Find y' for y = 1/x
Find $y'$ for $y = \dfrac{1}{x}$.
Explanation of Method
Differentiate a reciprocal by writing it as a negative power.
Step 1: Rewrite as a power function
\[ y = \frac{1}{x} = x^{-1}. \]
Step 2: Apply the power rule
Using $(x^{n})' = nx^{n-1}$ with $n = -1$: $y' = (-1)\cdot x^{-1-1} = -x^{-2}$.
Step 3: Rewrite in fraction form
\[ y' = -\frac{1}{x^{2}}. \]
Step 4: Verification
By the quotient rule with numerator $1$ and denominator $x$: \[y' = \frac{(1)'\cdot x - 1\cdot(x)'}{x^{2}} = \frac{0\cdot x - 1}{x^{2}} = -\frac{1}{x^{2}} \;\;\checkmark\]
\[x = 1:\quad y'(1) = -1, \text{ which matches the secant slope of Problem 1 as } M_2 \to M_1 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = -\dfrac{1}{x^{2}}, \qquad x \neq 0\,}\]
Remark
Remark. The derivative is negative for every $x \neq 0$, which says that each of the two branches of the hyperbola $y = 1/x$ is strictly decreasing. Note that this does not make the function decreasing on its whole domain, since the domain is not an interval.