If a differentiable function f defined for `x gt0` satisfies the relation `f(x^(2))=x^(3),x gt 0`, then what is the value of `f'(4)?`

Options

A. 2
B. 3
C. 4
D. none of these

Explanation

To solve the problem, we need to find the value of \( f'(4) \) given that the function \( f \) satisfies the relation \( f(x^2) = x^3 \) for \( x > 0 \).

### Step-by-Step Solution:

1. **Substitute \( x^2 = y \)**:
Since we have \( f(x^2) = x^3 \), we can let \( y = x^2 \). Then, \( x = \sqrt{y} \). We can rewrite the equation as:
\[ f(y) = (\sqrt{y})^3 = y^{3/2} \]
This means that \( f(y) = y^{3/2} \) for \( y > 0 \).

2. **Differentiate \( f(y) \)**:
Now we differentiate \( f(y) \) with respect to \( y \):
\[ f'(y) = \frac{d}{dy}(y^{3/2}) = \frac{3}{2}y^{1/2} \]

3. **Evaluate \( f'(4) \)**:
We need to find \( f'(4) \). Substitute \( y = 4 \) into the derivative:
\[ f'(4) = \frac{3}{2}(4)^{1/2} = \frac{3}{2} \cdot 2 = 3 \]

Thus, the value of \( f'(4) \) is \( \boxed{3} \).