Difficulty: Introductory to Intermediate | Prerequisites: Limits (the 09/19 recitation material), basic algebra and exponent rules
This topic covers the mechanics of finding derivatives: first from the limit definition, then using the shortcut rules (power, chain, product, quotient). These rules are the core toolkit for the rest of the course. If limits felt comfortable, this is the natural next step; if they did not, revisit limits first.
Derivatives measure instantaneous rate of change. You can find them the hard way (limit definition) or the fast way (rules: power, chain, product, quotient). This recitation walks through both approaches with concrete examples, from seventh roots to composite and combined functions.
Derivative
The instantaneous rate of change of f(x) with respect to x, defined as f'(x) = lim(h→0) [f(x+h) − f(x)] / h. Think of it as the slope of the function at a single point, rather than between two points.
Limit definition of the derivative
The formal definition above. Computing derivatives this way is called "differentiating from first principles." In simple terms, you build the difference quotient, simplify, then take the limit as h→0.
Power rule
If f(x) = xⁿ, then f'(x) = n · xⁿ⁻¹. Works for any real exponent, including fractions and negatives. Think of it as: bring the exponent down as a multiplier, then reduce the exponent by one.
Chain rule
If y = f(g(x)), then dy/dx = f'(g(x)) · g'(x). Used whenever one function sits inside another. In simple terms: differentiate the outer function (leaving the inner one alone), then multiply by the derivative of the inner function.
Product rule
If P(x) = f(x) · g(x), then P'(x) = f'(x) · g(x) + f(x) · g'(x). Think of it as: each factor takes a turn being differentiated while the other stays put, and you add the two results.
Quotient rule
If Q(x) = f(x) / g(x), then Q'(x) = [f'(x) · g(x) − f(x) · g'(x)] / [g(x)]². In simple terms: "low d-high minus high d-low, all over low squared." The mnemonic works.
The limit definition is the foundational approach. You will rarely use it after learning the rules, but exams test it and it is essential for understanding what a derivative actually is.
Example: g(t) = t^(1/7) (the seventh root of t)
Form the difference quotient: [g(t+h) − g(t)] / h = [(t+h)^(1/7) − t^(1/7)] / h
Rationalise by multiplying numerator and denominator by the conjugate expression: [(t+h)^(1/7) + t^(1/7)]
After simplification, the h cancels and you take lim(h→0)
Result: g'(t) = 1 / [7 · t^(6/7)]
This matches the power rule: d/dt [t^(1/7)] = (1/7) · t^(1/7 − 1) = (1/7) · t^(−6/7)
The power rule handles any function of the form xⁿ, including fractional exponents.
Example: f(x) = √x = x^(1/2), evaluate f'(4)
f'(x) = (1/2) · x^(−1/2) = 1 / (2√x)
f'(4) = 1 / (2√4) = 1 / (2 · 2) = 1/4
Used when you differentiate a composite function: an "outer" function wrapped around an "inner" function.
Example: G(x) = 9(x − 5)⁸
Outer function: 9u⁸, where u = x − 5
Derivative of outer: 9 · 8u⁷ = 72u⁷
Derivative of inner: d/dx (x − 5) = 1
G'(x) = 72(x − 5)⁷ · 1 = 72(x − 5)⁷
Used when two functions are multiplied together.
Example: g(x) = (x + 4)³ and f(x) = √(x − 1). Let P(x) = g(x) · f(x).
g'(x) = 3(x + 4)² (chain rule)
f'(x) = (1/2)(x − 1)^(−1/2) (power rule)
P'(x) = g'(x) · f(x) + g(x) · f'(x)
P'(x) = 3(x + 4)² · √(x − 1) + (x + 4)³ · (1/2)(x − 1)^(−1/2)
This can be simplified further by factoring, but the unsimplified form is typically accepted on exams.
Used when one function is divided by another.
Example: Q(x) = g(x) / f(x) = (x + 4)³ / √(x − 1)
Q'(x) = [g'(x) · f(x) − g(x) · f'(x)] / [f(x)]²
Q'(x) = [3(x + 4)² · √(x − 1) − (x + 4)³ · (1/2)(x − 1)^(−1/2)] / [(x − 1)]
The denominator is [√(x − 1)]² = (x − 1).
