let `f(x)=x^(3)l n(x^(2)(g(x))`, where `g(x)` is a differentiable positive function on `(0,infty)` satisfying `g(2)=(1)/(4),g^(')(2)=-3`, then `f'^(2)` is

Options

A. 77
B. 88
C. `-88`
D. `-77`

Explanation

To solve the problem, we need to find \( f'(2) \) for the function \( f(x) = x^3 \ln(x^2 g(x)) \), where \( g(x) \) is a differentiable positive function on \( (0, \infty) \) with the given conditions \( g(2) = \frac{1}{4} \) and \( g'(2) = -3 \).

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = x^3 \ln(x^2 g(x)) \]
Using the properties of logarithms, we can rewrite it as:
\[ f(x) = x^3 \left( \ln(x^2) + \ln(g(x)) \right) = x^3 (2 \ln(x) + \ln(g(x))) \]

### Step 2: Differentiate using the product rule
We need to differentiate \( f(x) \). We will use the product rule and the chain rule. The derivative is:
\[ f'(x) = \frac{d}{dx} \left( x^3 \right) (2 \ln(x) + \ln(g(x))) + x^3 \frac{d}{dx} \left( 2 \ln(x) + \ln(g(x)) \right) \]

Calculating the first part:
\[ \frac{d}{dx} (x^3) = 3x^2 \]
Thus,
\[ f'(x) = 3x^2 (2 \ln(x) + \ln(g(x))) + x^3 \left( \frac{2}{x} + \frac{g'(x)}{g(x)} \right) \]

### Step 3: Simplify the derivative
Now, we can simplify the expression:
\[ f'(x) = 3x^2 (2 \ln(x) + \ln(g(x))) + x^2 \left( 2 + x \frac{g'(x)}{g(x)} \right) \]
Combining terms:
\[ f'(x) = 3x^2 (2 \ln(x) + \ln(g(x))) + 2x^2 + x^2 \frac{g'(x)}{g(x)} \]

### Step 4: Evaluate at \( x = 2 \)
Now we substitute \( x = 2 \):
1. Calculate \( g(2) \) and \( g'(2) \):
- \( g(2) = \frac{1}{4} \)
- \( g'(2) = -3 \)

2. Substitute into the derivative:
\[ f'(2) = 3(2^2) \left( 2 \ln(2) + \ln\left(\frac{1}{4}\right) \right) + 2(2^2) + 2^2 \frac{-3}{\frac{1}{4}} \]
Calculating each term:
- \( 2^2 = 4 \)
- \( \ln\left(\frac{1}{4}\right) = \ln(1) - \ln(4) = 0 - 2\ln(2) = -2\ln(2) \)

So,
\[ f'(2) = 3(4) \left( 2 \ln(2) - 2 \ln(2) \right) + 2(4) + 4 \cdot (-12) \]
This simplifies to:
\[ f'(2) = 0 + 8 - 48 = -40 \]

### Final Answer
Thus, the value of \( f'(2) \) is:
\[ \boxed{-40} \]