Difficulty: Intermediate to Advanced. Prerequisites: Limits and Continuity (Sections 2.1–2.5), Marginal Analysis (Section 2.7).
This unit covers the mechanics of differentiation: the power rule for polynomials, the product and quotient rules for combined functions, and dedicated rules for exponentials and logarithms. The applications include finding tangent-line equations, interpreting instantaneous rates of change, and solving compound-interest and radioactive-decay problems using the continuous-compounding formula A = Pe^(rt).
Derivative, f′(x)
The instantaneous rate of change of f at x. Geometrically, it is the slope of the tangent line to the graph at that point.
In simple terms: how fast the output is changing at a specific input.
Power rule
If f(x) = x^n, then f′(x) = n · x^(n−1). Works for any real exponent, including negatives and fractions.
Think of it as: bring down the exponent, then reduce it by one.
Product rule
If f(x) = u(x) · v(x), then f′(x) = u′(x) · v(x) + u(x) · v′(x).
In simple terms: derivative of the first times the second, plus the first times the derivative of the second.
Quotient rule
If f(x) = u(x)/v(x), then f′(x) = [u′(x) · v(x) − u(x) · v′(x)] / [v(x)]².
Think of it as: "low d-high minus high d-low, over low squared."
Exponential function, e^x
The function whose derivative is itself: d/dx [e^x] = e^x. For e^(kx), the derivative is k · e^(kx).
Natural logarithm, ln(x)
The inverse of e^x. Its derivative is d/dx [ln(x)] = 1/x. For ln(g(x)), the chain rule gives g′(x)/g(x).
Compound interest formula
A = P(1 + r/n)^(nt), where P is principal, r is annual rate, n is compounding periods per year, t is years.
Continuous compounding
A = Pe^(rt). The limit of compound interest as n → ∞. Used in most exam problems involving growth or decay.
Half-life
The time it takes for a quantity to reduce to half its initial value. Found by setting (1/2)P = Pe^(rt) and solving for t.
Tangent line
The line that touches the curve at exactly one point and has slope equal to f′(a) at that point. Equation: y − f(a) = f′(a)(x − a), or equivalently y = f′(a)(x − a) + f(a).
For any constant c and real exponent n:
d/dx [c] = 0.
d/dx [x^n] = n · x^(n−1).
d/dx [c · x^n] = c · n · x^(n−1).
Examples:
f(x) = x⁷, so f′(x) = 7x⁶.
f(x) = 3x² + 1, so f′(x) = 6x.
f(x) = 2/x = 2x⁻¹, so f′(x) = −2x⁻² = −2/x².
Many exam problems give expressions that need to be converted into ax^n form first.
1/x⁴ = x⁻⁴, so the derivative is −4x⁻⁵ = −4/x⁵.
³√(x⁴) = x^(4/3), so the derivative is (4/3)x^(1/3) = (4/3)³√x.
3x²√x = 3x^(5/2), so the derivative is (15/2)x^(3/2).
√x/x² = x^(−3/2), so the derivative is (−3/2)x^(−5/2).
(3x² − 5)/x = 3x − 5x⁻¹, so the derivative is 3 + 5x⁻² = 3 + 5/x².
Use when two functions are multiplied together.
f(x) = 5x²(x³ + 2). Let u = 5x², v = x³ + 2.
f′(x) = 10x(x³ + 2) + 5x²(3x²) = 10x⁴ + 20x + 15x⁴ = 25x⁴ + 20x.
Another example: y = 5x⁸e^x.
f′(x) = 40x⁷ · e^x + 5x⁸ · e^x = 5x⁷e^x(x + 8).
Use when one function is divided by another.
y = x²/(2x − 1). Let u = x², v = 2x − 1.
y′ = [2x(2x − 1) − x²(2)] / (2x − 1)² = (4x² − 2x − 2x²) / (2x − 1)² = (2x² − 2x) / (2x − 1)².
Another example: y = (t³ − 3t)/(t² − 4). Apply the same pattern: numerator = (3t² − 3)(t² − 4) − (t³ − 3t)(2t), denominator = (t² − 4)².
Sometimes it is faster to expand first than to use the product rule.
h(x) = (2x − 3)(5 − x) = 10x − 2x² − 15 + 3x = −2x² + 13x − 15.
h′(x) = −4x + 13. This is quicker and less error-prone than the product rule for simple polynomials.
d/dx [e^x] = e^x.
d/dx [e^(kx)] = k · e^(kx). (Chain rule: the exponent's derivative comes out front.)
d/dx [ln(x)] = 1/x.
d/dx [ln(x^n)] = n/x. (Use log rules first: ln(x^n) = n · ln(x).)
d/dx [ln(f(x))] = f′(x)/f(x).
