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The given function is:

\[
R = f(x) = 24.2 \left( 1 + \frac{x - 13}{\sqrt{x^2 - 26x + 352}} \right)
\]

To find the derivative of \( R \) with respect to \( x \), we will use the product rule, chain rule, and the derivative of a square root function.

### Step 1: Differentiate the outer function
The outer function is:
\[
R = 24.2 \left( 1 + \frac{x - 13}{\sqrt{x^2 - 26x + 352}} \right)
\]

Since the constant multiplier \( 24.2 \) does not affect the differentiation, we only need to differentiate the expression inside the parentheses:
\[
f(x) = 1 + \frac{x - 13}{\sqrt{x^2 - 26x + 352}}
\]

### Step 2: Differentiate the inner function
The derivative of the constant \(1\) is zero. Now, differentiate the fraction:
\[
\frac{d}{dx}\left(\frac{x - 13}{\sqrt{x^2 - 26x + 352}}\right)
\]

This is a quotient, so we use the quotient rule:
\[
\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}
\]
where \( u(x) = x - 13 \) and \( v(x) = \sqrt{x^2 - 26x + 352} \).

### Step 3: Find \( u'(x) \) and \( v'(x) \)
- \( u(x) = x - 13 \), so \( u'(x) = 1 \).
- \( v(x) = \sqrt{x^2 - 26x + 352} \), so we first find \( v'(x) \) using the chain rule:
  \[
  v'(x) = \frac{1}{2\sqrt{x^2 - 26x + 352}} \cdot (2x - 26) = \frac{x - 13}{\sqrt{x^2 - 26x + 352}}
  \]

### Step 4: Apply the quotient rule
\[
\frac{d}{dx} \left( \frac{x - 13}{\sqrt{x^2 - 26x + 352}} \right) = \frac{(1) \cdot \sqrt{x^2 - 26x + 352} - (x - 13) \cdot \frac{x - 13}{\sqrt{x^2 - 26x + 352}}}{x^2 - 26x + 352}
\]

Simplify the expression:
\[
= \frac{\sqrt{x^2 - 26x + 352} - \frac{(x - 13)^2}{\sqrt{x^2 - 26x + 352}}}{x^2 - 26x + 352}
\]
\[
= \frac{x^2 - 26x + 352 - (x - 13)^2}{(x^2 - 26x + 352)^{3/2}}
\]

Now, substitute back into the derivative of \( R \):

\[
R'(x) = 24.2 \times \frac{x^2 - 26x + 352 - (x - 13)^2}{(x^2 - 26x + 352)^{3/2}}
\]

This is the derivative of the given function \( R(x) \).

Would you like more details on any part of this solution, or do you have any specific questions?

Here are five related questions to further explore the topic:

1. How can we simplify the expression further in this derivative?
2. What are some practical applications of this type of derivative in physics or engineering?
3. How does the derivative of a square root function generally behave?
4. How would we evaluate this derivative at a specific value of \( x \)?
5. How does the product rule apply when multiple functions are multiplied?

**Tip:** When differentiating complex functions, it's often useful to break them down into smaller parts and apply differentiation rules step by step.

Explore the detailed solution for finding the derivative of the function R = 24.2 (1 + (x - 13)/√(x² - 26x + 352)). This problem involves applying the quotient rule, chain rule, and the derivative of a square root function. Suitable for undergraduate calculus students.

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