If `f(t)= int_(t^(2))^(t^(3)) (1)/(log x) dx`, then f'(t)= ….

Answer

To find \( f'(t) \) for the function defined by the integral

\[ f(t) = \int_{t^2}^{t^3} \frac{1}{\log x} \, dx, \]

we will use the Leibniz rule for differentiating under the integral sign. The Leibniz rule states that if you have an integral of the form

\[ I(t) = \int_{a(t)}^{b(t)} f(x) \, dx, \]

then the derivative with respect to \( t \) is given by:

\[ I'(t) = f(b(t)) \cdot b'(t) - f(a(t)) \cdot a'(t). \]

### Step-by-step Solution:

1. **Identify the components**:
- Here, \( a(t) = t^2 \) and \( b(t) = t^3 \).
- The function inside the integral is \( f(x) = \frac{1}{\log x} \).

2. **Calculate the derivatives of the limits**:
- The derivative of the upper limit \( b(t) \) is:
\[ b'(t) = \frac{d}{dt}(t^3) = 3t^2. \]
- The derivative of the lower limit \( a(t) \) is:
\[ a'(t) = \frac{d}{dt}(t^2) = 2t. \]

3. **Evaluate the function at the limits**:
- Evaluate \( f(b(t)) \):
\[ f(b(t)) = f(t^3) = \frac{1}{\log(t^3)} = \frac{1}{3 \log t}. \]
- Evaluate \( f(a(t)) \):
\[ f(a(t)) = f(t^2) = \frac{1}{\log(t^2)} = \frac{1}{2 \log t}. \]

4. **Apply the Leibniz rule**:
- Substitute into the Leibniz rule:
\[ f'(t) = f(b(t)) \cdot b'(t) - f(a(t)) \cdot a'(t). \]
- This gives:
\[ f'(t) = \left(\frac{1}{3 \log t}\right)(3t^2) - \left(\frac{1}{2 \log t}\right)(2t). \]

5. **Simplify the expression**:
- Simplifying the first term:
\[ \frac{3t^2}{3 \log t} = \frac{t^2}{\log t}. \]
- Simplifying the second term:
\[ \frac{2t}{2 \log t} = \frac{t}{\log t}. \]
- Therefore, we have:
\[ f'(t) = \frac{t^2}{\log t} - \frac{t}{\log t}. \]

6. **Combine the terms**:
- Combine the fractions:
\[ f'(t) = \frac{t^2 - t}{\log t} = \frac{t(t - 1)}{\log t}. \]

### Final Answer:
\[ f'(t) = \frac{t(t - 1)}{\log t}. \]