If F(x)`=(1)/(x^(2))overset(x)underset(4)int [4t^(2)-2F'(t)]dt` then F'(4) equals

Options

A. `(15)/(9)`
B. `(32)/(9)`
C. `(37)/(9)`
D. None of these

Answer

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

\[ F(x) = \frac{1}{x^2} \int_4^x \left( 4t^2 - 2F'(t) \right) dt. \]

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

We start by differentiating both sides with respect to \( x \):

\[ F'(x) = \frac{d}{dx} \left( \frac{1}{x^2} \int_4^x \left( 4t^2 - 2F'(t) \right) dt \right). \]

### Step 2: Apply the Product Rule

Using the product rule, where \( u = \frac{1}{x^2} \) and \( v = \int_4^x \left( 4t^2 - 2F'(t) \right) dt \):

\[ F'(x) = u'v + uv'. \]

### Step 3: Calculate \( u' \) and \( v' \)

1. **Calculate \( u' \)**:
\[ u' = \frac{d}{dx} \left( \frac{1}{x^2} \right) = -\frac{2}{x^3}. \]

2. **Calculate \( v' \)** using the Leibniz rule:
\[ v' = 4x^2 - 2F'(x). \]

### Step 4: Substitute \( u' \) and \( v' \) into the equation

Now substituting \( u' \) and \( v' \) back into the equation for \( F'(x) \):

\[ F'(x) = -\frac{2}{x^3} \int_4^x \left( 4t^2 - 2F'(t) \right) dt + \frac{1}{x^2} \left( 4x^2 - 2F'(x) \right). \]

### Step 5: Simplify the expression

This simplifies to:

\[ F'(x) = -\frac{2}{x^3} \int_4^x \left( 4t^2 - 2F'(t) \right) dt + \frac{4}{x^2} - \frac{2F'(x)}{x^2}. \]

### Step 6: Evaluate at \( x = 4 \)

Now we substitute \( x = 4 \):

\[ F'(4) = -\frac{2}{4^3} \int_4^4 \left( 4t^2 - 2F'(t) \right) dt + \frac{4}{4^2} - \frac{2F'(4)}{4^2}. \]

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

\[ F'(4) = 0 + 1 - \frac{2F'(4)}{16}. \]

### Step 7: Solve for \( F'(4) \)

This gives us:

\[ F'(4) + \frac{2F'(4)}{16} = 1. \]

Multiplying through by 16 to eliminate the fraction:

\[ 16F'(4) + 2F'(4) = 16. \]

This simplifies to:

\[ 18F'(4) = 16. \]

Thus,

\[ F'(4) = \frac{16}{18} = \frac{8}{9}. \]

### Step 8: Final Answer

The final answer is:

\[ F'(4) = \frac{8}{9}. \]