Difficulty: Intermediate | Prerequisites: Exponent rules, basic derivative concepts (Module 3), exponential/log properties
Derivatives measure how fast something is changing at a given instant. Antiderivatives (integrals) reverse that process: they accumulate a rate of change to find a total. In business calculus, derivatives tell you things like marginal cost (the cost of producing one more unit), and integrals tell you total cost, total depreciation, or total population change over an interval. This module pulls together the derivative rules from earlier chapters and pairs them with integration, so you need the power, product, quotient, and chain rules firmly in place before diving in.
You have four main derivative rules (power, product, quotient, chain) and three main antiderivative rules (power, log, exponential). Most exam problems combine these: differentiate a composite or product function, or integrate a rate function over an interval to find a total. The applied problems (depreciation, marginal cost, population growth) all follow the same pattern: set up the integral with the correct bounds and evaluate.
Power rule (derivatives)
If y = x^n, then y' = n x^(n-1). In simple terms, bring the exponent down as a coefficient and reduce it by one.
Product rule
If f(x) = w * v, then f'(x) = w'v + wv'. Think of it as: "derivative of the first times the second, plus the first times the derivative of the second."
Quotient rule
If y = f/g, then y' = (g f' - f g') / g^2. In simple terms, "bottom times derivative of top, minus top times derivative of bottom, all over the bottom squared." Some students remember this as "lo d-hi minus hi d-lo over lo-lo."
Chain rule
If y = (u)^n or any composite function, differentiate the outer function and multiply by the derivative of the inner function. Think of it as: "outside prime times inside prime." In the notation from the review: y' = n(u)^(n-1) * u'.
Antiderivative (indefinite integral)
The reverse of differentiation. If F'(x) = f(x), then F(x) is an antiderivative of f(x). Always add + C (the constant of integration) for indefinite integrals.
Definite integral
An integral evaluated between two bounds, giving a numerical value rather than a function. The integral from a to b of f(x) dx = F(b) - F(a). In simple terms, this is the accumulated total of f(x) over the interval [a, b].
Marginal cost, C'(x)
The derivative of the total cost function. It represents the cost of producing one additional unit. Integrating marginal cost recovers the total cost function (up to a constant).
Additional rules for e and ln (from the review)
(e^u)' = e^u * u' (derivative of e to a function equals e to that function times the derivative of the function)
(ln u)' = u'/u
(ln x)' = 1/x
ln(1) = 0
If a = e^x, then ln(a) = x
y = x^n implies y' = n x^(n-1).
This is the foundation. Every other rule builds on it. Fractional and negative exponents follow the same pattern:
y = x^(1/2) (i.e. sqrt(x)): y' = (1/2) x^(-1/2) = 1/(2 sqrt(x))
y = x^(-1) (i.e. 1/x): y' = -x^(-2) = -1/x^2
f'(x) = w'v + wv'. Use whenever two functions are multiplied together.
Worked Example 5: y = e^(sqrt(x)) * ln(x^2 + 1)
Let w = e^(sqrt(x)), v = ln(x^2 + 1)
w' = e^(sqrt(x)) * 1/(2 sqrt(x)) (chain rule on the exponent sqrt(x))
v' = 2x / (x^2 + 1) (chain rule on the inside of ln)
y' = w'v + wv' = [e^(sqrt(x)) / (2 sqrt(x))] * ln(x^2 + 1) + e^(sqrt(x)) * [2x / (x^2 + 1)]
Notice how the chain rule is used inside both the product rule pieces. This layering of rules is typical of exam questions.
y = f/g implies y' = (g f' - f g') / g^2.
Used less often in this module's review problems, but still examinable. The key is keeping track of which function is on top (f) and which is on the bottom (g).
The most frequently tested rule in this module. Differentiate the outer function, leave the inner function alone, then multiply by the derivative of the inner function.
Worked Example 4: y = (e^(5x) + 5)^6
Outer function: ( )^6. Inner function: e^(5x) + 5.
Step 1: Bring down the 6: 6(e^(5x) + 5)^5
Step 2: Multiply by the derivative of the inner: the derivative of e^(5x) + 5 is 5e^(5x) (chain rule again on e^(5x), and the derivative of the constant 5 is zero).
Result: y' = 6(e^(5x) + 5)^5 * 5e^(5x) = 30 e^(5x) (e^(5x) + 5)^5
Quiz Example 5: y = e^(1/x) * ln(x^3 + 1)
This combines the product rule and chain rule. The derivative of e^(1/x) requires rewriting 1/x as x^(-1), then:
d/dx [e^(x^(-1))] = e^(x^(-1)) * (-x^(-2)) = -e^(1/x) / x^2
d/dx [ln(x^3 + 1)] = 3x^2 / (x^3 + 1)
Apply the product rule to assemble the full derivative.
The integral of x^n dx = x^(n+1) / (n+1) + C, provided n is not equal to -1.