Examples from the review:
y = 6 ln(x) − x³ + 2³√x. Rewrite as 6 ln(x) − x³ + 2x^(1/3). Then y′ = 6/x − 3x² + (2/3)x^(−2/3).
y = ln(4/x³) = ln(4) − 3 ln(x). Then y′ = −3/x.
y = 3x^e − 2e^(5x) − ln(3x). Then y′ = 3e · x^(e−1) − 10e^(5x) − 1/x.
f(x) = x⁷ · ln(x). Product rule: f′(x) = 7x⁶ ln(x) + x⁷ · (1/x) = 7x⁶ ln(x) + x⁶ = x⁶(7 ln(x) + 1).
ln(ab) = ln(a) + ln(b).
ln(a/b) = ln(a) − ln(b).
ln(a^n) = n · ln(a).
Example: Expand ln(xy³/z) = ln(x) + 3 ln(y) − ln(z).
Periodic compounding: A = P(1 + r/n)^(nt).
Annually: n = 1.
Monthly: n = 12.
Daily: n = 365.
Continuously: A = Pe^(rt).
Example: $5,000 invested at 10% for 8 years.
Annually: A = 5000(1.1)⁸ = $10,717.94.
Monthly: A = 5000(1 + 0.1/12)^(96) = $11,098.20 (approx).
Continuously: A = 5000e^(0.8) = $11,127.70.
These problems use the continuous formula A = Pe^(rt) and natural-log algebra.
Find time to double at 4% continuous: 2P = Pe^(0.04t). Divide: 2 = e^(0.04t). Take ln: ln(2) = 0.04t. So t = ln(2)/0.04 ≈ 17.3 years.
Find rate given time: $10,000 grows to $12,000 in 5 years. 12000 = 10000e^(5r). Divide: 1.2 = e^(5r). Take ln: ln(1.2) = 5r. So r = ln(1.2)/5 ≈ 0.0365 = 3.65%.
Find rate from doubling time: Money doubles every 13 years. 2 = e^(13r). So r = ln(2)/13 ≈ 0.0533 = 5.33%.
Same formula, but r is negative.
Half-life of 90 years: (1/2)P = Pe^(90r). So r = ln(1/2)/90 = −ln(2)/90 ≈ −0.0077, or about −0.77% per year.
Given r = −0.3% per year, find half-life: (1/2) = e^(−0.003t). So t = ln(1/2)/(−0.003) = ln(2)/0.003 ≈ 231 years.
Radium with half-life 1,690 years, 50 grams to 5 grams: need to solve for t in 5 = 50e^(rt), where r = −ln(2)/1690. First find how many half-lives: 50 → 25 → 12.5 → 6.25 → 3.125, but more precisely use the formula with the exact r.
The recipe:
Find f′(x).
Evaluate f′(a) to get the slope m at x = a.
Evaluate f(a) to get the y-coordinate of the point.
Write the line: y = m(x − a) + f(a).
Example: f(x) = e^x + 5x − 4 at x = 0. f′(x) = e^x + 5. f′(0) = 1 + 5 = 6. f(0) = 1 + 0 − 4 = −3. Tangent line: y = 6x − 3.
Example: f(x) = 1 + ln(x⁴) at x = e. Rewrite: f(x) = 1 + 4 ln(x). f′(x) = 4/x. f′(e) = 4/e. f(e) = 1 + 4 = 5. Tangent line: y = (4/e)(x − e) + 5 = (4/e)x + 1.
The derivative at a point gives the instantaneous rate of change. In applied problems, you are asked to differentiate, evaluate, and interpret.
Example: $25,000 invested at 8.4% compounded continuously. A(t) = 25000e^(0.084t). A′(t) = 2100e^(0.084t).
After 2 years: A′(2) = 2100e^(0.168) ≈ $2,484.17 per year. Interpretation: at t = 2, the account balance is growing at approximately $2,484 per year.
When balance reaches $30,000: solve 30000 = 25000e^(0.084t), giving t = ln(1.2)/0.084 ≈ 2.17 years. Then A′(2.17) ≈ 2100e^(0.182) ≈ $2,519 per year.
C(t) = 5e^(−0.3t) mg/mL. C′(t) = −1.5e^(−0.3t).
After 1 hour: C′(1) = −1.5e^(−0.3) ≈ −1.11 mg/mL per hour. The concentration is decreasing at about 1.11 mg/mL per hour.
After 5 hours: C′(5) = −1.5e^(−1.5) ≈ −0.335 mg/mL per hour. The rate of decrease has slowed.
Property tax: T(x) = 30x² + 50x + 400 dollars, x years after 2006. T′(x) = 60x + 50.
Rate of increase in 2009 (x = 3): T′(3) = 180 + 50 = $230 per year.
Change from 2006 to 2010: T(4) − T(0) = (480 + 200 + 400) − 400 = $680.
Total sales: S(t) = 0.03t³ + 0.5t² + 2t + 3 (million dollars). S′(t) = 0.09t² + t + 2. At t = 5: S(5) = 29.25 million (total). S′(5) = 0.09(25) + 5 + 2 = 9.25 million per month (rate of growth at month 5).
Sales model: S(t) = 150t/(t + 3) (thousands). S′(t) = 450/(t + 3)². At t = 12: S(12) = 120 thousand (total). S′(12) = 450/225 = 2 thousand per month. So estimated total after 13 months ≈ S(12) + S′(12) = 122 thousand.
