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Let's begin solving part (a) of the problem using the formal definition of the derivative.

The given function is:
\[
f(x) = \frac{6}{x}
\]

The formal definition of the derivative is:
\[
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\]

We are tasked with finding the derivative at \(x = 5\), so we need to compute \(f'(5)\):
\[
f'(5) = \lim_{h \to 0} \frac{f(5+h) - f(5)}{h}
\]

First, let's substitute \(f(x) = \frac{6}{x}\) into this limit:

\[
f'(5) = \lim_{h \to 0} \frac{\frac{6}{5+h} - \frac{6}{5}}{h}
\]

Now let's simplify the expression inside the limit:
\[
f'(5) = \lim_{h \to 0} \frac{6 \left(\frac{1}{5+h} - \frac{1}{5}\right)}{h}
\]

To combine the terms inside the parentheses, we find a common denominator:
\[
\frac{1}{5+h} - \frac{1}{5} = \frac{5 - (5+h)}{5(5+h)} = \frac{-h}{5(5+h)}
\]

Substituting this into the limit expression:
\[
f'(5) = \lim_{h \to 0} \frac{6 \cdot \frac{-h}{5(5+h)}}{h}
\]

Simplifying:
\[
f'(5) = \lim_{h \to 0} \frac{-6h}{h \cdot 5(5+h)}
\]

Cancel out the \(h\) in the numerator and denominator:
\[
f'(5) = \lim_{h \to 0} \frac{-6}{5(5+h)}
\]

Now take the limit as \(h \to 0\):
\[
f'(5) = \frac{-6}{5(5)} = \frac{-6}{25}
\]

Thus, the derivative at \(x = 5\) is:
\[
f'(5) = \frac{-6}{25}
\]

Would you like to continue to part (b) or ask questions? Let me know if you need more details on this solution!

Here are 5 related questions:
1. How would the derivative change if the function were \( f(x) = \frac{c}{x} \) where \(c\) is a constant?
2. Can you use the product rule to differentiate \( f(x) = \frac{6}{x} \) instead of using the definition of the derivative?
3. What is the significance of the sign of \(f'(5)\)? What does it tell you about the slope of the tangent line at \(x = 5\)?
4. How would you find the second derivative of \(f(x) = \frac{6}{x}\)?
5. How would the graph of the function \(y = \frac{6}{x}\) change if the numerator were a negative constant?

**Tip:** The derivative at a point gives you the slope of the tangent line to the curve at that point.

Use the function f(x) = 6/x to answer parts (a)-(c). (a) Use the formal definition to find the derivative of y = f(x) at x = 5. (b) Find f(5) and find the equation of the normal line at the point (5, f(5)). (c) Graph y = f(x) and the tangent line at the point (5, f(5)) in the same coordinate system.

Learn how to find the derivative of the function f(x) = 6/x using the formal definition of the derivative at x = 5. This problem covers concepts in calculus, including the slope of the tangent and normal lines, and graphing the function along with its tangent line. Suitable for students taking AP Calculus or undergraduate calculus courses.

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