When n = -1, the integral of x^(-1) dx = the integral of 1/x dx = ln|x| + C. This is the log rule, and it is a separate case that students frequently forget.
The integral of 1/x dx = ln|x| + C. This appears any time you have x^(-1) in the integrand.
The integral of e^x dx = e^x + C.
Constants can be factored out of integrals. For example, the integral of 5 * x^(-1) dx = 5 ln|x| + C.
Worked Example 3: integral of (1/(3x) + 1/(5x)) dx
Rewrite as (1/3) x^(-1) + (1/5) x^(-1). Now integrate each term:
The integral of (1/3) x^(-1) dx. The power rule for x^(-1) gives ln|x|, not x^0/0 (which is undefined). So this equals (1/3) ln|x|.
Wait: the review notes show this problem involves (3x^(1/2))' + (5x)^(-1), which splits differently. Let us follow the review's working more closely.
The integral of 3^(-1) x^(-1/2) dx: rewrite as (1/3) * x^(1/2) / (1/2) = (2/3) x^(1/2). But the review also shows a term where the power rule gives an undefined result (dividing by zero), so the log rule is used instead.
Result: -1/(3x) + (1/5) ln|x| + C
The key lesson: always check whether your exponent is -1 before applying the power rule. If it is, switch to the log rule.
Worked Example 6: definite integral from 1 to 2 of (x + 4)/x dx
Split the fraction: (x + 4)/x = 1 + 4/x = 1 + 4x^(-1).
Integrate:
The integral of 1 dx = x
The integral of 4x^(-1) dx = 4 ln|x|
Combined: x + 4 ln|x|
Evaluate from 1 to 2:
F(2) = 2 + 4 ln(2)
F(1) = 1 + 4 ln(1) = 1 + 0 = 1 (since ln(1) = 0)
Answer: (2 + 4 ln 2) - 1 = 1 + 4 ln 2
Quiz Example 7: definite integral from 0 to ln 2 of (e^x - 1) dx
Split into two integrals:
The integral of e^x dx from 0 to ln 2: evaluates to e^(ln 2) - e^0 = 2 - 1 = 1
The integral of 1 dx from 0 to ln 2: evaluates to ln 2 - 0 = ln 2
Combined: 1 - ln 2
Notice the use of e^(ln 2) = 2. This identity is tested constantly.
Quiz Example 6: definite integral from 1 to 4 of (sqrt(x) + 3)/x dx
Split: sqrt(x)/x + 3/x = x^(-1/2) + 3x^(-1).
The integral of x^(-1/2) dx = x^(1/2) / (1/2) = 2 sqrt(x)
The integral of 3x^(-1) dx = 3 ln|x| (log rule, since the exponent is -1)
Evaluate from 1 to 4: [2 sqrt(4) + 3 ln 4] - [2 sqrt(1) + 3 ln 1] = [4 + 3 ln 4] - [2 + 0] = 2 + 3 ln 4
Simplify: since ln 4 = 2 ln 2, this can also be written 2 + 6 ln 2. The review writes it as 2 sqrt(2) - 2 + 3 ln 2 - 3, which simplifies to the same value.
These problems give you a rate function and ask for a total over an interval. The pattern is always the same: set up the definite integral from the start of the period to the end, integrate, and evaluate.
Worked Example 8: Surfboard depreciation
A surfboard depreciates at a rate of 100(2x - 4) per year. How much does it depreciate during the third year?
The third year runs from x = 2 to x = 3. Set up the integral from 2 to 3 of 100(2x - 4) dx.
Expand: 200x - 400
Integrate: 100x^2 - 400x
Evaluate: [100(9) - 400(3)] - [100(4) - 400(2)] = [900 - 1200] - [400 - 800] = (-300) - (-400) = 100
The surfboard depreciates by $100 during the third year.
Note: "during the third year" means from year 2 to year 3 (not from 0 to 3). This is a common place to lose marks.
Quiz Example 4: T-shirt printing
Marginal cost is C'(x) = 25 / sqrt(x) = 25 x^(-1/2). It costs $75 to print one shirt. Find the cost of printing 9 shirts.
Integrate: C(x) = integral of 25 x^(-1/2) dx = 25 * x^(1/2) / (1/2) = 50 x^(1/2) + K = 50 sqrt(x) + K
Use the initial condition C(1) = 75: 75 = 50 sqrt(1) + K = 50 + K, so K = 25.
Therefore C(x) = 50 sqrt(x) + 25.
C(9) = 50 sqrt(9) + 25 = 50(3) + 25 = 175.
The cost of printing 9 shirts is $175.
The constant K is found from the given condition. Without it, you would have infinitely many possible cost functions. This is why indefinite integrals always need + C.
Quiz Example 8: Community population growth
Population grows at a rate of (10 + 2x) persons per month. How much does it increase during the third month?
The third month runs from x = 2 to x = 3.
