1. **State the problem:** We have a function for the number of bicycles assembled per day after $d$ days of training: $$M(d) = \frac{101 d^2}{2 d^2 + 7}$$ We need to find: (a) The rate of change function $M'(d)$. (b) The values and interpretation of $M'(2)$ and $M'(5)$. 2. **Find the rate of change function $M'(d)$:** Use the quotient rule for derivatives: If $M(d) = \frac{f(d)}{g(d)}$, then $$M'(d) = \frac{f'(d) g(d) - f(d) g'(d)}{(g(d))^2}$$ Here, $f(d) = 101 d^2$ and $g(d) = 2 d^2 + 7$. Calculate derivatives: $$f'(d) = 202 d$$ $$g'(d) = 4 d$$ Apply quotient rule: $$M'(d) = \frac{202 d (2 d^2 + 7) - 101 d^2 (4 d)}{(2 d^2 + 7)^2}$$ 3. **Simplify numerator:** $$202 d (2 d^2 + 7) = 202 d \times 2 d^2 + 202 d \times 7 = 404 d^3 + 1414 d$$ $$101 d^2 (4 d) = 404 d^3$$ So numerator: $$404 d^3 + 1414 d - 404 d^3 = 1414 d$$ 4. **Final derivative:** $$M'(d) = \frac{1414 d}{(2 d^2 + 7)^2}$$ 5. **Evaluate $M'(2)$:** Calculate denominator: $$2 (2)^2 + 7 = 2 \times 4 + 7 = 8 + 7 = 15$$ Square it: $$15^2 = 225$$ Calculate numerator: $$1414 \times 2 = 2828$$ So: $$M'(2) = \frac{2828}{225} \approx 12.57$$ Interpretation: At day 2, the number of bicycles assembled per day is increasing at approximately 12.57 bicycles per day per day. 6. **Evaluate $M'(5)$:** Calculate denominator: $$2 (5)^2 + 7 = 2 \times 25 + 7 = 50 + 7 = 57$$ Square it: $$57^2 = 3249$$ Calculate numerator: $$1414 \times 5 = 7070$$ So: $$M'(5) = \frac{7070}{3249} \approx 2.18$$ Interpretation: At day 5, the number of bicycles assembled per day is increasing at approximately 2.18 bicycles per day per day, showing the rate of improvement is slowing down. **Final answers:** (a) $$M'(d) = \frac{1414 d}{(2 d^2 + 7)^2}$$ (b) $$M'(2) \approx 12.57$$ (rapid increase), $$M'(5) \approx 2.18$$ (slower increase)