Loav · VCE Specialist Mathematics · Units 3 & 4
S34-D01 · Recommended sequence: Unit 3
1. Extend familiar differentiation rules
Advanced differentiation uses the same chain, product and quotient rules as earlier calculus, but applies them to a wider family of functions and to relationships that do not explicitly give y in terms of x. The central question remains: how does a small change in one variable affect another? Before differentiating, identify the independent variable, the dependent variable and the interval on which the relationship is meaningful.
The reciprocal circular derivatives follow directly from ordinary rules. Since sec x = (cos x)⁻¹, differentiation gives sec x tan x. Similarly, the derivative of cosec x is −cosec x cot x, and that of cot x is −cosec²x. These formulas apply only where the original functions are defined, and their conventional forms assume radian measure.
d[cosec u]/dx = −cosec u cot u · u′
d[cot u]/dx = −cosec²u · u′
The factor u′ is still required for a composite input. For example, d[sec(3x)]/dx = 3sec(3x)tan(3x), with cos(3x) ≠ 0. Replacing sec with its reciprocal representation gives an independent route to the same answer. A memorised outer derivative without the inner derivative is only part of the chain rule.
2. Derive inverse circular formulas from their branches
Let y = arcsin x, so sin y = x and −π/2 ≤ y ≤ π/2. Differentiating the relationship gives cos y · y′ = 1. Inside the branch, cos y is positive, and cos y = √(1 − sin²y) = √(1 − x²). Therefore y′ = 1/√(1 − x²) for −1 < x < 1. The function exists at the endpoints, but the displayed finite derivative does not.
For y = arccos x, differentiation of cos y = x gives −sin y · y′ = 1. On the interior of [0, π], sine is positive, giving y′ = −1/√(1 − x²). For y = arctan x, differentiating tan y = x gives sec²y · y′ = 1. Since sec²y = 1 + x², its derivative is 1/(1 + x²) for every real x.
d[arccos u]/dx = −u′/√(1 − u²)
d[arctan u]/dx = u′/(1 + u²)
The arcsine and arccosine formulas require |u| < 1 where they are applied directly. Endpoint behaviour may need separate analysis rather than substitution into a zero denominator. In a transformed function, solve the inequality for x before announcing a derivative domain. The positive square root in the derivation is justified by the chosen branch, not by a general rule that every cosine is positive.
Worked example 1: Differentiate a composite inverse function
Let f(x) = arcsin(2x − 1) + arctan(x²). Its real function domain is 0 ≤ x ≤ 1: the first term requires −1 ≤ 2x − 1 ≤ 1, and the second adds no restriction. For 0 < x < 1, the chain rule gives f′(x) = 2/√[1 − (2x − 1)²] + 2x/(1 + x⁴).
At x = 1/2, the slope is 2 + 1/(17/16) = 50/17. The point on the curve is (1/2, arctan(1/4)), because the arcsine term is zero. Hence the tangent is y − arctan(1/4) = (50/17)(x − 1/2).
The derivative calculation does not remove the function's endpoints, but its finite formula is valid only in the interval interior. Keep the distinction between where the function exists and where this derivative expression is finite. A decimal angle is unnecessary in the exact tangent equation.
3. Differentiate a relationship without solving it first
An equation such as x² + xy + y² = 7 defines a curve implicitly. Near a suitable point, y can be regarded as a differentiable function of x even when solving explicitly is awkward. Differentiate every term with respect to x. The derivative of x² is 2x, that of xy is y + xy′ by the product rule, and that of y² is 2yy′ by the chain rule.
Collect all terms containing y′ and solve for it. The result is a slope formula involving x and y. Substitute the coordinates of the specified point only after differentiating the full relationship. Substituting too early turns variable terms into constants and destroys the derivative information. Check that the specified point actually lies on the curve before using it.
An implicit curve can have several local branches, and a denominator zero in the solved slope formula needs interpretation. If the numerator is nonzero while the denominator is zero, a vertical tangent may occur at a regular point. If both vanish, the formula alone gives no slope information; do not label the point vertical by default. The local curve geometry must be examined separately.
Worked example 2: Finish both requested line equations
Find the tangent and normal to x² + xy + y² = 7 at (1, 2). The point check is 1 + 2 + 4 = 7. Differentiating gives 2x + y + xy′ + 2yy′ = 0, so y′ = −(2x + y)/(x + 2y). At the point, y′ = −4/5.
The tangent is y − 2 = −(4/5)(x − 1), equivalently 4x + 5y = 14. A normal is perpendicular to this tangent, so its slope is 5/4 and its equation is y − 2 = (5/4)(x − 1), equivalently 5x − 4y = −3.
A response ending at −4/5 has found a gradient but has not answered a request for a tangent equation. The negative-reciprocal rule for normal slope applies to finite nonzero tangent slope. A horizontal tangent has a vertical normal, and a vertical tangent has a horizontal normal; those cases are better expressed directly as line equations.
4. Reverse a derivative only when a local inverse exists
Suppose x = f(y). If f is locally one-to-one and differentiable with a finite nonzero derivative at the point, then dy/dx = 1/(dx/dy). This reciprocal relationship describes inverse local rates. It does not mean that an arbitrary derivative can be inverted without conditions, or that d²y/dx² is the reciprocal of d²x/dy².
For x = y³ + y, we have dx/dy = 3y² + 1 > 0. Thus dy/dx = 1/(3y² + 1), and at y = 1, where x = 2, the slope is 1/4. The positive derivative also supports a single-valued inverse on the entire real line. By contrast, x = y² is not one-to-one on all real y. On the branch y > 0, dy/dx = 1/(2y), but at y = 0 the reciprocal formula cannot give a finite slope.