Integral from 2 to 3 of (10 + 2x) dx
Integrate: 10x + x^2
Evaluate: [10(3) + 9] - [10(2) + 4] = 39 - 24 = 15
The population increases by 15 people during the third month.
Same pattern as the depreciation problem: identify the correct bounds, integrate the rate, and subtract.
Derivative Rules
\text{Power: } y = x^n \implies y' = nx^{n-1}\text{Product: } f'(x) = w'v + wv'\text{Quotient: } y' = \frac{gf' - fg'}{g^2}\text{Chain: } y = [u(x)]^n \implies y' = n[u(x)]^{n-1} \cdot u'(x)(e^u)' = e^u \cdot u'(\ln u)' = \frac{u'}{u}Antiderivative (Integration) Rules
\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)\int \frac{1}{x} \, dx = \ln|x| + C\int e^x \, dx = e^x + CDefinite Integral (Fundamental Theorem of Calculus)
\int_a^b f(x)\, dx = F(b) - F(a)Marginal cost functions tell businesses exactly how much the next unit will cost to produce, which is essential for pricing decisions. Definite integrals of rate functions calculate totals: total revenue over a quarter, total pollution emitted over a period, or total depreciation of an asset.
Students apply the power rule to x^(-1) and write x^0 / 0, which is undefined. When the exponent is -1, you must use the log rule: the integral of x^(-1) dx = ln|x| + C.
Students forget the chain rule when differentiating e^(f(x)). The derivative of e^(5x) is 5e^(5x), not e^(5x). You must multiply by the derivative of the inner function.
When using the product rule, students sometimes differentiate both factors simultaneously (writing w' * v') instead of w'v + wv'. This is wrong.
For definite integrals representing "during the third year" or "during the third month," students integrate from 0 to 3 instead of from 2 to 3. The third period runs from the end of the second to the end of the third.
Chain rule problems with e^(something) and ln(something) are almost guaranteed on the exam. Practise differentiating nested exponentials and logs until it is automatic.
At least one definite integral problem will require splitting a fraction before integrating (as in Examples 6 and Quiz 6). Recognise the pattern: (a + b)/x = a/x + b/x.
The marginal-cost-to-total-cost problem type (integrate, then use an initial condition to find C) is a classic exam question. Know the steps cold.
ln(1) = 0 and e^(ln a) = a are identities that simplify final answers. They appear in nearly every definite integral involving e or ln.
Fill in the blank: The integral of x^(-1) dx = ______. (ln|x| + C)
True or False: The derivative of e^(3x) is e^(3x). (False: it is 3e^(3x).)
Fill in the blank: In the product rule, f'(x) = ______ + ______. (w'v + wv')
True or False: To find the total change during the third month, integrate from 0 to 3. (False: integrate from 2 to 3.)
Fill in the blank: e^(ln 2) = ______. (2)
Q: Find the derivative of y = (3x^2 + 1)^4.
A: Chain rule. y' = 4(3x^2 + 1)^3 * 6x = 24x(3x^2 + 1)^3.
Q: Evaluate the integral from 1 to e of (2/x) dx.
A: The integral of 2/x dx = 2 ln|x|. Evaluate: 2 ln(e) - 2 ln(1) = 2(1) - 2(0) = 2.
Q: A factory's marginal cost is C'(x) = 12 x^(1/2). If C(4) = 80, find C(x).
A: Integrate: C(x) = 12 * x^(3/2) / (3/2) + K = 8 x^(3/2) + K. Use C(4) = 80: 8(8) + K = 64 + K = 80, so K = 16. C(x) = 8 x^(3/2) + 16.
Q: Find dy/dx for y = e^(2x) * ln(x).
A: Product rule. w = e^(2x), v = ln(x). w' = 2e^(2x), v' = 1/x. y' = 2e^(2x) ln(x) + e^(2x)/x.
Q: A machine loses value at a rate of 50(x - 2) dollars per year. How much value does it lose during year 3 (from x = 2 to x = 3)?
A: Integral from 2 to 3 of 50(x - 2) dx = integral of (50x - 100) dx = [25x^2 - 100x] from 2 to 3. F(3) = 225 - 300 = -75. F(2) = 100 - 200 = -100. Answer: -75 - (-100) = 25. The machine loses $25 in value during year 3.
Derivative rules connect back to the exponential growth and decay material: the fact that (e^(rt))' = r e^(rt) is precisely why B = Pe^(rt) solves the differential equation dB/dt = rB. Integration connects forward to consumer and producer surplus in economics, and to probability distributions in statistics.
Power rule, product rule, quotient rule, chain rule, antiderivative, indefinite integral, definite integral, fundamental theorem of calculus, marginal cost, total cost, e^x derivative, ln x derivative, ln x integral, 1/x integral, exponential derivative, composite function, nested function, FIU Calculus for Business, Module 4, Professor Fox, depreciation integral, population growth rate, applied integration