https://file.aiquickdraw.com/mathbot/file/1729137603488wdw3pwtp.jpg

Let's analyze and solve parts (a), (e), and (f) from the image you uploaded. I will focus on these specific derivatives using the provided table of values.

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### **(a)** \( \frac{d}{dx} f(g(g(x))) \bigg|_{x=1} \)

This is a composite function where the chain rule will be applied.

The chain rule states:  
\[
\frac{d}{dx} f(g(h(x))) = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)
\]

Here, we have \( f(g(g(x))) \), so we apply the chain rule twice:
\[
\frac{d}{dx} f(g(g(x))) = f'(g(g(x))) \cdot g'(g(x)) \cdot g'(x)
\]

Now, substitute \( x = 1 \):

- From the table, \( g(1) = 2 \), and \( g'(1) = \frac{1}{2} \).
- Next, \( g(2) = 1 \), and \( g'(2) = 1 \).
- Finally, \( f'(1) = 1 \).

Now calculate:
\[
f'(g(g(1))) \cdot g'(g(1)) \cdot g'(1) = f'(1) \cdot g'(2) \cdot g'(1)
\]
Substitute the values:
\[
1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}
\]

So, the answer for (a) is \( \frac{1}{2} \).

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### **(e)** \( \frac{d}{dx} \sqrt{g(x)} \bigg|_{x=2} \)

This requires applying the chain rule to \( \frac{d}{dx} \sqrt{g(x)} \), which is equivalent to \( g(x)^{1/2} \).

Differentiate:
\[
\frac{d}{dx} \sqrt{g(x)} = \frac{1}{2} g(x)^{-1/2} \cdot g'(x)
\]

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

Now calculate:
\[
\frac{1}{2} \cdot 1^{-1/2} \cdot 1 = \frac{1}{2}
\]

So, the answer for (e) is \( \frac{1}{2} \).

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### **(f)** \( \frac{d}{dx} \frac{1}{g(x)} \bigg|_{x=2} \)

This is a derivative of a reciprocal function, where we use the quotient rule or rewrite it as a negative exponent:
\[
\frac{d}{dx} \left( g(x)^{-1} \right) = -g(x)^{-2} \cdot g'(x)
\]

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

Now calculate:
\[
-1^{-2} \cdot 1 = -1
\]

So, the answer for (f) is \( -1 \).

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Would you like more detailed steps or explanations for any of these derivatives? Also, here are some related questions to further your understanding:

1. How is the chain rule applied in multi-level composite functions?
2. Why is the derivative of \( \sqrt{g(x)} \) proportional to \( g(x)^{-1/2} \)?
3. How does the quotient rule compare to using negative exponents for derivatives?
4. What are some common errors to watch out for when applying the chain rule?
5. Can you find a situation where \( g'(x) = 0 \) and analyze how it affects the derivative?

**Tip**: Always identify inner and outer functions when applying the chain rule to avoid missing terms.

Solve (a), (e), and (f) from the image, which involve derivatives of composite functions, square roots, and reciprocal functions.

Learn how to differentiate composite, square root, and reciprocal functions using the chain rule and quotient rule. This problem covers derivatives of f(g(g(x))), sqrt(g(x)), and 1/g(x) with specific examples and table-based values. Suitable for undergraduate calculus students.

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