If `f(x) = int_(0)^(x^(2)) sqrt(cost)dt`, find `f'(x)`

Answer

To find \( f'(x) \) for the function defined as

\[ f(x) = \int_{0}^{x^2} \sqrt{\cos t} \, dt, \]

we will use the Fundamental Theorem of Calculus, which states that if \( F(x) = \int_{a}^{g(x)} f(t) \, dt \), then

\[ F'(x) = f(g(x)) \cdot g'(x). \]

### Step-by-step Solution:

1. **Identify the components**:
- Here, \( a = 0 \), \( g(x) = x^2 \), and \( f(t) = \sqrt{\cos t} \).

2. **Differentiate using the Fundamental Theorem of Calculus**:
- According to the theorem, we need to find \( f(g(x)) \) and \( g'(x) \).
- First, compute \( g'(x) \):
\[ g'(x) = \frac{d}{dx}(x^2) = 2x. \]

3. **Evaluate \( f(g(x)) \)**:
- We need to evaluate \( f(g(x)) = f(x^2) \):
\[ f(x^2) = \sqrt{\cos(x^2)}. \]

4. **Combine the results**:
- Now, we can apply the formula:
\[ f'(x) = f(g(x)) \cdot g'(x) = \sqrt{\cos(x^2)} \cdot 2x. \]

5. **Final expression for \( f'(x) \)**:
- Thus, we have:
\[ f'(x) = 2x \sqrt{\cos(x^2)}. \]

### Final Answer:
\[ f'(x) = 2x \sqrt{\cos(x^2)}. \]