Let `f:R to R` and `h:R to R` be differentiable functions such that `f(x)=x^(3)+3x+2,g(f(x))=x and h(g(g(x)))=x` for all `x in R`. Then, h'(1) equals.

Explanation

To find \( h'(1) \), we start with the given functions and relationships:

1. **Given Functions**:
- \( f(x) = x^3 + 3x + 2 \)
- \( g(f(x)) = x \)
- \( h(g(g(x))) = x \)

2. **Finding \( h(x) \)**:
- From \( h(g(g(x))) = x \), we can set \( x = f(x) \):
\[ h(g(g(f(x)))) = f(x) \]
- Since \( g(f(x)) = x \), we can replace \( g(f(x)) \) with \( x \):
\[ g(g(f(x))) = g(x) \]
- Thus, we have:
\[ h(g(x)) = f(x) \]

3. **Finding \( h(x) \)**:
- Now, we set \( x = f(x) \) again:
\[ h(g(f(x))) = f(f(x)) \]
- Using \( g(f(x)) = x \) again:
\[ h(x) = f(f(x)) \]

4. **Differentiating \( h(x) \)**:
- To find \( h'(x) \), we apply the chain rule:
\[ h'(x) = f'(f(x)) \cdot f'(x) \]

5. **Calculating \( h'(1) \)**:
- First, we need \( f(1) \):
\[ f(1) = 1^3 + 3 \cdot 1 + 2 = 1 + 3 + 2 = 6 \]
- Now we calculate \( f'(x) \):
\[ f'(x) = 3x^2 + 3 \]
- Next, we find \( f'(1) \):
\[ f'(1) = 3 \cdot 1^2 + 3 = 3 + 3 = 6 \]
- Now we find \( f'(f(1)) = f'(6) \):
\[ f'(6) = 3 \cdot 6^2 + 3 = 3 \cdot 36 + 3 = 108 + 3 = 111 \]

6. **Putting it all together**:
- Now we can calculate \( h'(1) \):
\[ h'(1) = f'(f(1)) \cdot f'(1) = f'(6) \cdot f'(1) = 111 \cdot 6 = 666 \]

Thus, the final answer is:
\[ \boxed{666} \]