Let `f (x) =(1)/(x ^(2)) int _(4)^(x) (4t ^(2) -2 f '(t) dt, ` find `9f'(4)`

Options

A. 16
B. 4
C. 8
D. 32

Answer

To solve the problem, we need to find \( 9f'(4) \) given that

\[ f(x) = \frac{1}{x^2} \int_{4}^{x} (4t^2 - 2f'(t)) \, dt. \]

### Step 1: Differentiate \( f(x) \)

We start by differentiating both sides with respect to \( x \). We will use the product rule and the Fundamental Theorem of Calculus.

\[ f'(x) = \frac{d}{dx} \left( \frac{1}{x^2} \right) \int_{4}^{x} (4t^2 - 2f'(t)) \, dt + \frac{1}{x^2} \cdot (4x^2 - 2f'(x)). \]

### Step 2: Apply the Product Rule

The derivative of \( \frac{1}{x^2} \) is \( -\frac{2}{x^3} \). Thus, we can write:

\[ f'(x) = -\frac{2}{x^3} \int_{4}^{x} (4t^2 - 2f'(t)) \, dt + \frac{1}{x^2} (4x^2 - 2f'(x)). \]

### Step 3: Simplify the Expression

Now, we can simplify the expression:

\[ f'(x) = -\frac{2}{x^3} \int_{4}^{x} (4t^2 - 2f'(t)) \, dt + \frac{4}{x^2} - \frac{2f'(x)}{x^2}. \]

### Step 4: Rearranging the Equation

Rearranging gives us:

\[ f'(x) + \frac{2f'(x)}{x^2} = \frac{4}{x^2} - \frac{2}{x^3} \int_{4}^{x} (4t^2 - 2f'(t)) \, dt. \]

Factoring out \( f'(x) \):

\[ f'(x) \left( 1 + \frac{2}{x^2} \right) = \frac{4}{x^2} - \frac{2}{x^3} \int_{4}^{x} (4t^2 - 2f'(t)) \, dt. \]

### Step 5: Solve for \( f'(x) \)

Now we can solve for \( f'(x) \):

\[ f'(x) = \frac{\frac{4}{x^2} - \frac{2}{x^3} \int_{4}^{x} (4t^2 - 2f'(t)) \, dt}{1 + \frac{2}{x^2}}. \]

### Step 6: Substitute \( x = 4 \)

Next, we substitute \( x = 4 \) to find \( f'(4) \):

\[ f'(4) = \frac{\frac{4}{4^2} - \frac{2}{4^3} \int_{4}^{4} (4t^2 - 2f'(t)) \, dt}{1 + \frac{2}{4^2}}. \]

Since the integral from 4 to 4 is zero, we have:

\[ f'(4) = \frac{\frac{4}{16}}{1 + \frac{2}{16}} = \frac{\frac{1}{4}}{1 + \frac{1}{8}} = \frac{\frac{1}{4}}{\frac{9}{8}} = \frac{1}{4} \cdot \frac{8}{9} = \frac{2}{9}. \]

### Step 7: Calculate \( 9f'(4) \)

Finally, we calculate \( 9f'(4) \):

\[ 9f'(4) = 9 \cdot \frac{2}{9} = 2. \]

### Final Answer

Thus, the answer is:

\[ \boxed{2}. \]