Difficulty: Intermediate | Prerequisites: Limits and Continuity notes, basic derivative definition
Differentiation rules are the mechanical toolkit of calculus. Once you know how to handle products, quotients, and compositions of functions, you can differentiate almost anything built from polynomials, trig functions, exponentials, and logarithms. This material is tested heavily throughout the course, and later topics (optimisation, related rates, integration by substitution) all assume fluency with these rules.
The product rule, quotient rule, and chain rule let you differentiate combinations of functions without expanding everything out first. Most exam problems layer two or more of these together (for instance, a quotient whose numerator involves a trig composition), so practise recognising which rule applies at the outermost level and work inward.
Product rule
The derivative of f(x) · g(x) is f'(x) · g(x) + f(x) · g'(x). Think of it as: each function takes a turn being differentiated while the other stays put, and you add the results.
Quotient rule
The derivative of f(x)/g(x) is [f'(x) · g(x) − f(x) · g'(x)] / [g(x)]². In simple terms, it is "low d-high minus high d-low, over the square of what's below."
Chain rule
The derivative of a composite function f(g(x)) is f'(g(x)) · g'(x). Think of it as: differentiate the outer function, leave the inner function alone, then multiply by the derivative of the inner function.
Composite function
A function built by feeding one function's output into another, written f(g(x)). The chain rule exists precisely to handle these.
Derivative of arctan(u)
The derivative is 1/(1 + u²) · u'. In simple terms, arctan flattens out as u grows, and the 1/(1 + u²) captures that.
Derivative of ln(u)
The derivative is (1/u) · u'. The natural log's rate of change is the reciprocal of its input, scaled by whatever is happening inside.
Given tables of f(x), f'(x), g(x), g'(x) at selected x values
d/dx [f(x) · g(x)] = f'(x) · g(x) + f(x) · g'(x)
Example at x = −2: f(−2) = 3, f'(−2) = −2, g(−2) = 4, g'(−2) = 3
(−2)(4) + (3)(3) = −8 + 9 = 1
Answer: 1
d/dx [f(x)/g(x)] = [f'(x) · g(x) − f(x) · g'(x)] / [g(x)]²
Example at x = 0: f(0) = 5, f'(0) = 0, g(0) = 6, g'(0) = 1
[(0)(6) − (5)(1)] / (6)² = −5/36
Answer: −5/36
For f(x) = x + sin(2x), the derivative is 1 + cos(2x) · 2 = 1 + 2cos(2x)
At x = 0: f'(0) = 1 + 2cos(0) = 1 + 2 = 3
Answer: 3 (choice f)
f(x) = (x² − 1)^(1/3). Rewrite cube root as a power of 1/3
f'(x) = (1/3)(x² − 1)^(−2/3) · 2x
At x = 3: x² − 1 = 8, so (8)^(−2/3) = 1/4
f'(3) = (1/3)(1/4)(6) = 6/12 = 1/2
Answer: 1/2
f(x) = arctan(3 ln x)
f'(x) = [1 / (1 + (3 ln x)²)] · (3/x)
At x = e: ln e = 1, so 3 ln e = 3
f'(e) = [1 / (1 + 9)] · (3/e) = (1/10)(3/e) = 3/(10e)
Answer: 3/(10e)
\frac{d}{dx}[f \cdot g] = f' \cdot g + f \cdot g'Product rule.
\frac{d}{dx}\left[\frac{f}{g}\right] = \frac{f' \cdot g - f \cdot g'}{g^2}Quotient rule.
\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)Chain rule.
Key derivatives to know cold:
d/dx [sin(u)] = cos(u) · u'
d/dx [cos(u)] = −sin(u) · u'
d/dx [arctan(u)] = u' / (1 + u²)
d/dx [ln(u)] = u' / u
d/dx [eᵘ] = eᵘ · u'
d/dx [uⁿ] = n · uⁿ⁻¹ · u' (generalised power rule)
Students often think the derivative of f · g is just f' · g'. It is not. You must use the product rule. The derivative of a product is never the product of the derivatives.
In the quotient rule, students frequently get the subtraction order wrong. It is f' · g minus f · g' in the numerator, not the other way round. Swapping them flips the sign of the answer.
When applying the chain rule, students forget to multiply by the derivative of the inner function. If you differentiate sin(2x) and get cos(2x) without the factor of 2, you have missed the chain rule.
Students sometimes apply the quotient rule when simple algebra would be faster. For instance, x²/x can be simplified to x before differentiating, which avoids the quotient rule entirely.
⚠️ Table-based product and quotient rule problems are a favourite exam format. You are given a table of values and must read off f, f', g, g' at the right x value and plug them in. Careful reading is everything.
⚠️ Chain rule problems regularly combine two or three layers, such as arctan(3 ln x). Identify the outermost function first, then work inward.
⚠️ Rewriting radicals as fractional exponents before differentiating is expected. The cube root of (x² − 1) is (x² − 1)^(1/3), and you apply the generalised power rule.
⚠️ Derivatives of inverse trig functions (especially arctan) appear on most Calculus 1 finals. Memorise d/dx [arctan(u)] = u'/(1 + u²).
True or false: The derivative of f(x) · g(x) is f'(x) · g'(x). (False. It is f' · g + f · g'.)
Fill in the blank: In the quotient rule, the numerator is f' · g ______ f · g'. (minus)
True or false: The chain rule requires you to multiply by the derivative of the inner function. (True.)
Fill in the blank: d/dx [arctan(u)] = ______ . (u' / (1 + u²))
True or false: To differentiate the cube root of (x² − 1), rewrite it as (x² − 1)^(1/3) and apply the power rule with the chain rule. (True.)
Q: Using the product rule and the data table, find d/dx [f(x) · g(x)] at x = −2 given f(−2) = 3, f'(−2) = −2, g(−2) = 4, g'(−2) = 3.
A: f' · g + f · g' = (−2)(4) + (3)(3) = −8 + 9 = 1.
Q: Using the quotient rule, find d/dx [f(x)/g(x)] at x = 0 given f(0) = 5, f'(0) = 0, g(0) = 6, g'(0) = 1.
A: [f' · g − f · g'] / g² = [(0)(6) − (5)(1)] / 36 = −5/36.
Q: Find f'(0) for f(x) = x + sin(2x).
A: f'(x) = 1 + 2cos(2x). At x = 0: 1 + 2cos(0) = 1 + 2 = 3.
Q: Find f'(3) for f(x) = (x² − 1)^(1/3).
A: f'(x) = (1/3)(x² − 1)^(−2/3) · 2x. At x = 3: (1/3)(8)^(−2/3)(6) = (1/3)(1/4)(6) = 1/2.
Q: Find f'(e) for f(x) = arctan(3 ln x).
A: f'(x) = [3/x] / [1 + (3 ln x)²]. At x = e: [3/e] / [1 + 9] = 3/(10e).
The product and quotient rules build directly on the limit definition of the derivative. The chain rule is essential for u-substitution in integration (the reverse of the chain rule). Derivatives of inverse trig functions come back in integration when you encounter forms like 1/(1 + x²).
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