Let's go through each part of the problem step-by-step.

Given:

1. $f(u) = u^2 + u$
2. $g(x) = \cos(x)$

We need to find the following expressions and derivatives.

Solution

1. $f(g(x))$
   Substitute $g(x) = \cos(x)$ into $f(u)$. $f(g(x)) = (\cos(x))^2 + \cos(x) = \cos^2(x) + \cos(x)$

2. $f'(u)$
   Differentiate $f(u) = u^2 + u$ with respect to $u$. $f'(u) = 2u + 1$

3. $f'(g(x))$
   Using $f'(u) = 2u + 1$, substitute $g(x)$ for $u$: $f'(g(x)) = 2\cos(x) + 1$

4. $g'(x)$
   Differentiate $g(x) = \cos(x)$ with respect to $x$. $g'(x) = -\sin(x)$

5. $(f \circ g)'(x)$
   We need to find the derivative of $f(g(x))$, which is $(f \circ g)'(x)$. Use the chain rule: $(f \circ g)'(x) = f'(g(x)) \cdot g'(x)$ Substitute $f'(g(x)) = 2\cos(x) + 1$ and $g'(x) = -\sin(x)$: $(f \circ g)'(x) = (2\cos(x) + 1)(-\sin(x)) = -\sin(x)(2\cos(x) + 1)$

Summary of Answers

1. $f(g(x)) = \cos^2(x) + \cos(x)$
2. $f'(u) = 2u + 1$
3. $f'(g(x)) = 2\cos(x) + 1$
4. $g'(x) = -\sin(x)$
5. $(f \circ g)'(x) = -\sin(x)(2\cos(x) + 1)$

Would you like further details on any of these steps?

Here are some related questions to explore:

1. How would the solution change if $g(x) = \sin(x)$ instead of $\cos(x)$?
2. What is the second derivative of $f(g(x))$ with respect to $x$?
3. How would you interpret $f(g(x))$ graphically?
4. What if $f(u)$ included a higher power term, such as $u^3$?
5. Can you find $(g \circ f)'(u)$ given the same functions?

Tip: When applying the chain rule, always identify the outer and inner functions clearly before differentiating.