Let f(x) `=sinxandg(x)=x^(3)`
`(d)/(dx)f(g(x))` at `x=(pi)/(2)` is ________.

Options

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

Answer

To solve the problem, we need to find the derivative of the composition of the functions \( f(x) = \sin x \) and \( g(x) = x^3 \) at \( x = \frac{\pi}{2} \). We will use the chain rule for differentiation.

### Step-by-step Solution:

1. **Identify the Functions**:
- Let \( f(x) = \sin x \)
- Let \( g(x) = x^3 \)

2. **Apply the Chain Rule**:
- According to the chain rule, the derivative of \( f(g(x)) \) is given by:
\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \]

3. **Find \( f'(x) \)**:
- The derivative of \( f(x) = \sin x \) is:
\[ f'(x) = \cos x \]

4. **Find \( g'(x) \)**:
- The derivative of \( g(x) = x^3 \) is:
\[ g'(x) = 3x^2 \]

5. **Substitute \( g(x) \) into \( f' \)**:
- We need to evaluate \( f'(g(x)) \):
\[ f'(g(x)) = f'(x^3) = \cos(x^3) \]

6. **Combine the Derivatives**:
- Now substitute \( f'(g(x)) \) and \( g'(x) \) into the chain rule formula:
\[ \frac{d}{dx} f(g(x)) = \cos(g(x)) \cdot g'(x) = \cos(x^3) \cdot 3x^2 \]

7. **Evaluate at \( x = \frac{\pi}{2} \)**:
- Substitute \( x = \frac{\pi}{2} \):
\[ g\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^3 = \frac{\pi^3}{8} \]
- Therefore:
\[ \cos\left(g\left(\frac{\pi}{2}\right)\right) = \cos\left(\frac{\pi^3}{8}\right) \]
- Now calculate \( g'(\frac{\pi}{2}) \):
\[ g'\left(\frac{\pi}{2}\right) = 3\left(\frac{\pi}{2}\right)^2 = \frac{3\pi^2}{4} \]

8. **Final Calculation**:
- Substitute these values back into the derivative:
\[ \frac{d}{dx} f(g(x))\bigg|_{x=\frac{\pi}{2}} = \cos\left(\frac{\pi^3}{8}\right) \cdot \frac{3\pi^2}{4} \]

### Final Answer:
The value of \( \frac{d}{dx} f(g(x)) \) at \( x = \frac{\pi}{2} \) is:
\[ \frac{3\pi^2}{4} \cos\left(\frac{\pi^3}{8}\right) \]