Difficulty: Intermediate to Advanced | Prerequisites: Power rule, basic derivatives, properties of ln and e, Sections 2.7 and 3.1 material
Sections 3.1 through 3.3 expand your differentiation toolkit beyond polynomials. You now need to differentiate functions involving ln(x) and e^x, and combine them with the product and quotient rules. The applied problems in this section ask you to find tangent lines, compute instantaneous rates of change for financial accounts and drug concentration models, and use marginal analysis with more complex functions. This is the technical core of Exam 2: nearly every problem requires at least one of these derivative rules.
The derivative of ln(x) is 1/x; the derivative of e^x is e^x. Use the product rule when two functions are multiplied and the quotient rule when one is divided by another. For applied problems, the derivative gives you the slope of the tangent line (for tangent-line equations) or the instantaneous rate of change (for interpreting financial or scientific models).
Product rule
If f(x) = u(x) · v(x), then f'(x) = u'(x) · v(x) + u(x) · v'(x).
Think of it as: differentiate the first and multiply by the second, then add 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.
In simple terms: "low d-high minus high d-low, over the square of what's below."
Tangent line
A straight line that touches a curve at exactly one point and has the same slope as the curve at that point. Its equation is y − f(a) = f'(a)(x − a).
Instantaneous rate of change
The derivative evaluated at a specific point. It tells you how quickly the function's output is changing at that exact input value.
Think of it as the speedometer reading at one precise moment, as opposed to the average speed over a journey.
Slope-intercept form
The equation y = mx + b, where m is the slope and b is the y-intercept. Tangent line problems often ask for the answer in this form.
Logarithmic differentiation
Using properties of ln to simplify a function before differentiating. For example, rewriting ln(4/x³) as ln(4) − 3ln(x) before taking the derivative.
d/dx [ln(x)] = 1/x
Use log properties first to simplify before differentiating
ln(xy³/z) = ln(x) + 3ln(y) − ln(z), then differentiate term by term
ln(4/x³) = ln(4) − 3ln(x), derivative is −3/x
Example: y = 6ln(x) − x³ + 2∛x. Then y' = 6/x − 3x² + (2/3)x^(−2/3)
Example: y = 4√x + ln(x⁹). Rewrite as 4x^(1/2) + 9ln(x). Then y' = 2/√x + 9/x
d/dx [e^x] = e^x (the exponential function is its own derivative)
d/dx [e^(kx)] = ke^(kx) by the chain rule
Example: y = 3x^e − 2e^(5x) − ln(3x). Then y' = 3ex^(e−1) − 10e^(5x) − 1/x
Use when two functions are multiplied: (fg)' = f'g + fg'
Example: f(x) = 5x²(x³ + 2). Expand first or use the product rule: f'(x) = 10x(x³ + 2) + 5x²(3x²) = 10x⁴ + 20x + 15x⁴ = 25x⁴ + 20x
Example: y = 5x⁸e^x. Product rule gives y' = 40x⁷e^x + 5x⁸e^x = 5x⁷e^x(8 + x)
Example: f(x) = x⁷ · ln(x). Product rule: f'(x) = 7x⁶ · ln(x) + x⁷ · (1/x) = 7x⁶ln(x) + x⁶ = x⁶(7ln(x) + 1)
Use when one function is divided by another: (u/v)' = (u'v − uv') / v²
Example: y = x²/(2x − 1). Then y' = [2x(2x − 1) − x²(2)] / (2x − 1)² = (2x² − 2x) / (2x − 1)²
Example: y = (t³ − 3t)/(t² − 4). Apply the rule with u = t³ − 3t, v = t² − 4
Example: f(x) = ln(x)/(1 + x). Then f'(x) = [(1/x)(1 + x) − ln(x)(1)] / (1 + x)²
Step 1: Find f(a) to get the point (a, f(a))
Step 2: Find f'(a) to get the slope
Step 3: Use point-slope form: y − f(a) = f'(a)(x − a), then rearrange to slope-intercept form
Example: f(x) = e^x + 5x − 4 at x = 0. f(0) = 1 + 0 − 4 = −3. f'(x) = e^x + 5, so f'(0) = 6. Tangent line: y − (−3) = 6(x − 0), so y = 6x − 3
The derivative at a point gives the rate at which the function is changing at that instant
For a continuously compounded account A = 25000e^(0.084t), the rate of change is A'(t) = 25000(0.084)e^(0.084t) = 2100e^(0.084t)
After 2 years: A'(2) = 2100e^(0.168) ≈ $2,484.17 per year
Interpretation: at the 2-year mark, the balance is growing at roughly $2,484 per year
For a drug concentration C(t) = 5e^(−0.3t), the rate of change is C'(t) = −1.5e^(−0.3t)
After 1 hour: C'(1) = −1.5e^(−0.3) ≈ −1.11 mg/mL per hour (concentration is falling)
After 5 hours: C'(5) = −1.5e^(−1.5) ≈ −0.335 mg/mL per hour (still falling, but more slowly)
For S(t) = 150t/(t + 3), use the quotient rule to find S'(t) = 450/(t + 3)²
S(12) = 1800/15 = 120 (thousand), so total sales after 12 months are $120,000
S'(12) = 450/225 = 2 (thousand per month), estimating sales in month 13
Total sales after 13 months ≈ S(12) + S'(12) = 120 + 2 = $122,000
\frac{d}{dx}[\ln(x)] = \frac{1}{x}\frac{d}{dx}[e^x] = e^x\frac{d}{dx}[e^{kx}] = ke^{kx}(fg)' = f'g + fg' \quad \text{(product rule)}\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \quad \text{(quotient rule)}y - f(a) = f'(a)(x - a) \quad \text{(tangent line, point-slope form)}Log properties used before differentiating:
ln(ab) = ln(a) + ln(b)
ln(a/b) = ln(a) − ln(b)
ln(a^n) = n · ln(a)
Tangent lines are the foundation of linear approximation, which engineers and scientists use to estimate function values near a known point without recalculating the full expression. Drug concentration models like C(t) = 5e^(−0.3t) are used in pharmacokinetics to determine dosing schedules: the rate of change tells clinicians how quickly a drug is clearing and when the next dose is needed.