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \quad \text{(limit definition)}\frac{d}{dx} x^n = n \cdot x^{n-1} \quad \text{(power rule)}\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \quad \text{(chain rule)}(fg)' = f'g + fg' \quad \text{(product rule)}\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \quad \text{(quotient rule)}Students often forget the chain rule entirely. If you see a composite function like (x − 5)⁸ and write 8(x − 5)⁷ without the inner derivative, you have technically made an error, even though the inner derivative happens to be 1 in that case. For something like (3x + 1)⁸, forgetting the chain rule gives 8(3x + 1)⁷ instead of the correct 24(3x + 1)⁷.
In the quotient rule, students frequently get the subtraction order wrong. It is f'g − fg', not fg' − f'g. The numerator's derivative comes first. The mnemonic "low d-high minus high d-low" keeps the order straight.
Students sometimes apply the product rule when they should use the chain rule, or vice versa. The product rule is for f(x) · g(x) (two separate functions multiplied). The chain rule is for f(g(x)) (one function inside another). 9(x − 5)⁸ is a chain rule problem, not a product rule problem.
When differentiating √(x − 1), students sometimes write the exponent as −1/2 instead of converting correctly. √(x − 1) = (x − 1)^(1/2), so its derivative is (1/2)(x − 1)^(−1/2). The negative exponent appears after differentiation, not before.
⚠️ "Find the derivative from the definition" is a standard exam question. You will be asked to show the full limit process, not just apply the power rule. Practise the algebra of simplifying difference quotients.
⚠️ The chain rule is tested in nearly every exam problem that involves composition. If the argument of a function is anything other than plain x, the chain rule applies.
⚠️ Combined product-and-chain or quotient-and-chain problems (like the P(x) and Q(x) examples above) are common on exams. You are expected to identify which rules apply and use them together.
⚠️ Simplification is often worth marks. After applying the product or quotient rule, factor out common terms where possible. Even when the unsimplified form is accepted, a clean answer shows command of the material.
Fill in the blank: d/dx [x⁷] = ___ Answer: 7x⁶
True or False: The derivative of (2x + 3)⁵ is 5(2x + 3)⁴. Answer: False. By the chain rule it is 10(2x + 3)⁴.
Fill in the blank: f(x) = x^(1/3), so f'(x) = ___ Answer: (1/3)x^(−2/3)
True or False: The product rule states (fg)' = f' · g'. Answer: False. It is f'g + fg'.
Fill in the blank: In the quotient rule, the denominator of the result is ___. Answer: [g(x)]²
Q: Use the limit definition to find the derivative of f(x) = x² at x = 3.
A: [f(3+h) − f(3)] / h = [(3+h)² − 9] / h = [9 + 6h + h² − 9] / h = (6h + h²) / h = 6 + h. As h→0, this gives f'(3) = 6.
Q: Differentiate g(x) = (4x − 7)⁶.
A: Chain rule. g'(x) = 6(4x − 7)⁵ · 4 = 24(4x − 7)⁵.
Q: Differentiate h(x) = x³ · √(x + 2). State which rule(s) you use.
A: Product rule (with chain/power rule on the second factor). h'(x) = 3x² · √(x + 2) + x³ · (1/2)(x + 2)^(−1/2).
Q: Find f'(x) for f(x) = (x² + 1) / (x − 4).
A: Quotient rule. f'(x) = [2x(x − 4) − (x² + 1)(1)] / (x − 4)² = (2x² − 8x − x² − 1) / (x − 4)² = (x² − 8x − 1) / (x − 4)².
Q: Evaluate the derivative of f(x) = √x at x = 9.
A: f'(x) = (1/2)x^(−1/2) = 1/(2√x). At x = 9: f'(9) = 1/(2 · 3) = 1/6.
Derivatives build directly on limits (the 09/19 recitation). The limit definition of the derivative is a limit, and every shortcut rule is proved from it.
The product and quotient rules reappear in integration by parts and in solving differential equations. The chain rule is central to implicit differentiation, related rates, and later to multivariable calculus (partial derivatives, the chain rule for several variables).
Applications: derivatives give velocity from position, marginal cost from total cost, and slope from any curve. Engineers use them to find maxima and minima (optimisation), which is typically the next major topic after differentiation rules.
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