Loav · VCE Mathematical Methods · Units 3 & 4
D02 · Differential calculus · Current curriculum
Choose a rule from the structure of the expression, combine rules carefully, and keep the derivative attached to the original domain.
1. Read the structure before calculating
Differentiation becomes manageable when you identify the final operation used to build an expression. In (3x² + 1)⁴, the final operation is taking a fourth power, so a composition rule is needed. In x²e−x, two variable expressions are multiplied, so use the product rule. In ln(x)/(x + 1), one variable expression is divided by another, so use the quotient rule. The outermost structure determines the first step; smaller structures inside may require additional rules.
Some expressions are easier after rewriting. For example, 5/x² is 5x−2, and √x is x1/2. A quotient by a non-zero constant is simply a constant multiple. Before differentiating, decide whether expanding, factorising or rewriting powers will reduce the work. Do not expand a high power automatically: a chain-rule expression often stays shorter and is easier to check when left factored.
2. The basic derivatives and their conditions
| Function | Derivative | Condition to remember |
|---|---|---|
| Constant c | 0 | The graph has zero gradient. |
| xⁿ | nxn−1 | Use where the real function is differentiable. |
| eˣ | eˣ | Defined for every real x. |
| ln x | 1/x | The original real domain is x > 0. |
| sin x | cos x | x is measured in radians. |
| cos x | −sin x | x is measured in radians. |
| tan x | 1/cos²x, also written sec²x | cos x ≠ 0; use radians. |
Derivatives distribute over sums and constant multiples: the derivative of af(x) + bg(x) is af′(x) + bg′(x). This explains why a polynomial can be differentiated term by term. The derivative of x² + 7 is 2x because shifting a graph vertically does not change its tangent gradients. The constant term affects function values but disappears from the derivative.
The power rule must be interpreted together with real domains. For √x, the original domain is x ≥ 0, while the finite derivative 1/(2√x) is defined for x > 0. Negative powers exclude zero. For unrestricted rational powers, examine the real root involved rather than assuming all inputs are allowed. Working on x > 0 is a straightforward common setting for real powers, but a particular function may also be differentiable at negative inputs.
3. The chain rule accounts for the inside changing
If y = F(u) and u = g(x), then dy/dx = F′(g(x))g′(x), wherever the needed derivatives exist. Differentiate the outside while keeping the entire inside unchanged, then multiply by the derivative of the inside. The factor g′(x) measures how quickly the inner input changes. Omitting it treats g(x) as though it changes at the same rate as x, which is usually false.
For a linear inside ax + b, familiar consequences are d/dx[eax+b] = aeax+b, d/dx[ln(ax + b)] = a/(ax + b), and d/dx[sin(ax + b)] = a cos(ax + b). The logarithm additionally requires ax + b > 0. A derivative formula alone does not contain every restriction inherited from the original expression.
Worked example 1: a power of a quadratic
Differentiate f(x) = (3x² + 1)⁴. Take u = 3x² + 1. The outside derivative is 4u³, and du/dx = 6x. Therefore f′(x) = 4(3x² + 1)³ × 6x = 24x(3x² + 1)³.
At x = 1, f′(1) = 24 × 4³ = 1536. The original function and its derivative are defined for every real x. Leaving the result factored reveals that its only zero is x = 0, since 3x² + 1 is always positive. Expanding the fourth power would create more arithmetic without adding information.
The incorrect answer 4(3x² + 1)³ has differentiated only with respect to the inner quantity u. Multiplying by 6x completes the conversion to differentiation with respect to x.
4. The product rule includes two contributions
When both factors vary, either factor can contribute to the change in their product. Consequently, d(uv)/dx = u′v + uv′. Differentiate the first factor while holding the second unchanged, then add the unchanged first factor times the derivative of the second. The derivative is not u′v′. That mistaken expression misses both first-order contributions and is already contradicted by x × x = x²: it would predict 1 rather than 2x.
