If `f (x) =int _(0)^(g(x))(dt)/(sqrt(1+t ^(3))),g (x) = int _(0)^(cos x ) (1+ sint ) ^(2) dt, ` then the value of `f'((pi)/(2))` is equal to:

Options

A. 1
B. `-1`
C. 0
D. `1/2`

Answer

To solve the problem, we need to find the value of \( f'(\frac{\pi}{2}) \) given the functions \( f(x) \) and \( g(x) \).

### Step-by-Step Solution:

1. **Define the Functions**:
\[ f(x) = \int_{0}^{g(x)} \frac{dt}{\sqrt{1 + t^3}} \]
\[ g(x) = \int_{0}^{\cos x} (1 + \sin t)^2 \, dt \]

2. **Differentiate \( f(x) \)**:
Using Leibniz's rule for differentiation under the integral sign, we have:
\[ f'(x) = \frac{1}{\sqrt{1 + (g(x))^3}} \cdot g'(x) \]
Here, \( h(x) = 0 \) contributes nothing to the derivative.

3. **Calculate \( g'(x) \)**:
Again, applying Leibniz's rule:
\[ g'(x) = (1 + \sin(\cos x))^2 \cdot \frac{d}{dx}(\cos x) = (1 + \sin(\cos x))^2 \cdot (-\sin x) \]

4. **Evaluate \( g'(\frac{\pi}{2}) \)**:
\[ g'(\frac{\pi}{2}) = (1 + \sin(\cos(\frac{\pi}{2})))^2 \cdot (-\sin(\frac{\pi}{2})) \]
Since \( \cos(\frac{\pi}{2}) = 0 \):
\[ g'(\frac{\pi}{2}) = (1 + \sin(0))^2 \cdot (-1) = 1^2 \cdot (-1) = -1 \]

5. **Calculate \( g(\frac{\pi}{2}) \)**:
\[ g(\frac{\pi}{2}) = \int_{0}^{\cos(\frac{\pi}{2})} (1 + \sin t)^2 \, dt = \int_{0}^{0} (1 + \sin t)^2 \, dt = 0 \]

6. **Substitute Values into \( f'(\frac{\pi}{2}) \)**:
Now we substitute \( g(\frac{\pi}{2}) \) and \( g'(\frac{\pi}{2}) \) into the expression for \( f'(\frac{\pi}{2}) \):
\[ f'(\frac{\pi}{2}) = \frac{1}{\sqrt{1 + (0)^3}} \cdot (-1) = \frac{1}{\sqrt{1}} \cdot (-1) = -1 \]

### Final Answer:
Thus, the value of \( f'(\frac{\pi}{2}) \) is:
\[ \boxed{-1} \]