Let `f(t)="ln"(t)`. Then, `(d)/(dx)(int_(x^(2))^(x^(3))f(t)" dt")`

Options

A. has value 0 when x=0
B. has value 0 when x = 1 and `x=4//9`
C. has value 9 `e^(2)-4e` when x=e`
D. has differential coefficient `27e-8` for x=e

Explanation

To solve the problem, we need to find the derivative of the integral of the function \( f(t) = \ln(t) \) from \( x^2 \) to \( x^3 \). We will apply the Leibniz rule for differentiation under the integral sign.

### Step-by-Step Solution:

1. **Identify the function and limits**:
\[ f(t) = \ln(t) \]
The integral we need to differentiate is:
\[ \int_{x^2}^{x^3} \ln(t) \, dt \]

2. **Apply the Leibniz rule**:
The Leibniz rule states that if you have an integral from \( u(x) \) to \( v(x) \) of a function \( f(t) \), then:
\[ \frac{d}{dx} \left( \int_{u(x)}^{v(x)} f(t) \, dt \right) = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x) \]
Here, \( u(x) = x^2 \) and \( v(x) = x^3 \).

3. **Calculate \( f(v(x)) \) and \( f(u(x)) \)**:
- For the upper limit \( v(x) = x^3 \):
\[ f(v(x)) = f(x^3) = \ln(x^3) = 3 \ln(x) \]
- For the lower limit \( u(x) = x^2 \):
\[ f(u(x)) = f(x^2) = \ln(x^2) = 2 \ln(x) \]

4. **Calculate the derivatives of the limits**:
- The derivative of the upper limit:
\[ v'(x) = \frac{d}{dx}(x^3) = 3x^2 \]
- The derivative of the lower limit:
\[ u'(x) = \frac{d}{dx}(x^2) = 2x \]

5. **Substitute into the Leibniz rule**:
Now we substitute into the Leibniz rule:
\[ \frac{d}{dx} \left( \int_{x^2}^{x^3} \ln(t) \, dt \right) = f(x^3) \cdot v'(x) - f(x^2) \cdot u'(x) \]
This gives:
\[ = (3 \ln(x)) \cdot (3x^2) - (2 \ln(x)) \cdot (2x) \]

6. **Simplify the expression**:
\[ = 9x^2 \ln(x) - 4x \ln(x) \]
Factor out \( \ln(x) \):
\[ = \ln(x) (9x^2 - 4x) \]

### Final Answer:
\[ \frac{d}{dx} \left( \int_{x^2}^{x^3} \ln(t) \, dt \right) = \ln(x) (9x^2 - 4x) \]