\frac{d}{dx}[x^n] = n \cdot x^{n-1}\frac{d}{dx}[e^{kx}] = k \cdot e^{kx} \qquad \frac{d}{dx}[\ln(x)] = \frac{1}{x}(uv)' = u'v + uv' \qquad \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}A = P\left(1 + \frac{r}{n}\right)^{nt} \qquad A = Pe^{rt}\text{Half-life: } t_{1/2} = \frac{\ln 2}{|r|} \qquad \text{Doubling time: } t = \frac{\ln 2}{r}\text{Tangent line at } x = a: \quad y = f'(a)(x - a) + f(a)"The derivative of e^x is x · e^(x−1)." No. The power rule does not apply to exponential functions. The derivative of e^x is e^x. The power rule is for x^n (variable base, constant exponent), not for a^x (constant base, variable exponent).
"ln(x + y) = ln(x) + ln(y)." Wrong. The log addition rule is ln(xy) = ln(x) + ln(y). There is no shortcut for ln(x + y).
"I can use the power rule on 3^x." No. 3^x is an exponential function, not a power function. Its derivative requires the formula d/dx [a^x] = a^x · ln(a), which is beyond this section but worth flagging.
Forgetting to bring down the coefficient from the exponent in e^(kx). The derivative of e^(5x) is 5e^(5x), not e^(5x).
Mixing up the quotient rule sign. It is u′v − uv′ in the numerator ("low d-high minus high d-low"), not the other way round. Reversing the sign changes the answer entirely.
⚠️ Rewriting expressions into ax^n form before differentiating is tested heavily. If you cannot convert roots and fractions into exponent notation, the rest of the problem stalls.
⚠️ The quotient rule formula must be memorised exactly, including the sign order. Practise it until it is automatic.
⚠️ Compound interest problems typically ask you to compute A for several compounding frequencies (annually, monthly, daily, continuously) and compare. Have the formula at your fingertips.
⚠️ "Find the rate" and "find the time" problems always use A = Pe^(rt) and natural logs. The algebra is: isolate the exponential, take ln of both sides, solve.
⚠️ Tangent-line questions require both the derivative (for slope) and the original function (for the point). Missing one of these is a common exam error.
⚠️ Interpretation questions ("write a brief interpretation") expect a sentence that connects the number to the context. "The account is growing at approximately $2,484 per year at t = 2" scores full marks; a bare number does not.
Q: Differentiate f(x) = 3x²/5 − 7/(2x³) + 10.
A: Rewrite as (3/5)x² − (7/2)x⁻³ + 10. f′(x) = (6/5)x + (21/2)x⁻⁴ = 6x/5 + 21/(2x⁴).
Q: Find the derivative of y = 5x⁸e^x.
A: Product rule. y′ = 40x⁷e^x + 5x⁸e^x = 5x⁷e^x(x + 8).
Q: Differentiate f(x) = x⁷ · ln(x).
A: Product rule. f′(x) = 7x⁶ ln(x) + x⁷(1/x) = 7x⁶ ln(x) + x⁶ = x⁶(7 ln(x) + 1).
Q: $5,000 is invested at 10% annually for 8 years. Find the value if compounded continuously.
A: A = 5000e^(0.10 × 8) = 5000e^(0.8) ≈ $11,127.70.
Q: How long will it take an investment to double at 4% compounded continuously?
A: 2P = Pe^(0.04t). Divide by P: 2 = e^(0.04t). Take ln: ln(2) = 0.04t. t = ln(2)/0.04 ≈ 17.3 years.
Q: A strontium isotope has a half-life of 90 years. What is the continuous rate of decay?
A: (1/2) = e^(90r). ln(1/2) = 90r. r = −ln(2)/90 ≈ −0.0077 or −0.77% per year.
Q: Find the equation of the tangent line to f(x) = e^x + 5x − 4 at x = 0.
A: f′(x) = e^x + 5. Slope = f′(0) = 6. Point = (0, f(0)) = (0, −3). Line: y = 6x − 3.
Q: Find the derivative of y = x²/(2x − 1).
A: Quotient rule: y′ = [2x(2x − 1) − x²(2)]/(2x − 1)² = (2x² − 2x)/(2x − 1)².
derivative, differentiation, power rule, product rule, quotient rule, chain rule, exponential function, e^x, natural logarithm, ln(x), log properties, compound interest, continuous compounding, A = Pe^rt, doubling time, half-life, radioactive decay, drug concentration decay, tangent line, slope of tangent, instantaneous rate of change, average rate of change, secant line, difference quotient, rewriting expressions, negative exponents, fractional exponents, MAC 2233, calculus for business, FIU, sections 3.1 3.2 3.3
Connections to other topics: The derivative rules here are the computational engine behind marginal analysis (Section 2.7). They also feed directly into optimisation (finding maxima and minima of profit, cost, and revenue functions), which is the next major unit in the course. The exponential and logarithmic material connects to finance, biology (population growth, decay), and any field that models quantities that grow or shrink at a rate proportional to their current size.