Given that f(x)=−5/(x−2) and c=5c=5, find f′(c)f′(c) by forming the difference quotient, f(c+h)−f(c)hf(c+h)−f(c)h, and taking the limit as h→0


To find the derivative of \( f(x) = -\frac{5}{x - 2} \) at \( c = 5 \), we'll form the difference quotient:

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

### Step 1: Compute \( f(c+h) \) and \( f(c) \)

We are given that \( c = 5 \) and \( f(x) = -\frac{5}{x - 2} \). So,

\[
f(c+h) = f(5+h) = -\frac{5}{(5+h) - 2} = -\frac{5}{h + 3}
\]
and
\[
f(c) = f(5) = -\frac{5}{5 - 2} = -\frac{5}{3}
\]

### Step 2: Form the difference quotient

Now substitute \( f(c+h) \) and \( f(c) \) into the difference quotient formula:

\[
f'(5) = \lim_{h \to 0} \frac{-\frac{5}{h+3} - \left( -\frac{5}{3} \right)}{h}
= \lim_{h \to 0} \frac{-\frac{5}{h+3} + \frac{5}{3}}{h}
\]

### Step 3: Simplify the expression

We need to combine the fractions in the numerator:

\[
f'(5) = \lim_{h \to 0} \frac{5 \left( \frac{1}{3} - \frac{1}{h+3} \right)}{h}
= \lim_{h \to 0} \frac{5 \left( \frac{(h+3) - 3}{3(h+3)} \right)}{h}
= \lim_{h \to 0} \frac{5 \left( \frac{h}{3(h+3)} \right)}{h}
\]

### Step 4: Simplify further and take the limit

Now, simplify the expression:

\[
f'(5) = \lim_{h \to 0} \frac{5h}{3h(h+3)} = \lim_{h \to 0} \frac{5}{3(h+3)}
\]

Finally, as \( h \to 0 \), we substitute \( h = 0 \):

\[
f'(5) = \frac{5}{3(0+3)} = \frac{5}{9}
\]

### Final Answer:
\[
f'(5) = \frac{5}{9}
\]

Would you like more details on any of these steps or another explanation?

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**Relative Questions**:
1. What is the geometric meaning of the derivative \( f'(5) \)?
2. Can you find the derivative of \( f(x) = -\frac{5}{x-2} \) using the power rule instead?
3. What happens to the limit if we choose a different \( c \) value in the function?
4. How does the difference quotient method generalize for any \( f(x) \)?
5. How would the process change if we had a different denominator in the function?

**Tip**: When simplifying expressions involving fractions, always look for a common denominator to combine terms efficiently.

Given that f(x) = -5/(x - 2) and c = 5, find f′(c) by forming the difference quotient, f(c+h)−f(c)/h, and taking the limit as h → 0.

Learn how to find the derivative of f(x) = -5/(x - 2) at c = 5 using the difference quotient method. This step-by-step guide covers key calculus concepts such as limits and rational functions, ideal for students in Grades 10-12.

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