Find the derivative with respect to x of the following functions :
(a) `F(x) = int_(x^(2))^(x^(3)) "In t dt " (x gt 0)`
(b) `f(x) = int_(1//x)^(sqrt(x)) cos (t^(2)) dt (x gt 0)`

Answer

To find the derivatives of the given functions, we will use the Fundamental Theorem of Calculus and the chain rule. Let's solve each part step by step.

### Part (a):
Given:
\[ F(x) = \int_{x^2}^{x^3} \ln(t) \, dt \]

**Step 1: Identify the limits of integration and the integrand.**
- The lower limit is \( h(x) = x^2 \)
- The upper limit is \( g(x) = x^3 \)
- The integrand is \( f(t) = \ln(t) \)

**Step 2: Apply the Leibniz rule for differentiation under the integral sign.**
According to the Leibniz rule:
\[ F'(x) = f(g(x)) \cdot g'(x) - f(h(x)) \cdot h'(x) \]

**Step 3: Compute \( g'(x) \) and \( h'(x) \).**
- \( g'(x) = \frac{d}{dx}(x^3) = 3x^2 \)
- \( h'(x) = \frac{d}{dx}(x^2) = 2x \)

**Step 4: Evaluate \( f(g(x)) \) and \( f(h(x)) \).**
- \( f(g(x)) = f(x^3) = \ln(x^3) = 3\ln(x) \)
- \( f(h(x)) = f(x^2) = \ln(x^2) = 2\ln(x) \)

**Step 5: Substitute these values into the Leibniz rule.**
\[ F'(x) = (3\ln(x)) \cdot (3x^2) - (2\ln(x)) \cdot (2x) \]
\[ F'(x) = 9x^2 \ln(x) - 4x \ln(x) \]

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

### Final Answer for Part (a):
\[ F'(x) = (9x^2 - 4x) \ln(x) \]

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### Part (b):
Given:
\[ f(x) = \int_{\frac{1}{x}}^{\sqrt{x}} \cos(t^2) \, dt \]

**Step 1: Identify the limits of integration and the integrand.**
- The lower limit is \( h(x) = \frac{1}{x} \)
- The upper limit is \( g(x) = \sqrt{x} \)
- The integrand is \( f(t) = \cos(t^2) \)

**Step 2: Apply the Leibniz rule for differentiation under the integral sign.**
\[ f'(x) = f(g(x)) \cdot g'(x) - f(h(x)) \cdot h'(x) \]

**Step 3: Compute \( g'(x) \) and \( h'(x) \).**
- \( g'(x) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \)
- \( h'(x) = \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} \)

**Step 4: Evaluate \( f(g(x)) \) and \( f(h(x)) \).**
- \( f(g(x)) = f(\sqrt{x}) = \cos((\sqrt{x})^2) = \cos(x) \)
- \( f(h(x)) = f\left(\frac{1}{x}\right) = \cos\left(\left(\frac{1}{x}\right)^2\right) = \cos\left(\frac{1}{x^2}\right) \)

**Step 5: Substitute these values into the Leibniz rule.**
\[ f'(x) = \cos(x) \cdot \frac{1}{2\sqrt{x}} - \cos\left(\frac{1}{x^2}\right) \cdot \left(-\frac{1}{x^2}\right) \]
\[ f'(x) = \frac{\cos(x)}{2\sqrt{x}} + \frac{1}{x^2} \cos\left(\frac{1}{x^2}\right) \]

### Final Answer for Part (b):
\[ f'(x) = \frac{\cos(x)}{2\sqrt{x}} + \frac{1}{x^2} \cos\left(\frac{1}{x^2}\right) \]

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