1. **State the problem:** Jerry invests 6000 at 7.5% per annum compounded annually. We want to find the instantaneous rate of change of the investment value at 10 years if the interest were compounded semi-annually instead. 2. **Formulas and explanation:** - Compound interest formula: $$A = P\left(1 + \frac{r}{n}\right)^{nt}$$ where $P$ is principal, $r$ is annual interest rate (decimal), $n$ is number of compounding periods per year, and $t$ is time in years. - Instantaneous rate of change is the derivative of $A$ with respect to $t$. 3. **Given values:** - $P = 6000$ - $r = 0.075$ - $n = 2$ (semi-annual compounding) 4. **Write the function for amount $A(t)$:** $$A(t) = 6000\left(1 + \frac{0.075}{2}\right)^{2t} = 6000\left(1.0375\right)^{2t}$$ 5. **Find the derivative $A'(t)$:** Using chain rule, $$A'(t) = 6000 \cdot \left(1.0375\right)^{2t} \cdot \ln(1.0375) \cdot 2$$ 6. **Calculate $A'(10)$:** $$A'(10) = 6000 \cdot (1.0375)^{20} \cdot \ln(1.0375) \cdot 2$$ Calculate intermediate values: - $(1.0375)^{20} \approx 2.030856$ - $\ln(1.0375) \approx 0.03682$ So, $$A'(10) \approx 6000 \cdot 2.030856 \cdot 0.03682 \cdot 2 = 6000 \cdot 2.030856 \cdot 0.07364$$ $$= 6000 \cdot 0.14956 = 897.36$$ 7. **Interpretation:** The instantaneous rate of change of the investment value at 10 years with semi-annual compounding is approximately 897.36 per year. **Final answer:** $$\boxed{897.36}$$