)} = \frac{a^{kx}}{a^{ky}} = a^{k(x - y)} = f(x - y) \]

This shows that our assumption satisfies the functional equation.

### Step 3: Differentiate \( f(x) \)
Next, we find the derivative \( f'(x) \):

\[ f'(x) = \frac{d}{dx}(a^{kx}) = k \ln(a) \cdot a^{kx} \]

### Step 4: Evaluate \( f'(0) \) and \( f'(5) \)
Now we can find \( f'(0) \) and \( f'(5) \):

1. \( f'(0) = k \ln(a) \cdot a^{0} = k \ln(a) \)
2. \( f'(5) = k \ln(a) \cdot a^{5} \)

Given that \( f'(0) = p \) and \( f'(5) = q \), we have:

\[ p = k \ln(a) \quad \text{(1)} \]
\[ q = k \ln(a) \cdot a^{5} \quad \text{(2)} \]

### Step 5: Find \( f'(-5) \)
Now we need to find \( f'(-5) \):

\[ f'(-5) = k \ln(a) \cdot a^{-5} \]

### Step 6: Express \( f'(-5) \) in terms of \( p \) and \( q \)
From equation (1), we know \( k \ln(a) = p \). Thus:

\[ f'(-5) = p \cdot a^{-5} \]

### Step 7: Relate \( a^{-5} \) to \( p \) and \( q \)
From equation (2), we can express \( a^{5} \):

\[ q = k \ln(a) \cdot a^{5} = p \cdot a^{5} \]

So, we can express \( a^{5} \) in terms of \( p \) and \( q \):

\[ a^{5} = \frac{q}{p} \]

Thus, \( a^{-5} = \frac{p}{q} \).

### Step 8: Substitute back to find \( f'(-5) \)
Now substituting \( a^{-5} \) back into the expression for \( f'(-5) \):

\[ f'(-5) = p \cdot a^{-5} = p \cdot \frac{p}{q} = \frac{p^2}{q} \]

### Final Result
Thus, we find that:

\[ f'(-5) = \frac{p^2}{q} \]