Find the derivative with respect to x of the following functions :
(a) `F(x) = int_(0)^(2x) (sin t)/(t)` dt ,
(b) `f(x) = int_(x)^(0) sqrt(1 + t^(4)) dt`

Answer

To find the derivatives of the given functions, we will use the Fundamental Theorem of Calculus and the chain rule.

### (a) \( F(x) = \int_{0}^{2x} \frac{\sin t}{t} \, dt \)

1. **Identify the function and its limits**: We have \( F(x) = \int_{0}^{2x} \frac{\sin t}{t} \, dt \). Here, the upper limit is \( 2x \) and the lower limit is a constant \( 0 \).

2. **Apply the Fundamental Theorem of Calculus**: According to the theorem, if \( F(x) = \int_{a}^{g(x)} f(t) \, dt \), then:
\[ F'(x) = f(g(x)) \cdot g'(x) \]
In our case, \( f(t) = \frac{\sin t}{t} \) and \( g(x) = 2x \).

3. **Calculate \( g'(x) \)**: The derivative of \( g(x) = 2x \) is:
\[ g'(x) = 2 \]

4. **Evaluate \( f(g(x)) \)**: Now, substitute \( g(x) \) into \( f(t) \):
\[ f(g(x)) = f(2x) = \frac{\sin(2x)}{2x} \]

5. **Combine the results**: Now, we can find \( F'(x) \):
\[ F'(x) = f(g(x)) \cdot g'(x) = \frac{\sin(2x)}{2x} \cdot 2 = \frac{\sin(2x)}{x} \]

Thus, the derivative \( F'(x) \) is:
\[ \boxed{F'(x) = \frac{\sin(2x)}{x}} \]