Let's analyze the problem step by step based on the given graph and information.

The question is asking to find derivatives of the functions defined by:

• $u(x) = f(g(x))$,
• $v(x) = g(f(x))$,
• $w(x) = g(g(x))$.

(a) Finding $u'(1) = \frac{d}{dx}[f(g(x))]$ at $x = 1$

By the chain rule, the derivative of $u(x)$ is: $u'(x) = f'(g(x)) \cdot g'(x)$ Now, let's plug in $x = 1$: $u'(1) = f'(g(1)) \cdot g'(1)$ From the graph:

• $g(1) = 5$ (from the blue curve at $x = 1$).
• We now need $f'(5)$ and $g'(1)$:
  • $f'(5)$: At $x = 5$, the red graph (representing $f(x)$) has a sharp corner, indicating that the derivative does not exist at $x = 5$.
  • $g'(1)$: From the blue graph at $x = 1$, the slope appears to be 0, as it is a flat horizontal line.

Thus, $u'(1)$ does not exist because $f'(5)$ does not exist.

(b) Finding $v'(1) = \frac{d}{dx}[g(f(x))]$ at $x = 1$

Using the chain rule again: $v'(x) = g'(f(x)) \cdot f'(x)$ Now, evaluate at $x = 1$: $v'(1) = g'(f(1)) \cdot f'(1)$ From the graph:

• $f(1) = 1$ (from the red curve at $x = 1$),
• Therefore, we need $g'(1)$ and $f'(1)$:
  • $g'(1) = 0$ (as previously noted, the blue graph is flat at $x = 1$).
  • $f'(1)$: At $x = 1$, the red graph has a positive slope of 1.

Thus, since $g'(1) = 0$, $v'(1) = 0$.

(c) Finding $w'(1) = \frac{d}{dx}[g(g(x))]$ at $x = 1$

Again, by the chain rule: $w'(x) = g'(g(x)) \cdot g'(x)$ At $x = 1$: $w'(1) = g'(g(1)) \cdot g'(1)$ From the graph:

• $g(1) = 5$,
• We need $g'(5)$ and $g'(1)$:
  • $g'(1) = 0$ (flat blue curve at $x = 1$),
  • $g'(5)$: The blue graph has a sharp corner at $x = 5$, so the derivative $g'(5)$ does not exist.

Thus, $w'(1)$ does not exist because $g'(5)$ does not exist.

Final Answers:

• (a) $u'(1)$ does not exist because $f'(5)$ does not exist.
• (b) $v'(1) = 0$.
• (c) $w'(1)$ does not exist because $g'(5)$ does not exist.

Let me know if you want further details on any part of the solution!

Here are 5 related questions:

1. How is the chain rule applied to composite functions like $f(g(x))$?
2. What does it mean for a derivative to not exist at a corner?
3. How can we estimate the slope from a graph visually?
4. Why does $g'(1) = 0$ despite the function being defined?
5. Can you describe a situation where a derivative exists but equals zero?

Tip: Sharp corners in a function's graph are key indicators that the derivative does not exist at that point.