To analyze this expression:

$\left| \frac{f(g(t)) - f(g(x))}{t - x} - f'(g(x)) g'(x) \right|$

we'll use the functions $u$ and $v$ defined in the problem, as well as the inequalities provided.

Step-by-Step Solution:

1. Define $u(t)$:
   For $t \neq x$, $u(t) = \frac{g(t) - g(x)}{t - x} - g'(x).$ Since $g$ is differentiable at $x$, we expect that as $t \to x$, $u(t) \to 0$.

   Given Bound:
   $|u(t)| \leq |t - x|^2.$

2. Define $v(s)$:
   For $s \neq g(x)$, $v(s) = \frac{f(s) - f(g(x))}{s - g(x)} - f'(g(x)).$ Since $f$ is differentiable at $g(x)$, we expect that as $s \to g(x)$, $v(s) \to 0$.

   Given Bound:
   $|v(s)| \leq 3 |s - g(x)|.$

3. Rewrite the Main Expression Using $u$ and $v$:
   We want to evaluate $\left| \frac{f(g(t)) - f(g(x))}{t - x} - f'(g(x)) g'(x) \right|.$ Expanding $f(g(t)) - f(g(x))$ around $g(x)$ and using the definition of $v$, we get $\frac{f(g(t)) - f(g(x))}{t - x} = \frac{(f(g(t)) - f(g(x)))}{g(t) - g(x)} \cdot \frac{g(t) - g(x)}{t - x}.$ Let’s substitute: $= \left( f'(g(x)) + v(g(t)) \right) \cdot \left( g'(x) + u(t) \right).$

4. Expand and Simplify:
   Expanding this product, $= f'(g(x)) g'(x) + f'(g(x)) u(t) + g'(x) v(g(t)) + v(g(t)) u(t).$ Therefore, $\frac{f(g(t)) - f(g(x))}{t - x} - f'(g(x)) g'(x) = f'(g(x)) u(t) + g'(x) v(g(t)) + v(g(t)) u(t).$

5. Apply the Given Bounds:
   We know:

   • $|u(t)| \leq |t - x|^2$.
   • $|v(g(t))| \leq 3 |g(t) - g(x)|$.
   • $|g(t) - g(x)| \leq |t - x|^3$.

   Using these bounds:

   • $|f'(g(x)) u(t)| \leq 4 |t - x|^2$.
   • $|g'(x) v(g(t))| \leq 2 \cdot 3 |t - x|^3 = 6 |t - x|^3$.
   • $|v(g(t)) u(t)| \leq 3 |t - x|^3 \cdot |t - x|^2 = 3 |t - x|^5$.

6. Combine the Results:
   Summing these, we get: $\left| \frac{f(g(t)) - f(g(x))}{t - x} - f'(g(x)) g'(x) \right| \leq 4 |t - x|^2 + 6 |t - x|^3 + 3 |t - x|^5.$ For $|t - x| \leq 0.1$, higher powers of $|t - x|$ (like $|t - x|^3$ and $|t - x|^5$) become very small, so the dominant term here is $4 |t - x|^2$.

Final Result:

$\left| \frac{f(g(t)) - f(g(x))}{t - x} - f'(g(x)) g'(x) \right| \leq 4 |t - x|^2 + 6 |t - x|^3 + 3 |t - x|^5.$

Thus, for small $|t - x| \leq 0.1$, we can approximate:

$\left| \frac{f(g(t)) - f(g(x))}{t - x} - f'(g(x)) g'(x) \right| \approx 4 |t - x|^2.$

Would you like further clarification on any part of this solution?

Related Questions to Explore:

1. What would happen if $f'(g(x))$ or $g'(x)$ were zero?
2. How would the result change if we altered the bounds on $u(t)$ or $v(s)$?
3. Can we extend this method to other compositions of differentiable functions?
4. What is the significance of each term in the expansion in Step 4?
5. How does this result relate to the precise definition of the chain rule?

Tip:

In approximation problems, identifying the dominant term helps simplify analysis, especially when dealing with higher-order small quantities.