Worked example 2: a polynomial times an exponential
Let g(x) = x²e−x. Choose u = x² and v = e−x. Then u′ = 2x, while the chain rule gives v′ = −e−x. The product rule gives g′(x) = 2xe−x − x²e−x = xe−x(2 − x).
All real inputs are permitted. The exponential factor never equals zero, so g′(x) = 0 precisely when x = 0 or x = 2. At x = 1 the derivative is 1/e. Factoring helps both substitution and later sign analysis, but the unfactored expression is mathematically equivalent.
Notice that the product rule and chain rule are both required. Correctly identifying a product does not remove the need to differentiate its exponential factor carefully. Writing the small component derivatives first reduces sign errors.
5. The quotient rule and the inherited exclusions
For y = u/v, with v ≠ 0, the quotient rule is y′ = (u′v − uv′)/v². Keep the subtraction order fixed. Parentheses around the entire numerator are important because every term in uv′ must be subtracted. Squaring the denominator does not make excluded inputs admissible; the original quotient remains undefined wherever v is zero.
This rule is consistent with writing u/v as uv−1 and applying the product and chain rules. That connection explains why the denominator is squared and why a negative term appears. When a quotient simplifies before differentiation, record any excluded inputs first. For instance, cancelling a factor from a rational expression does not fill a hole in the original function.
Worked example 3: a logarithmic quotient
Find h′(x) for h(x) = ln(x)/(x + 1). The logarithm requires x > 0. On this domain x + 1 is already positive, so there is no additional restriction. Set u = ln x, u′ = 1/x, v = x + 1 and v′ = 1.
Thus h′(x) = [(x + 1)/x − ln x]/(x + 1)² = [x + 1 − x ln x]/[x(x + 1)²], for x > 0. At x = 1, h′(1) = 2/4 = 1/2 because ln 1 = 0.
The final rational-looking denominator might tempt you to state x ≠ 0, −1 as the domain. That would wrongly permit negative x even though ln x was never real there. Domain restrictions come from the original function as well as the derivative expression.
6. Circular functions need radians and careful notation
The derivatives of sine and cosine above assume radian measure. A calculator left in degree mode can produce misleading numerical checks even when the written differentiation is correct. For sin(2x), the derivative is 2 cos(2x). For cos(3x − 1), it is −3 sin(3x − 1). The negative sign comes from cosine; the factor 3 comes from the chain rule.
The notation sin²x means (sin x)², so its derivative is 2 sin x cos x. It does not mean sin(x²), whose derivative is 2x cos(x²). Parentheses clarify which quantity is squared. Read the expression aloud if necessary: “square of the sine” and “sine of the square” are different compositions with different derivatives.
For tan x = sin x/cos x, the quotient rule gives [cos²x + sin²x]/cos²x = 1/cos²x. This is also written sec²x. The derivative exists only where cos x ≠ 0, matching the original tangent function’s exclusions. Applying the chain rule, d/dx[tan(2x)] = 2/cos²(2x), with x ≠ π/4 + kπ/2 for integer k. This positive derivative describes increasing branches; it does not justify joining those branches across their vertical asymptotes.
Other exponential bases can be rewritten using e. For a > 0, a ≠ 1, the identity aˣ = ex ln a gives derivative aˣ ln a. Similarly, logax = ln x/ln a has derivative 1/(x ln a), with x > 0. These follow from the same chain and constant-multiple rules rather than requiring a new method. The fixed base a is a constant, while x is the changing variable.
7. A reliable working routine
Start by writing the domain and identifying the outer operation. Name components when the structure is complicated. Differentiate each component, assemble the correct rule, and only then simplify. This separates calculus decisions from algebraic manipulation. If you simplify during every intermediate step, it becomes harder to locate the source of a missing factor or sign.
