If `f(x)=sinx,g(x)=x^(2)andh(x)=logx.` IF `F(x)=h(f(g(x))),` then `F'(x)` is

Options

A. `2x cot x^(2)`
B. `2 cosec^(3)x`
C. `-2cosec^(2)x`
D. None of these

Answer

To solve the problem, we need to find the derivative \( F'(x) \) where \( F(x) = h(f(g(x))) \) given the functions \( f(x) = \sin x \), \( g(x) = x^2 \), and \( h(x) = \log x \).

### Step-by-Step Solution:

1. **Identify the Functions**:
- \( f(x) = \sin x \)
- \( g(x) = x^2 \)
- \( h(x) = \log x \)

2. **Find \( F(x) \)**:
- We need to compute \( F(x) = h(f(g(x))) \).
- First, find \( g(x) \):
\[ g(x) = x^2 \]
- Now, find \( f(g(x)) \):
\[ f(g(x)) = f(x^2) = \sin(x^2) \]
- Now, substitute \( f(g(x)) \) into \( h(x) \):
\[ F(x) = h(f(g(x))) = h(\sin(x^2)) = \log(\sin(x^2)) \]

3. **Differentiate \( F(x) \)**:
- We need to find \( F'(x) \):
\[ F'(x) = \frac{d}{dx}[\log(\sin(x^2))] \]
- Using the chain rule:
\[ F'(x) = \frac{1}{\sin(x^2)} \cdot \frac{d}{dx}[\sin(x^2)] \]
- Now, differentiate \( \sin(x^2) \):
\[ \frac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot \frac{d}{dx}[x^2] = \cos(x^2) \cdot 2x \]
- Substitute back into the derivative of \( F(x) \):
\[ F'(x) = \frac{1}{\sin(x^2)} \cdot (2x \cos(x^2)) \]
- Simplifying gives:
\[ F'(x) = \frac{2x \cos(x^2)}{\sin(x^2)} = 2x \cot(x^2) \]

4. **Final Result**:
- Therefore, the derivative \( F'(x) \) is:
\[ F'(x) = 2x \cot(x^2) \]