Let `f(x)=int_0^g(x) dt/sqrt(1+t^2)` where `g(x) =int_0^cosx (1+sint^2) dt.` Also `h(x) =e^(-|x|)` and `l(x)=x^2 sin(1/x)` if `x != 0` and `l(0)=0` then `f'(pi/2)` equals:

Options

A. (a) `l'(0)`
B. (b) `h'(0^(-))`
C. (c) `h'(0^(+))`
D. (d) `underset (x rarr 0 )(lim) (1- cos x) /(x sin x)`

Answer

To solve the problem step by step, we will analyze the functions provided and compute the necessary derivatives.

### Step 1: Define the Functions
We have:
- \( f(x) = \int_0^{g(x)} \frac{dt}{\sqrt{1+t^2}} \)
- \( g(x) = \int_0^{\cos x} (1 + \sin^2 t) \, dt \)
- \( h(x) = e^{-|x|} \)
- \( l(x) = \begin{cases} x^2 \sin(1/x) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \)

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

### Step 3: Compute \( g'(x) \)
Using the Fundamental Theorem of Calculus:
\[ g'(x) = \frac{d}{dx} \left( \int_0^{\cos x} (1 + \sin^2 t) \, dt \right) = (1 + \sin^2(\cos x)) \cdot (-\sin x) \]

### Step 4: Evaluate \( g'\left(\frac{\pi}{2}\right) \)
Now, we evaluate \( g'\left(\frac{\pi}{2}\right) \):
\[ g'\left(\frac{\pi}{2}\right) = (1 + \sin^2(0)) \cdot (-\sin\left(\frac{\pi}{2}\right)) = (1 + 0) \cdot (-1) = -1 \]

### Step 5: Compute \( f'(x) \)
Using the Fundamental Theorem of Calculus again:
\[ f'(x) = \frac{1}{\sqrt{1 + (g(x))^2}} \cdot g'(x) \]

### Step 6: Evaluate \( f'\left(\frac{\pi}{2}\right) \)
Now we can evaluate \( f'\left(\frac{\pi}{2}\right) \):
\[ f'\left(\frac{\pi}{2}\right) = \frac{1}{\sqrt{1 + (g\left(\frac{\pi}{2}\right))^2}} \cdot g'\left(\frac{\pi}{2}\right) = \frac{1}{\sqrt{1 + 0^2}} \cdot (-1) = 1 \cdot (-1) = -1 \]

### Step 7: Compare with Given Functions
Now we need to find which of the given options equals \(-1\):
1. \( l'(0) \)
2. \( h'(0) \)
3. \(-h'(0)\)
4. \( \lim_{x \to 0} \frac{1 - \cos x}{x \sin x} \)

### Step 8: Compute \( h'(0) \)
For \( h(x) = e^{-|x|} \):
- For \( x > 0 \), \( h'(x) = -e^{-x} \)
- For \( x < 0 \), \( h'(x) = e^{x} \)

At \( x = 0 \):
- Left-hand derivative: \( \lim_{x \to 0^-} h'(x) = e^{0} = 1 \)
- Right-hand derivative: \( \lim_{x \to 0^+} h'(x) = -e^{0} = -1 \)

Thus, \( h'(0) \) does not exist as left and right derivatives do not match.

### Step 9: Compute \( l'(0) \)
Using the definition of the derivative:
\[ l'(0) = \lim_{x \to 0} \frac{l(x) - l(0)}{x} = \lim_{x \to 0} \frac{x^2 \sin(1/x)}{x} = \lim_{x \to 0} x \sin(1/x) \]
Since \( |\sin(1/x)| \leq 1 \), this limit approaches \( 0 \).

### Step 10: Compute the Limit
For \( \lim_{x \to 0} \frac{1 - \cos x}{x \sin x} \):
Using L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x \sin x} = \lim_{x \to 0} \frac{\sin x}{\sin x + x \cos x} = \frac{0}{0 + 0} = 0 \]

### Conclusion
The only option that equals \(-1\) is \(-h'(0)\), which is \(-(-1) = 1\). Thus, we conclude:
\[ f'\left(\frac{\pi}{2}\right) = -1 \]

### Final Answer
The value of \( f'\left(\frac{\pi}{2}\right) \) equals \(-1\).