TASSpecialist MathematicsSyllabus dot point

How do we differentiate relations that are not given as explicit functions of x?

Apply implicit and parametric differentiation and use related rates to solve problems.

Implicit differentiation, parametric differentiation and related rates of change, with fully worked examples and common pitfalls for TCE Mathematics Specialised.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

Earlier calculus assumed $y = f(x)$ written explicitly. Many important curves, such as the circle $x^2 + y^2 = 25$ , cannot be written that way over their whole domain. Implicit differentiation handles these by differentiating both sides of the equation with respect to $x$ , treating $y$ as a function of $x$ .

The engine is the chain rule. For example,

\frac{d}{dx}\big(y^3\big) = 3y^2 \frac{dy}{dx}, \qquad \frac{d}{dx}\big(\sin y\big) = \cos y \, \frac{dy}{dx}.

Products of

x

and

y

need the product rule, for instance

\tfrac{d}{dx}(xy) = y + x\tfrac{dy}{dx}

Parametric differentiation

Sometimes a curve is given as $x = x(t)$ and $y = y(t)$ . By the chain rule,

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \text{provided } \frac{dx}{dt} \neq 0.

The second derivative is found by differentiating

\tfrac{dy}{dx}

with respect to

t

and dividing again by

\tfrac{dx}{dt}

\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}.

Tangent to an implicit curve

Worked example

Find the equation of the tangent to the curve $x^2 + xy + y^2 = 7$ at the point $(1, 2)$ .

First check the point lies on the curve: $1 + (1)(2) + 4 = 7$ . Correct.

Differentiate each term with respect to $x$ :

\frac{d}{dx}(x^2) = 2x, \qquad \frac{d}{dx}(xy) = y + x\frac{dy}{dx}, \qquad \frac{d}{dx}(y^2) = 2y\frac{dy}{dx}, \qquad \frac{d}{dx}(7) = 0.

2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0.

Collect the

\tfrac{dy}{dx}

terms:

(x + 2y)\frac{dy}{dx} = -(2x + y), \qquad \frac{dy}{dx} = -\frac{2x + y}{x + 2y}.

(1, 2)

\frac{dy}{dx} = -\frac{2(1) + 2}{1 + 2(2)} = -\frac{4}{5}.

The tangent line is

y - 2 = -\frac{4}{5}(x - 1) \quad\Longrightarrow\quad 5y - 10 = -4x + 4 \quad\Longrightarrow\quad 4x + 5y = 14.

Related rates

Related rates problems link the rate of change of two quantities through an equation and the chain rule. Differentiate the relationship with respect to time $t$ , then substitute known values.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 TASC5 marksConsider the curve given implicitly by (x^2 + y^2)^3 = 8x^2 y^2. Determine the equations of the tangent and normal at the point (1, 1).

Show worked answer →

First confirm (1, 1) is on the curve: LHS = (1 + 1)^3 = 8, RHS = 8(1)(1) = 8, so it lies on the curve.

Differentiate implicitly with respect to x. Left side: 3(x^2 + y^2)^2 (2x + 2y y'). Right side: 8(2x y^2 + x^2 . 2y y').

Substitute x = 1, y = 1: 3(2)^2 (2 + 2y') = 8(2 + 2y'), i.e. 12(2 + 2y') = 8(2 + 2y'). Then (12 - 8)(2 + 2y') = 0, so 2 + 2y' = 0 and y' = -1.

Tangent: slope -1 through (1, 1): y - 1 = -1(x - 1), i.e. y = -x + 2. Normal: slope is the negative reciprocal, +1: y - 1 = 1(x - 1), i.e. y = x. Markers reward correct implicit differentiation, the value y' = -1, and both line equations.

2024 TASC5 marksIf y = (x + sqrt(x^2 + 1))^p, where p is real, (a) show that sqrt(x^2 + 1) dy/dx = p y. (b) Hence deduce that (1 + x^2) d^2y/dx^2 + x dy/dx - p^2 y = 0.

Show worked answer →

(a) Let u = x + sqrt(x^2 + 1), so y = u^p. Then du/dx = 1 + x/sqrt(x^2 + 1) = (sqrt(x^2 + 1) + x)/sqrt(x^2 + 1) = u/sqrt(x^2 + 1). By the chain rule dy/dx = p u^(p-1) . du/dx = p u^(p-1) . u/sqrt(x^2 + 1) = p u^p / sqrt(x^2 + 1) = p y / sqrt(x^2 + 1). Multiplying both sides by sqrt(x^2 + 1) gives sqrt(x^2 + 1) dy/dx = p y. (2 marks)

(b) Square the result: (x^2 + 1)(dy/dx)^2 = p^2 y^2. Differentiate with respect to x: 2x(dy/dx)^2 + (x^2 + 1) . 2 (dy/dx)(d^2y/dx^2) = p^2 . 2y (dy/dx). Divide through by 2 dy/dx (nonzero): x(dy/dx) + (1 + x^2)(d^2y/dx^2) = p^2 y. Rearranging gives (1 + x^2) d^2y/dx^2 + x dy/dx - p^2 y = 0. (3 marks)

2023 TASC5 marksIf y = [arcsin(x)]^2 show that (1 - x^2) d^2y/dx^2 - x dy/dx = 2.

Show worked answer →

Differentiate once: dy/dx = 2 arcsin(x) . 1/sqrt(1 - x^2). So sqrt(1 - x^2) dy/dx = 2 arcsin(x).

Differentiate this product again with respect to x. Left side, by the product rule: (-x/sqrt(1 - x^2)) dy/dx + sqrt(1 - x^2) d^2y/dx^2. Right side: 2 . 1/sqrt(1 - x^2).

Multiply every term by sqrt(1 - x^2): -x dy/dx + (1 - x^2) d^2y/dx^2 = 2. Rearranged, (1 - x^2) d^2y/dx^2 - x dy/dx = 2, as required. Markers reward forming sqrt(1 - x^2) dy/dx = 2 arcsin(x) before the second differentiation, which avoids a messy quotient.