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 find the value of \( f'\left(\frac{\pi}{2}\right) \) given the functions \( f(x) \) and \( g(x) \), we will follow these steps:

### Step 1: Define the Functions
We have:
\[ f(x) = \int_0^{g(x)} \frac{dt}{\sqrt{1 + t^3}} \]
and
\[ g(x) = \int_0^{\cos x} (1 + \sin t^2) \, dt \]

### Step 2: Differentiate \( f(x) \)
Using the Fundamental Theorem of Calculus and the chain rule, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx} \left( \int_0^{g(x)} \frac{dt}{\sqrt{1 + t^3}} \right) = \frac{1}{\sqrt{1 + (g(x))^3}} \cdot g'(x) \]

### Step 3: Differentiate \( g(x) \)
Now we differentiate \( g(x) \):
\[ g'(x) = \frac{d}{dx} \left( \int_0^{\cos x} (1 + \sin t^2) \, dt \right) = (1 + \sin(\cos x)^2) \cdot \frac{d}{dx}(\cos x) = (1 + \sin(\cos x)^2)(-\sin x) \]

### Step 4: Evaluate \( g'\left(\frac{\pi}{2}\right) \)
Substituting \( x = \frac{\pi}{2} \):
\[ g'\left(\frac{\pi}{2}\right) = (1 + \sin(\cos(\frac{\pi}{2}))^2)(-\sin(\frac{\pi}{2})) = (1 + \sin(0)^2)(-1) = 1 \cdot (-1) = -1 \]

### Step 5: Evaluate \( g\left(\frac{\pi}{2}\right) \)
Now we need to find \( g\left(\frac{\pi}{2}\right) \):
\[ g\left(\frac{\pi}{2}\right) = \int_0^{\cos(\frac{\pi}{2})} (1 + \sin t^2) \, dt = \int_0^0 (1 + \sin t^2) \, dt = 0 \]

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

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