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Determine f'(x) from first principles if f(x) = x³

Mathematics

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Here's the solution to Question 7 and Question 8.

QUESTION 7

7.1 Calculate the scrap value of the server at the end of 6 years.
The value depreciates using the diminishing-balance method.

  1. Use the diminishing-balance depreciation formula A=P(1−i)nA = P(1 - i)^n.

    Given: P=R500 000P = \text{R}500\,000, i=20%=0.20i = 20\% = 0.20, n=6n = 6 years.

    A=500 000(1−0.20)6A = 500\,000(1 - 0.20)^6 A=500 000(0.8)6A = 500\,000(0.8)^6 A=500 000(0.262144)A = 500\,000(0.262144) A=131 072A = 131\,072

    The scrap value of the server is R131 072.00\boxed{\text{R}131\,072.00}.

    7.2 Calculate the cost of the new server if the inflation rate is at 7% per annum.

  2. Use the compound interest formula for inflation A=P(1+i)nA = P(1 + i)^n.

    Given: P=R500 000P = \text{R}500\,000, i=7%=0.07i = 7\% = 0.07, n=6n = 6 years.

    A=500 000(1+0.07)6A = 500\,000(1 + 0.07)^6 A=500 000(1.07)6A = 500\,000(1.07)^6 A=500 000(1.50073035)A = 500\,000(1.50073035) A=750 365.175A = 750\,365.175

    The cost of the new server is R750 365.18\boxed{\text{R}750\,365.18}.

    7.3.1 Calculate the value of the monthly installment into the sinking fund.
    The sinking fund needs to accumulate the cost of the new server (from 7.2), which is R750 365.18\text{R}750\,365.18.
    Interest is 8% per annum compounded monthly. Payments are made immediately, and the last payment is at the end of the 6-year period. This implies an annuity due with 72 payments.

  3. Identify the variables.

    FV=R750 365.18FV = \text{R}750\,365.18
    i=0.0812i = \frac{0.08}{12} per month
    n=6 years×12 months/year=72 paymentsn = 6 \text{ years} \times 12 \text{ months/year} = 72 \text{ payments}

  4. Use the future value of an annuity due formula

    FV=PMT(1+i)n−1i(1+i)FV = PMT \frac{(1+i)^n - 1}{i} (1+i).

    750 365.18=PMT(1+0.0812)72−10.0812(1+0.0812)750\,365.18 = PMT \frac{\left(1 + \frac{0.08}{12}\right)^{72} - 1}{\frac{0.08}{12}} \left(1 + \frac{0.08}{12}\right)
  5. Calculate the factor.

    (1+0.0812)72−10.0812(1+0.0812)≈(1.006666667)72−10.006666667(1.006666667)\frac{\left(1 + \frac{0.08}{12}\right)^{72} - 1}{\frac{0.08}{12}} \left(1 + \frac{0.08}{12}\right) \approx \frac{(1.006666667)^{72} - 1}{0.006666667} (1.006666667) ≈1.61303639−10.006666667(1.006666667)\approx \frac{1.61303639 - 1}{0.006666667} (1.006666667) ≈0.613036390.006666667(1.006666667)\approx \frac{0.61303639}{0.006666667} (1.006666667) ≈91.9554585×1.006666667≈92.56914\approx 91.9554585 \times 1.006666667 \approx 92.56914
  6. Solve for PMTPMT.

    PMT=750 365.1892.56914PMT = \frac{750\,365.18}{92.56914} PMT≈8106.00PMT \approx 8106.00

    The monthly installment into the sinking fund is R8106.00\boxed{\text{R}8106.00}.

    7.4 After how many months will there be more than R1 000 000 in the fund?
    The business pays a monthly installment of R15 000. Assuming payments are made at the beginning of each month (annuity due, consistent with 7.3.1).

  7. Identify the variables.

    FV=R1 000 000FV = \text{R}1\,000\,000
    PMT=R15 000PMT = \text{R}15\,000
    i=0.0812i = \frac{0.08}{12} per month

  8. Use the future value of an annuity due formula

    FV=PMT(1+i)n−1i(1+i)FV = PMT \frac{(1+i)^n - 1}{i} (1+i).

    1 000 000=15 000(1+0.0812)n−10.0812(1+0.0812)1\,000\,000 = 15\,000 \frac{\left(1 + \frac{0.08}{12}\right)^n - 1}{\frac{0.08}{12}} \left(1 + \frac{0.08}{12}\right)
  9. Isolate the term with nn.