Check equivalence algebraically rather than comparing appearances. The forms 2 sin x cos x and sin(2x) represent the same derivative. A CAS may produce expanded, factored or trigonometric alternatives. In technology-active work, define the function with the intended brackets, request its derivative, and compare at a valid input as a secondary check. A single matching value alone does not prove that two formulas agree everywhere.
Keep exact constants until the last line. Values involving e, π, logarithms and surds are often the intended exact answers. Approximate only when asked, or when the context requires a numerical report. Finally, state the derivative’s valid domain and identify any endpoint or excluded input needing separate reasoning. These habits remain useful when the derivative becomes part of a tangent, optimisation or numerical-root problem.
Practice: seven original questions
- Differentiate f(x) = 5x−2 − 3√x + 4 and state where the finite derivative exists.
- Differentiate g(x) = (2x − 1)⁵ and evaluate g′(1).
- Find the derivative of h(x) = e3x sin(2x).
- Find the derivative and real domain of p(x) = ln(5 − 2x).
- Differentiate q(x) = sin²x and p(x) = tan(2x). Explain why sin²x differs from sin(x²), and give the exclusions for p.
- Use the quotient rule to differentiate r(x) = (x² + 1)/(x − 2). State the excluded input.
- For s(x) = xeˣ, find s′(x) and solve s′(x) = 0 exactly.
Worked answers and reasoning
- The common domain of the original terms is x > 0: the square root needs x ≥ 0 and x−2 excludes zero. The derivative is −10x−3 − (3/2)x−1/2, for x > 0. The constant contributes zero; the negative power changes from −2 to −3.
- The outside derivative is 5(2x − 1)⁴ and the inside derivative is 2, so g′(x) = 10(2x − 1)⁴. Hence g′(1) = 10. The formula is valid for all real x. Leaving out the factor 2 would give the derivative with respect to the inner expression instead.
- The product rule gives h′(x) = 3e3xsin(2x) + 2e3xcos(2x) = e3x[3 sin(2x) + 2 cos(2x)]. Each factor needs its own chain-rule multiplier. The domain is all real x, with circular arguments in radians.
- The logarithm requires 5 − 2x > 0, so x < 5/2. By the chain rule, p′(x) = −2/(5 − 2x) on this domain. A non-zero denominator alone would be too weak: inputs greater than 5/2 make the original real logarithm undefined.
- q(x) = (sin x)², so q′(x) = 2 sin x cos x. For sin(x²), the outer sine instead gives cos(x²), multiplied by the inner derivative 2x. Its derivative is 2x cos(x²). Both are defined for all real x, but their composition order differs. For p, the derivative is 2/cos²(2x). Exclude x = π/4 + kπ/2 for integer k, because these inputs make cos(2x) zero and the original tangent undefined.
- r′(x) = [2x(x − 2) − (x² + 1)]/(x − 2)² = (x² − 4x − 1)/(x − 2)², with x ≠ 2. The subtraction applies to both terms of x² + 1. The squared denominator does not restore the excluded input.
- s′(x) = eˣ + xeˣ = eˣ(1 + x). Since eˣ is positive for every real x, the derivative is zero exactly when 1 + x = 0, giving x = −1. Dividing by eˣ is valid because it never vanishes.
Curriculum and further study
Aligned to the current VCAA VCE Mathematics Study Design, Mathematical Methods Units 3 and 4, Calculus area of study and relevant outcome knowledge and skills. The VCAA Mathematical Methods FAQ, updated February 2026, clarifies inflection points and strict increasing or decreasing behaviour. Curriculum checked 15 October 2026. Unit 3 is the suggested teaching sequence; the official design specifies combined Units 3 and 4 coverage.
Further study: Evans, Lipson and Greenwood, Cambridge Senior Mathematics VCE: Mathematical Methods Units 3 & 4, second edition (2023), sections 9B–9C and 9E–9K. Independent Loav teaching material, not a VCAA or Cambridge publication. Explanations, worked examples, practice questions, answers and diagrams are original.
