What is wrong product rule on a mixed term?

ddx(xy) = y + xdydx; both terms are needed.

Find dydx for x^2 + y^2 = 16. [2 marks]

A curve has x = t^2, y = 2t. Find dydx in terms of t. [2 marks]

SEAB Original-style practice: A curve has equation x^2 + xy + y^2 = 7. Find dydx in terms of x and y, and the gradient at the point (1, 2).

Differentiate implicitly with respect to x, treating y as a function of x: 2x + (y + xdydx) + 2ydydx = 0 (product rule on xy). Collect: 2x + y + (x + 2y)dydx = 0, so dydx = -2x + yx + 2y. At (1, 2): dydx = -2 + 21 + 4 = -45. Markers reward differentiating each term implicitly (product rule on xy, chain rule on y^2), solving for dydx, and the numerical gradient.

SEAB Original-style practice: A curve is defined parametrically by x = t^2, y = t^3 - 3t. Find dydx in terms of t and the values of t at which the tangent is horizontal.

dxdt = 2t and dydt = 3t^2 - 3. dydx = dy/dtdx/dt = 3t^2 - 32t = 3(t^2 - 1)2t. The tangent is horizontal where dydx = 0, that is t^2 - 1 = 0 (and t ≠ 0), so t = 1. Markers reward differentiating both parametric equations, dividing to get dydx, and setting the numerator to zero for horizontal tangents.

SingaporeMathsSyllabus dot point

How do we differentiate when y is defined implicitly or through a parameter?

Differentiate relations defined implicitly and curves defined parametrically, and find gradients, tangents and second derivatives in each case

A focused answer to the H2 Mathematics outcome on implicit and parametric differentiation. Differentiating implicit relations, finding dy/dx parametrically via the chain rule, and obtaining tangents and second derivatives.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-06

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

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What this dot point is asking
The answer
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Try this

What this dot point is asking

SEAB wants you to differentiate relations where $y$ is not given explicitly (implicit differentiation) and curves given parametrically, in each case finding $\dfrac{dy}{dx}$ , gradients at points, tangent equations, and where needed the second derivative.

The answer

Implicit differentiation

When a curve is given as a relation $\mathrm{F}(x, y) = 0$ that you cannot or need not solve for $y$ , differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ . Every $y$ -term picks up a factor $\dfrac{dy}{dx}$ via the chain rule:

\frac{d}{dx}\big(y^n\big) = n y^{n-1}\frac{dy}{dx}, \qquad \frac{d}{dx}(xy) = y + x\frac{dy}{dx}.

Then collect the $\dfrac{dy}{dx}$ terms and solve.

Parametric differentiation

When $x = \mathrm{f}(t)$ and $y = \mathrm{g}(t)$ , the chain rule gives

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

Differentiate each equation with respect to $t$ and divide. The gradient is naturally a function of $t$ .

Second derivatives

For a parametric curve the second derivative is not the second $t$ -derivative ratio; it is

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

Differentiate $\dfrac{dy}{dx}$ with respect to $t$ , then divide by $\dfrac{dx}{dt}$ again.

Tangents

Once you have $\dfrac{dy}{dx}$ at a point, the tangent uses the usual point-gradient form. Horizontal tangents occur where the numerator of $\dfrac{dy}{dx}$ is zero; vertical tangents where the denominator is zero.

Worked example

A curve is defined parametrically by $x = 2\cos t$ , $y = 3\sin t$ . Find $\dfrac{dy}{dx}$ and the equation of the tangent at $t = \dfrac{\pi}{6}$ .

Step 1: Differentiate each equation with respect to t

\frac{dx}{dt} = -2\sin t, \qquad \frac{dy}{dt} = 3\cos t

Step 2: Form dy/dx

\frac{dy}{dx} = \frac{3\cos t}{-2\sin t} = -\frac{3}{2}\cot t

Step 3: Evaluate at t = pi/6

$\cot\dfrac{\pi}{6} = \sqrt{3}$ , so $\dfrac{dy}{dx} = -\dfrac{3}{2}\sqrt{3} = -\dfrac{3\sqrt{3}}{2}$ .

Step 4: Find the point

$x = 2\cos\dfrac{\pi}{6} = 2 \cdot \dfrac{\sqrt{3}}{2} = \sqrt{3}$ ; $y = 3\sin\dfrac{\pi}{6} = 3 \cdot \dfrac{1}{2} = \dfrac{3}{2}$ .

Step 5: Write the tangent

y - \frac{3}{2} = -\frac{3\sqrt{3}}{2}\left(x - \sqrt{3}\right)

Examples in context

Example 1. A circle's tangent. For $x^2 + y^2 = 25$ , implicit differentiation gives $\dfrac{dy}{dx} = -\dfrac{x}{y}$ , so the tangent at $(3, 4)$ has gradient $-\dfrac{3}{4}$ , perpendicular to the radius, the familiar geometric fact recovered by calculus.

Example 2. Cycloid motion. A point on a rolling wheel traces a cycloid $x = t - \sin t$ , $y = 1 - \cos t$ ; parametric differentiation gives $\dfrac{dy}{dx} = \dfrac{\sin t}{1 - \cos t}$ , describing the changing direction of motion as the wheel turns.

Try this

Q1. Find $\dfrac{dy}{dx}$ for $x^2 + y^2 = 16$ . [2 marks]

Cue. $2x + 2y\dfrac{dy}{dx} = 0$ , so $\dfrac{dy}{dx} = -\dfrac{x}{y}$ .

Q2. A curve has $x = t^2$ , $y = 2t$ . Find $\dfrac{dy}{dx}$ in terms of $t$ . [2 marks]

Cue. $\dfrac{2}{2t} = \dfrac{1}{t}$ .

Q3. State how to find a vertical tangent on a parametric curve. [1 mark]

Cue. Where $\dfrac{dx}{dt} = 0$ (the denominator of $\dfrac{dy}{dx}$ is zero) while $\dfrac{dy}{dt} \neq 0$ .

Exam-style practice questions

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

Original5 marksA curve has equation

x^2 + xy + y^2 = 7

. Find

\dfrac{dy}{dx}

in terms of

x

and

y

, and the gradient at the point

(1, 2)

Show worked answer →

Differentiate implicitly with respect to $x$ , treating $y$ as a function of $x$ :

$2x + \left(y + x\dfrac{dy}{dx}\right) + 2y\dfrac{dy}{dx} = 0$ (product rule on $xy$ ).

Collect: $2x + y + (x + 2y)\dfrac{dy}{dx} = 0$ , so $\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}$ .

At $(1, 2)$ : $\dfrac{dy}{dx} = -\dfrac{2 + 2}{1 + 4} = -\dfrac{4}{5}$ .

Markers reward differentiating each term implicitly (product rule on $xy$ , chain rule on $y^2$ ), solving for $\dfrac{dy}{dx}$ , and the numerical gradient.

Original5 marksA curve is defined parametrically by

x = t^2

y = t^3 - 3t

. Find

\dfrac{dy}{dx}

in terms of

t

and the values of

t

at which the tangent is horizontal.

Show worked answer →

$\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 3t^2 - 3$ .

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{3t^2 - 3}{2t} = \dfrac{3(t^2 - 1)}{2t}$ .

The tangent is horizontal where $\dfrac{dy}{dx} = 0$ , that is $t^2 - 1 = 0$ (and $t \neq 0$ ), so $t = \pm 1$ .

Markers reward differentiating both parametric equations, dividing to get $\dfrac{dy}{dx}$ , and setting the numerator to zero for horizontal tangents.