    1 000 000×0.081215 000×(1+0.0812)=(1+0.0812)n−1\frac{1\,000\,000 \times \frac{0.08}{12}}{15\,000 \times \left(1 + \frac{0.08}{12}\right)} = \left(1 + \frac{0.08}{12}\right)^n - 1 6666.6666715 000×1.006666667=(1+0.0812)n−1\frac{6666.66667}{15\,000 \times 1.006666667} = \left(1 + \frac{0.08}{12}\right)^n - 1 6666.6666715100=(1+0.0812)n−1\frac{6666.66667}{15100} = \left(1 + \frac{0.08}{12}\right)^n - 1 0.441501=(1+0.0812)n−10.441501 = \left(1 + \frac{0.08}{12}\right)^n - 1 1.441501=(1+0.0812)n1.441501 = \left(1 + \frac{0.08}{12}\right)^n
  10. Solve for nn using logarithms.

    ln⁡(1.441501)=nln⁡(1+0.0812)\ln(1.441501) = n \ln\left(1 + \frac{0.08}{12}\right) ln⁡(1.441501)=nln⁡(1.006666667)\ln(1.441501) = n \ln(1.006666667) 0.36569≈n×0.00664440.36569 \approx n \times 0.0066444 n≈0.365690.0066444≈55.04n \approx \frac{0.36569}{0.0066444} \approx 55.04

    Since the question asks "more than R1 000 000", we round up to the next whole month.
    After 56 months\boxed{56 \text{ months}} there will be more than R1 000 000 in the fund.

    QUESTION 8

    8.1 Determine f′(x)f'(x) from first principles if f(x)=x3f(x) = x^3.

  11. Write down the definition of the derivative from first principles.

    f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
  12. Substitute f(x)=x3f(x) = x^3 and f(x+h)=(x+h)3f(x+h) = (x+h)^3 into the formula.

    Recall (x+h)3=x3+3x2h+3xh2+h3(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3.

    f′(x)=lim⁡h→0(x3+3x2h+3xh2+h3)−x3hf'(x) = \lim_{h \to 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3) - x^3}{h}
  13. Simplify the numerator.

    f′(x)=lim⁡h→03x2h+3xh2+h3hf'(x) = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}
  14. Factor out hh from the numerator and cancel it with the denominator.

    f′(x)=lim⁡h→0(3x2+3xh+h2)f'(x) = \lim_{h \to 0} (3x^2 + 3xh + h^2)
  15. Apply the limit as h→0h \to 0.

    f′(x)=3x2+3x(0)+(0)2f'(x) = 3x^2 + 3x(0) + (0)^2 f′(x)=3x2f'(x) = \boxed{3x^2}

    8.2 Determine:

    8.2.1 f′(x)f'(x) if f(x)=(x+1x)2f(x) = \left(x + \frac{1}{x}\right)^2.

  16. Expand the function f(x)f(x).

    f(x)=(x+x−1)2f(x) = \left(x + x^{-1}\right)^2 f(x)=x2+2(x)(x−1)+(x−1)2f(x) = x^2 + 2(x)(x^{-1}) + (x^{-1})^2 f(x)=x2+2+x−2f(x) = x^2 + 2 + x^{-2}
  17. Differentiate f(x)f(x) term by term.

    f′(x)=ddx(x2)+ddx(2)+ddx(x−2)f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(2) + \frac{d}{dx}(x^{-2}) f′(x)=2x+0−2x−3f'(x) = 2x + 0 - 2x^{-3} f′(x)=2x−2x3f'(x) = \boxed{2x - \frac{2}{x^3}}

    8.2.2 dydx\frac{dy}{dx} if y=3x3−πxy = \frac{3}{x^3} - \pi x.

  18. Rewrite the function using negative exponents.

    y=3x−3−πxy = 3x^{-3} - \pi x
  19. Differentiate yy term by term with respect to xx.

    dydx=ddx(3x−3)−ddx(πx)\frac{dy}{dx} = \frac{d}{dx}(3x^{-3}) - \frac{d}{dx}(\pi x) dydx=3(−3)x−3−1−π(1)x1−1\frac{dy}{dx} = 3(-3)x^{-3-1} - \pi(1)x^{1-1} dydx=−9x−4−π\frac{dy}{dx} = -9x^{-4} - \pi dydx=−9x4−π\frac{dy}{dx} = \boxed{-\frac{9}{x^4} - \pi}
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