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(x))=x` for all `x in R`. Then, h'(1) equals.

Options

A. `g'(2)=(1)/(15)`
B. `h'(1)=666`
C. `h(0)=16`
D. `h(g(3))=36`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step-by-Step Solution:

1. **Identify the Functions**: We have two differentiable functions \( f: \mathbb{R} \to \mathbb{R} \) and \( h: \mathbb{R} \to \mathbb{R} \) given by:
\[ f(x) = x^3 + 3x + 2 \]
and the relationships:
\[ g(f(x)) = x \quad \text{and} \quad h(g(g(x))) = x \]

2. **Substituting \( f(x) \)**: We will replace \( x \) in the second equation with \( f(x) \):
\[ h(g(g(f(x)))) = f(x) \]
Since \( g(f(x)) = x \), we can simplify \( g(g(f(x))) \) to \( g(x) \):
\[ h(g(f(x))) = f(x) \]
Again, since \( g(f(x)) = x \), we have:
\[ h(x) = f(f(x)) \]

3. **Differentiate \( h(x) \)**: Now we differentiate \( h(x) \):
\[ h'(x) = f'(f(x)) \cdot f'(x) \]

4. **Evaluate \( h'(1) \)**: We need to find \( h'(1) \):
\[ h'(1) = f'(f(1)) \cdot f'(1) \]

5. **Calculate \( f(1) \)**:
\[ f(1) = 1^3 + 3 \cdot 1 + 2 = 1 + 3 + 2 = 6 \]

6. **Calculate \( f'(x) \)**: First, we find \( f'(x) \):
\[ f'(x) = 3x^2 + 3 \]

7. **Calculate \( f'(1) \)**:
\[ f'(1) = 3 \cdot 1^2 + 3 = 3 + 3 = 6 \]

8. **Calculate \( f'(f(1)) = f'(6) \)**:
\[ f'(6) = 3 \cdot 6^2 + 3 = 3 \cdot 36 + 3 = 108 + 3 = 111 \]

9. **Substituting back into \( h'(1) \)**:
\[ h'(1) = f'(f(1)) \cdot f'(1) = f'(6) \cdot f'(1) = 111 \cdot 6 = 666 \]

### Final Answer:
Thus, \( h'(1) = 666 \).

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