Sometimes treating x as a function of y is the simpler calculation. Choose the representation that avoids unnecessary algebra, but report the slope or equation asked for. A question requesting dy/dx is not answered by an unlabelled value of dx/dy. Variable order in derivative notation is part of its meaning.
5. Related rates connect a changing model
A related-rates question gives two or more quantities changing with time. Build an equation linking their current values, differentiate it with respect to time, and substitute the values at the requested instant. Draw or describe the geometry first. Distinguish fixed dimensions from changing dimensions and decide which variable carries the supplied rate.
The units should fit the chain rule. If V is measured in cm³ and h in cm, then dV/dh has units cm². Multiplying it by dh/dt in cm/s gives dV/dt in cm³/s. A shrinking quantity has a negative rate even when its current length or volume is positive. Rates report direction of change, whereas dimensions report current size.
Never substitute the instantaneous radius or height into a volume formula before differentiating if that variable is changing. Doing so replaces the evolving quantity with a constant. Instead derive a general rate equation and then evaluate it. If two geometric dimensions vary together, express one in terms of the other using the fixed shape before taking the derivative.
Worked example 3: A cone with a changing water level
A cone-shaped container has height 12 cm and top radius 4 cm. Water enters at 6π cm³/s. Find the rate at which the water depth rises when h = 6 cm. Similar triangles give r/h = 4/12, so r = h/3. The water volume is V = πr²h/3 = πh³/27.
Differentiating with respect to time gives dV/dt = (πh²/9)dh/dt. At h = 6, substitute 6π = 4π dh/dt, hence dh/dt = 3/2 cm/s. The sign is positive because the volume is increasing.
The rate is not constant as the container fills: dh/dt = 54/h² under the model. A wider water surface means the same incoming volume produces a smaller rise in height. The model applies while 0 < h < 12 and the assumed inflow persists. It should not be extended through an empty cone's tip using a finite-height-rate claim.
6. Check the model as well as the algebra
A meaningful technology check uses at least two representations. For an implicit tangent, plot the curve and your line near the point and verify the derivative symbolically. For an inverse circular expression, check the domain before comparing numerical slopes. For a rate model, predict the sign and units before using a calculator. Agreement with an expected sign does not prove the calculation, but disagreement can reveal a mistaken relationship quickly.
A non-routine extension is to ask when a related rate doubles. In the cone example, dh/dt is proportional to h⁻². Halving the depth quadruples the rise rate; to double the rate, divide the depth by √2. This conclusion follows from the rate relationship without resolving the entire time-dependent filling problem. It also shows why “twice the height means twice the rate” is unreliable when geometry is nonlinear.
Before finishing, verify the requested object. A derivative, a tangent line, a normal line and a numerical rate with units are different deliverables. Write enough reasoning to make the modelling choice and the chain-rule factor visible. The next guide, second derivatives and concavity, uses these first derivatives to investigate how the slopes themselves change.
Practise, then check
- Differentiate arccos(3x), stating the function domain and the interval where the displayed derivative is finite.
- Differentiate sec²x and state its excluded inputs.
- Find the tangent to x² + 4y² = 20 at (2, 2).
- If x = y³ + 2y, find dy/dx at y = 1 and identify the corresponding x-coordinate.
- A sphere's radius increases at 0.2 cm/s. Find dV/dt when r = 5 cm.
- A 5 m ladder has its foot 3 m from a vertical wall and moving outward at 0.4 m/s. Find the top's vertical rate at that instant.
- Find the normal to y = arctan(2x) at x = 0. Explain why a gradient alone is insufficient.
Fully explained answers
- The function requires −1/3 ≤ x ≤ 1/3. The chain rule gives −3/√(1 − 9x²), finite for −1/3 < x < 1/3. The negative sign reflects the decreasing arccosine branch; the numerator 3 comes from its inner function.
- Treat the expression as a square of sec x. Its derivative is 2sec x · sec x tan x = 2sec²x tan x. Exclude x = π/2 + kπ for integer k, where the original secant is undefined.
- Differentiation gives 2x + 8yy′ = 0, so y′ = −x/(4y). At (2, 2) the slope is −1/4. The tangent is y − 2 = −(x − 2)/4, or x + 4y = 10. The point satisfies the original ellipse equation.
- Here dx/dy = 3y² + 2, which is positive. At y = 1, it equals 5, so dy/dx = 1/5. The corresponding point is (3, 1), since x = 1 + 2. The nonzero derivative justifies the reciprocal.
- Differentiate V = 4πr³/3 to obtain dV/dt = 4πr² dr/dt. Substitution gives 4π × 25 × 0.2 = 20π cm³/s. The positive sign agrees with the increasing radius.
- Let x be horizontal distance and y vertical height. From x² + y² = 25, the instant has y = 4. Differentiation gives 2x dx/dt + 2y dy/dt = 0. Thus dy/dt = −3(0.4)/4 = −0.3 m/s: the top moves downward.
- The derivative is 2/(1 + 4x²), so the tangent slope at (0, 0) is 2. The normal slope is −1/2, giving y = −x/2. The requested object is a line through a specified point; the number −1/2 alone does not identify that line.
Curriculum and sources
VCE Specialist Mathematics Units 3 and 4, Areas of Study 4 as applicable, with Outcomes 1–3. Scope follows the VCAA Mathematics Study Design 2023–2027 and the current Specialist Mathematics FAQ. Curriculum checked 15 October 2026. Teaching reference: Cambridge Senior Mathematics VCE Specialist Mathematics 3&4, second edition (2023), sections 8A–8C (pp317–330);8F (pp346–353);8H–8I (pp363–370). All teaching prose, examples, practice questions and diagrams are original. Independent teaching material; not a VCAA publication.