Students often write the derivative of ln(x) as ln(1/x) or x/ln(x). The derivative is simply 1/x.
When using the quotient rule, students frequently get the numerator sign wrong. It is u'v minus uv', not the other way round. The mnemonic "low d-high minus high d-low" can help.
Students sometimes apply the product rule when they should expand first. If f(x) = 5x²(x³ + 2), you can either use the product rule or expand to 5x⁵ + 10x² and differentiate directly. Both work, but expanding is sometimes simpler.
For tangent line problems, students occasionally evaluate f'(x) but forget to also compute f(a) for the y-coordinate. You need both the slope and the point.
⚠️ Expect several "find the derivative" problems mixing ln, e^x, products and quotients. Speed and accuracy with these rules is essential.
⚠️ Tangent line problems almost always appear. Know the three-step process: find f(a), find f'(a), write the equation.
⚠️ "Interpret your results" means write a sentence in context. "The balance is growing at $2,484 per year" is what the exam wants, not just the number.
⚠️ The quotient rule for S(t) = 150t/(t + 3) is a classic exam setup. Practise it until the algebra is automatic.
Fill in the blank: The derivative of ln(x) is ______. (1/x)
True or False: The derivative of e^(5x) is e^(5x). (False, it is 5e^(5x).)
Fill in the blank: In the quotient rule, the numerator is u'v ______ uv'. (minus, i.e. u'v − uv')
True or False: To find the tangent line at x = a, you only need f'(a). (False, you also need f(a) for the point.)
Fill in the blank: If C(t) = 5e^(−0.3t), then C'(t) = ______. (−1.5e^(−0.3t))
Q: Expand ln(xy³/z) and state the result.
A: ln(x) + 3ln(y) − ln(z).
Q: Differentiate y = 6ln(x) − x³ + 2∛x.
A: y' = 6/x − 3x² + (2/3)x^(−2/3).
Q: Differentiate y = 5x⁸e^x.
A: Use the product rule: y' = 40x⁷e^x + 5x⁸e^x = 5x⁷e^x(8 + x).
Q: Differentiate f(x) = x⁷ · ln(x).
A: Product rule: f'(x) = 7x⁶ln(x) + x⁷(1/x) = x⁶(7ln(x) + 1).
Q: Find the derivative of y = x²/(2x − 1).
A: Quotient rule: y' = [2x(2x − 1) − x²(2)] / (2x − 1)² = (4x² − 2x − 2x²) / (2x − 1)² = (2x² − 2x) / (2x − 1)².
Q: Find the tangent line to f(x) = e^x + 5x − 4 at x = 0 in slope-intercept form.
A: f(0) = 1 + 0 − 4 = −3. f'(x) = e^x + 5, so f'(0) = 6. Tangent: y = 6x − 3.
Q: An account has A = 25,000e^(0.084t). Find and interpret the instantaneous rate of change when the balance reaches $30,000.
A: Solve 30000 = 25000e^(0.084t), giving t = ln(1.2)/0.084 ≈ 2.17 years. A'(t) = 2100e^(0.084t). At that moment, A'(2.17) = 2100e^(0.182) ≈ $2,519 per year. The balance is growing at approximately $2,519 per year when it hits $30,000.
Q: S(t) = 150t/(t + 3). Find total sales after 12 months and estimate sales in the 13th month.
A: S(12) = 1800/15 = 120 (thousand). S'(t) = 450/(t + 3)², so S'(12) = 450/225 = 2 (thousand). Estimated total after 13 months: 120 + 2 = 122 (thousand), or $122,000.
The derivative rules here build directly on the power rule from earlier sections and combine with the marginal analysis framework from Section 2.7. Product and quotient rule mastery is also essential for optimisation problems later in the course, where you will set derivatives equal to zero and solve for critical points. The exponential differentiation rule (d/dx of e^(kx) = ke^(kx)) is the reason continuous compounding works so cleanly in the finance problems from Section 3.2.
derivative of ln(x), derivative of e^x, product rule, quotient rule, tangent line equation, slope-intercept form, instantaneous rate of change, logarithmic differentiation, chain rule exponential, MAC 2233, calculus for business, FIU, Sections 3.1 3.2 3.3, drug concentration model, marginal sales, rate of change interpretation, point-slope form, linear approximation