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**Exercise 8:**
Let the function \( h(t) = \sqrt{4 + f(t)} \) with the given conditions:
- \( f(1) = 7 \)
- \( f'(1) = 4 \)

Calculate \( h'(1) \).

To find \( h'(t) \), we need to differentiate \( h(t) = \sqrt{4 + f(t)} \).

Using the chain rule, the derivative of \( h(t) \) with respect to \( t \) is:

\[
h'(t) = \frac{1}{2\sqrt{4 + f(t)}} \cdot f'(t)
\]

Now, substituting \( t = 1 \):

\[
h'(1) = \frac{1}{2\sqrt{4 + f(1)}} \cdot f'(1)
\]

Substituting the given values \( f(1) = 7 \) and \( f'(1) = 4 \):

\[
h'(1) = \frac{1}{2\sqrt{4 + 7}} \cdot 4 = \frac{1}{2\sqrt{11}} \cdot 4 = \frac{4}{2\sqrt{11}} = \frac{2}{\sqrt{11}}
\]

Simplifying further:

\[
h'(1) = \frac{2\sqrt{11}}{11}
\]

So, the value of \( h'(1) \) is:

\[
h'(1) = \frac{2\sqrt{11}}{11}
\]

Would you like further details or have any questions?

Here are some related questions you might find helpful:

1. How do you apply the chain rule in differentiation?
2. What are some common mistakes when differentiating functions involving square roots?
3. How can the chain rule be used in more complex functions?
4. What happens if \( f(t) \) is a more complex function? How would \( h(t) \) change?
5. How do you simplify expressions involving square roots and fractions?

**Tip:** Always simplify the expression inside the derivative as much as possible before substituting values to avoid calculation errors!

Sea la función h(t) = \sqrt{4 + f(t)} y f(1) = 7, f'(1) = 4. Calcule h'(1).

Learn how to calculate the derivative of the function h(t) = √(4 + f(t)) at a specific point t = 1, given that f(1) = 7 and f'(1) = 4. This problem covers key calculus concepts, including the application of the chain rule in differentiation. Suitable for students in Grades 11-12.

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