1. **State the problem:** We have two resistors with resistances $r$ and $s$ connected in parallel, and the combined resistance $R$ satisfies $$\frac{1}{R} = \frac{1}{r} + \frac{1}{s}.$$ We want to find the derivative $\frac{dR}{dr}$, determine if $R$ is increasing or decreasing as a function of $r$, and find the global maximum and minimum of $R$ on the interval $a \leq r \leq b$. 2. **Rewrite the formula for $R$:** From $$\frac{1}{R} = \frac{1}{r} + \frac{1}{s},$$ we get $$\frac{1}{R} = \frac{s + r}{rs} \implies R = \frac{rs}{r + s}.$$ This formula expresses $R$ explicitly in terms of $r$ and constant $s$. 3. **Find $\frac{dR}{dr}$:** Using the quotient rule for $$R = \frac{rs}{r + s},$$ $$\frac{dR}{dr} = \frac{(r + s) \cdot s - rs \cdot 1}{(r + s)^2} = \frac{s(r + s) - rs}{(r + s)^2} = \frac{sr + s^2 - rs}{(r + s)^2} = \frac{s^2}{(r + s)^2}.$$ 4. **Interpret the derivative:** Since $s > 0$ and $(r + s)^2 > 0$ for positive $r$ and $s$, we have $$\frac{dR}{dr} = \frac{s^2}{(r + s)^2} > 0.$$ This means $R$ is an **increasing** function of $r$. 5. **Find global maximum and minimum on $a \leq r \leq b$:** Since $R$ is increasing in $r$, the minimum value of $R$ on the interval is at the smallest $r$, which is $r = a$, and the maximum value is at the largest $r$, which is $r = b$. **Final answers:** $$\frac{dR}{dr} = \frac{s^2}{(r + s)^2}$$ $R$ is **increasing**. Maximum: $r = b$ Minimum: $r = a$