We are asked to find the derivative $(f \circ g)'(1)$, which is the derivative of the composition of two functions $f(u)$ and $g(x)$ at $x = 1$.

Step 1: Chain Rule

To solve this, we use the chain rule for the composition of functions: $(f \circ g)'(x) = f'(g(x)) \cdot g'(x)$

Step 2: Find the individual derivatives

• Given functions:

  • $f(u) = u^9 - 1$
  • $u = g(x) = \sqrt{x}$

• Derivative of $g(x)$: $g(x) = \sqrt{x} = x^{1/2}$ Differentiating $g(x)$: $g'(x) = \frac{1}{2}x^{-1/2}$

• Derivative of $f(u)$: $f(u) = u^9 - 1$ Differentiating $f(u)$ with respect to $u$: $f'(u) = 9u^8$

Step 3: Apply the chain rule

Now we apply the chain rule formula: $(f \circ g)'(x) = f'(g(x)) \cdot g'(x)$

Substitute $g(x) = \sqrt{x}$ and the derivatives: $(f \circ g)'(x) = 9(g(x))^8 \cdot g'(x)$ At $x = 1$, first find the values of $g(1)$ and $g'(1)$:

• $g(1) = \sqrt{1} = 1$
• $g'(1) = \frac{1}{2}(1)^{-1/2} = \frac{1}{2}$

Now substitute these into the formula: $(f \circ g)'(1) = 9(1)^8 \cdot \frac{1}{2} = 9 \cdot \frac{1}{2} = \frac{9}{2}$

Thus, $(f \circ g)'(1) = \frac{9}{2}$.

Final Answer:

$(f \circ g)'(1) = \frac{9}{2}$

Would you like more details or have any questions?

Here are some related questions for further exploration:

1. What is the general form of the chain rule for higher-order derivatives?
2. How would the solution change if $f(u)$ were a different polynomial?
3. Can the chain rule be extended to functions of multiple variables?
4. How does the derivative of the square root function change at different points?
5. What are the conditions under which the chain rule applies?

Tip: Always double-check the intermediate steps when applying the chain rule to avoid mistakes in calculating the inner and outer derivatives.