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To find the derivative of \( y = \frac{1}{x} \) at \( x = 2 \) using the formal definition of the derivative, we use the formula:

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

Here, \( f(x) = \frac{1}{x} \), so we need to calculate:

\[
f'(2) = \lim_{h \to 0} \frac{\frac{1}{2+h} - \frac{1}{2}}{h}
\]

Let's simplify the expression:

1. Find the difference in the numerators:
   \[
   \frac{1}{2+h} - \frac{1}{2} = \frac{2 - (2 + h)}{2(2+h)} = \frac{-h}{2(2+h)}
   \]

2. Now, plug this into the original derivative formula:
   \[
   f'(2) = \lim_{h \to 0} \frac{-h}{h \cdot 2(2+h)} = \lim_{h \to 0} \frac{-1}{2(2+h)}
   \]

3. As \( h \to 0 \), \( 2 + h \to 2 \), so:
   \[
   f'(2) = \frac{-1}{2 \cdot 2} = -\frac{1}{4}
   \]

Thus, the derivative of \( y = \frac{1}{x} \) at \( x = 2 \) is \( -\frac{1}{4} \).

Would you like more details, or do you have any questions?

Here are some related questions to expand your understanding:
1. How would the derivative change if the function were \( y = \frac{1}{x^2} \)?
2. What is the geometric interpretation of finding the derivative at a point?
3. How can the formal definition of the derivative be used to find derivatives of polynomial functions?
4. How does the derivative of \( y = \frac{1}{x} \) relate to the slope of the tangent line at a given point?
5. What are some alternative ways to find derivatives, besides using the formal definition?

**Tip:** Practicing with the formal definition strengthens your understanding of the limit process in calculus.

Use the formal definition to find the derivative of y = 1/x at x = 2.

Learn how to find the derivative of y = 1/x at x = 2 using the formal definition of the derivative. This problem covers key calculus concepts, including limits and the formal definition of derivatives. Ideal for students in Grades 11-12.

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