If `2f(x+y)=f(x).f(y)` for all real x, y. where `f'(0)=3 and f(4)=25`, then the value of `f'(4)` is equal to

Answer

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

\[ 2f(x+y) = f(x)f(y) \]

for all real \( x \) and \( y \). We also know that \( f'(0) = 3 \) and \( f(4) = 25 \). We need to find \( f'(4) \).

### Step 1: Analyze the functional equation

The equation \( 2f(x+y) = f(x)f(y) \) suggests that \( f \) might have an exponential form. We can rewrite it as:

\[ f(x+y) = \frac{1}{2} f(x)f(y) \]

This form hints that \( f \) could be related to the exponential function.

### Step 2: Substitute specific values

Let's substitute \( x = 0 \) and \( y = 0 \):

\[ 2f(0+0) = f(0)f(0) \]

This simplifies to:

\[ 2f(0) = f(0)^2 \]

Let \( f(0) = c \). Then we have:

\[ 2c = c^2 \]

Rearranging gives:

\[ c^2 - 2c = 0 \]
\[ c(c - 2) = 0 \]

Thus, \( c = 0 \) or \( c = 2 \). Since \( f'(0) = 3 \) implies \( f(0) \neq 0 \), we have:

\[ f(0) = 2 \]

### Step 3: Differentiate the functional equation

Now, we differentiate the original functional equation with respect to \( y \):

Using the product rule on the right side:

\[ 2f'(x+y) = f'(x)f(y) \]

Setting \( y = 0 \):

\[ 2f'(x) = f'(x)f(0) \]

Substituting \( f(0) = 2 \):

\[ 2f'(x) = 2f'(x) \]

This does not provide new information, but confirms consistency.

### Step 4: Find \( f'(4) \)

To find \( f'(4) \), we can use the limit definition of the derivative:

\[ f'(4) = \lim_{h \to 0} \frac{f(4+h) - f(4)}{h} \]

Using the functional equation again, we have:

\[ f(4+h) = \frac{1}{2} f(4)f(h) \]

Substituting this into the derivative limit:

\[ f'(4) = \lim_{h \to 0} \frac{\frac{1}{2} f(4)f(h) - f(4)}{h} \]

Factoring out \( f(4) \):

\[ f'(4) = f(4) \cdot \lim_{h \to 0} \frac{\frac{1}{2} f(h) - 1}{h} \]

### Step 5: Evaluate the limit

Now, we know \( f(0) = 2 \) and \( f'(0) = 3 \):

The limit can be rewritten as:

\[ \lim_{h \to 0} \frac{\frac{1}{2} f(h) - 1}{h} = \frac{1}{2} \lim_{h \to 0} \frac{f(h) - 2}{h} = \frac{1}{2} f'(0) = \frac{1}{2} \cdot 3 = \frac{3}{2} \]

### Step 6: Final calculation

Now substituting back:

\[ f'(4) = 25 \cdot \frac{3}{2} = \frac{75}{2} = 37.5 \]

Thus, the value of \( f'(4) \) is:

\[ \boxed{37.5} \]