If `f(x)= int_(x^(2))^(x^(3)) (dt)/(log t), x gt 0` then

Options

A. `f'(x)= -(1)/(6log x)`
B. f is an increasing function
C. f has minimum at x=1
D. f is an increasing function on `[1, oo]`

Answer

To solve the problem, we need to differentiate the function defined by the integral:

Given:
\[ f(x) = \int_{x^2}^{x^3} \frac{dt}{\log t} \]

### Step 1: Differentiate \( f(x) \)
We will use the Fundamental Theorem of Calculus and Leibniz's rule for differentiation under the integral sign. According to this theorem:

\[ f'(x) = \frac{d}{dx} \left( \int_{a(x)}^{b(x)} g(t) \, dt \right) = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x) \]

Here, \( a(x) = x^2 \), \( b(x) = x^3 \), and \( g(t) = \frac{1}{\log t} \).

### Step 2: Calculate \( b'(x) \) and \( a'(x) \)
- \( b'(x) = \frac{d}{dx}(x^3) = 3x^2 \)
- \( a'(x) = \frac{d}{dx}(x^2) = 2x \)

### Step 3: Evaluate \( g(b(x)) \) and \( g(a(x)) \)
- \( g(b(x)) = g(x^3) = \frac{1}{\log(x^3)} = \frac{1}{3 \log x} \) (using the property \( \log(a^b) = b \log a \))
- \( g(a(x)) = g(x^2) = \frac{1}{\log(x^2)} = \frac{1}{2 \log x} \)

### Step 4: Substitute into the derivative formula
Now we can substitute these values into the derivative formula:

\[ f'(x) = g(b(x)) \cdot b'(x) - g(a(x)) \cdot a'(x) \]

Substituting the values we calculated:

\[ f'(x) = \left( \frac{1}{3 \log x} \right) \cdot (3x^2) - \left( \frac{1}{2 \log x} \right) \cdot (2x) \]

### Step 5: Simplify the expression
\[ f'(x) = \frac{3x^2}{3 \log x} - \frac{2x}{2 \log x} \]
\[ f'(x) = \frac{x^2}{\log x} - \frac{x}{\log x} \]
\[ f'(x) = \frac{x^2 - x}{\log x} \]

### Step 6: Factor the numerator
\[ f'(x) = \frac{x(x - 1)}{\log x} \]

### Step 7: Analyze the sign of \( f'(x) \)
1. For \( x > 1 \), \( x - 1 > 0 \) and \( \log x > 0 \) so \( f'(x) > 0 \). Thus, \( f(x) \) is increasing for \( x > 1 \).
2. For \( 0 < x < 1 \), \( x - 1 < 0 \) and \( \log x < 0 \) so \( f'(x) > 0 \) (since a negative divided by a negative is positive). Thus, \( f(x) \) is also increasing for \( 0 < x < 1 \).

### Conclusion
The function \( f(x) \) is increasing for all \( x > 0 \).