1. **State the problem:** We have the sales function $$S(t) = \frac{125 t^2}{t^2 + 100}$$ We need to find the derivative $S'(t)$ and then find the values $S(10)$ and $S'(10)$ to interpret them. 2. **Find $S'(t)$:** Use the quotient rule for derivatives: $$\frac{d}{dt} \left( \frac{f(t)}{g(t)} \right) = \frac{f'(t) g(t) - f(t) g'(t)}{[g(t)]^2}$$ where $$f(t) = 125 t^2, \quad g(t) = t^2 + 100$$ Compute derivatives: $$f'(t) = 250 t, \quad g'(t) = 2 t$$ Apply quotient rule: $$S'(t) = \frac{250 t (t^2 + 100) - 125 t^2 (2 t)}{(t^2 + 100)^2} = \frac{250 t^3 + 25000 t - 250 t^3}{(t^2 + 100)^2}$$ Simplify numerator: $$250 t^3 - 250 t^3 = 0,$$ so numerator is $$25000 t$$ Hence, $$S'(t) = \frac{25000 t}{(t^2 + 100)^2}$$ 3. **Find $S(10)$:** $$S(10) = \frac{125 \times 10^2}{10^2 + 100} = \frac{125 \times 100}{100 + 100} = \frac{12500}{200} = 62.5$$ Interpretation: Sales at 10 months are 62.5 thousand games. 4. **Find $S'(10)$:** $$S'(10) = \frac{25000 \times 10}{(10^2 + 100)^2} = \frac{250000}{(100 + 100)^2} = \frac{250000}{200^2} = \frac{250000}{40000} = 6.25$$ Interpretation: At 10 months, sales are increasing at a rate of 6.25 thousand games per month. **Final answers:** $$S'(t) = \frac{25000 t}{(t^2 + 100)^2}$$ $$S(10) = 62.5$$ $$S'(10) = 6.25$$