Let f be a differentiable function satisfying
`[f(x)]^(n)=f(nx)" for all "x inR.`
Then, `f'(x)f(nx)`

Options

A. `f(x)`
B. 0
C. `f(x)f'(nx)`
D. None of these

Answer

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

\[ f(x)^n = f(nx) \]

for all \( x \in \mathbb{R} \).

### Step 1: Differentiate both sides with respect to \( x \)

We differentiate both sides of the equation:

\[ \frac{d}{dx}[f(x)^n] = \frac{d}{dx}[f(nx)] \]

### Step 2: Apply the chain rule and power rule

Using the chain rule and the power rule on the left side:

\[ n f(x)^{n-1} f'(x) \]

On the right side, we apply the chain rule:

\[ f'(nx) \cdot n \]

Thus, we have:

\[ n f(x)^{n-1} f'(x) = n f'(nx) \]

### Step 3: Simplify the equation

We can divide both sides by \( n \) (assuming \( n \neq 0 \)):

\[ f(x)^{n-1} f'(x) = f'(nx) \]

### Step 4: Substitute \( f(nx) \)

From the original equation, we know that \( f(x)^n = f(nx) \). Therefore, we can express \( f(x)^{n-1} \) as:

\[ f(x)^{n-1} = \frac{f(nx)}{f(x)} \]

Substituting this into our differentiated equation gives:

\[ \frac{f(nx)}{f(x)} f'(x) = f'(nx) \]

### Step 5: Rearranging the equation

Now, we can rearrange this equation to isolate \( f'(x) f(nx) \):

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

### Conclusion

Thus, we have shown that:

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

This means the correct answer is option (c):

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