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We are given the functions \( f(x) \) and \( g(x) \), along with their derivatives \( f'(x) \) and \( g'(x) \), and the table provides the corresponding values at specific points. The function \( h(x) \) is defined as:

\[
h(x) = f(g(x))
\]

To find \( h'(x) \), we will apply the chain rule. The derivative of \( h(x) = f(g(x)) \) with respect to \( x \) is:

\[
h'(x) = f'(g(x)) \cdot g'(x)
\]

Now, we are asked to find \( h'(1) \), so we need to compute this derivative at \( x = 1 \).

1. From the table, when \( x = 1 \):
   - \( g(1) = -1 \)
   - \( g'(1) = 2 \)

2. Now, using \( g(1) = -1 \), we find \( f'(g(1)) = f'(-1) \). From the table:
   - \( f'(-1) = 5 \)

3. Finally, using the chain rule formula:

\[
h'(1) = f'(g(1)) \cdot g'(1) = f'(-1) \cdot g'(1) = 5 \cdot 2 = 10
\]

Thus, \( h'(1) = 10 \).

The correct answer is:
\[
\boxed{10}
\]

Would you like more details or have any further questions?

Here are some related questions you can explore:
1. What is the chain rule, and how does it apply to composite functions?
2. How would you compute \( h'(2) \) for \( h(x) = f(g(x)) \) using the table?
3. What happens to the derivative of a function when one of its terms has a zero derivative?
4. Could you explain how the derivative of the composite function differs from simple differentiation?
5. How does the chain rule apply in multivariable calculus?

**Tip**: Always check the values of both the function and its derivative from a table before applying the chain rule to ensure accurate results!

The table above gives values of the differentiable functions f and g and of their derivatives f' and g' at selected values of x. If h(x) = f(g(x)), then h'(1) = ?

Learn how to apply the chain rule to compute h'(1) for the composite function h(x) = f(g(x)) using a table of values for f, g, and their derivatives. This calculus problem explores composite function differentiation with the chain rule. Suitable for high school students in Grades 11-12 or introductory university-level calculus.

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