Let `f(x)=1/x^2 int_4^x (4t^2-2f'(t))dt` then find `9f'(4)`

Explanation

To solve the problem step by step, we will follow the same logical flow as outlined in the video transcript.

### Step-by-Step Solution:

1. **Define the function**:
We are given the function:
\[ f(x) = \frac{1}{x^2} \int_4^x (4t^2 - 2f'(t)) \, dt \]

2. **Multiply both sides by \(x^2\)**:
To simplify the equation, we multiply both sides by \(x^2\):
\[ x^2 f(x) = \int_4^x (4t^2 - 2f'(t)) \, dt \]

3. **Differentiate both sides with respect to \(x\)**:
We will differentiate the left side using the product rule and the right side using the Leibniz rule:
\[ \frac{d}{dx}(x^2 f(x)) = 2x f(x) + x^2 f'(x) \]
For the right side:
\[ \frac{d}{dx} \left( \int_4^x (4t^2 - 2f'(t)) \, dt \right) = 4x^2 - 2f'(x) \]

4. **Set the derivatives equal**:
Now we equate both derivatives:
\[ 2x f(x) + x^2 f'(x) = 4x^2 - 2f'(x) \]

5. **Rearranging the equation**:
Rearranging gives:
\[ x^2 f'(x) + 2f'(x) = 4x^2 - 2x f(x) \]
This simplifies to:
\[ f'(x)(x^2 + 2) = 4x^2 - 2x f(x) \]

6. **Substituting \(x = 4\)**:
We need to find \(f'(4)\), so we substitute \(x = 4\):
\[ f'(4)(4^2 + 2) = 4(4^2) - 2(4)f(4) \]
Simplifying gives:
\[ f'(4)(16 + 2) = 64 - 8f(4) \]
Thus:
\[ 18f'(4) = 64 - 8f(4) \]

7. **Finding \(f(4)\)**:
We need to find \(f(4)\):
\[ f(4) = \frac{1}{4^2} \int_4^4 (4t^2 - 2f'(t)) \, dt = \frac{1}{16} \cdot 0 = 0 \]
Therefore, \(f(4) = 0\).

8. **Substituting \(f(4)\) back into the equation**:
Substituting \(f(4) = 0\) into the equation:
\[ 18f'(4) = 64 - 8(0) \implies 18f'(4) = 64 \]
Thus:
\[ f'(4) = \frac{64}{18} = \frac{32}{9} \]

9. **Finding \(9f'(4)\)**:
Finally, we compute \(9f'(4)\):
\[ 9f'(4) = 9 \cdot \frac{32}{9} = 32 \]

### Final Answer:
\[ \boxed{32} \]