Papers 2 and 3. Not every curve comes as \(y = \text{f}(x)\). Some are given by parameters, \(x = \text{f}(t)\) and \(y = \text{g}(t)\); others by an equation that mixes \(x\) and \(y\), such as \(x^2 + y^2 = xy + 7\). Both still have a gradient at every point, and you can find it without rearranging.
Parametric: \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\). Implicit: differentiate every term with respect to \(x\), treating \(y\) as a function of \(x\), so \(\dfrac{d}{dx}y^{n} = ny^{n-1}\dfrac{dy}{dx}\) and \(\dfrac{d}{dx}(xy) = x\dfrac{dy}{dx} + y\); then make \(\dfrac{dy}{dx}\) the subject.
Fig. 1: implicit differentiation of x² + y² = 25 gives dy/dx = −x/y, which is −3/4 at P(3, 4). The tangent 3x + 4y = 25 is perpendicular to the radius OP, whose gradient is 4/3, exactly as circle geometry says
Section 01
Parametric curves.
When \(x\) and \(y\) both depend on a parameter \(t\), differentiate each with respect to \(t\) and divide. This is the chain rule: \(\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx}\). The gradient usually comes out in terms of \(t\), so find the value of \(t\) at the point first.
Worked example A tangent
A curve has parametric equations \(x = 3t^2\), \(y = 2t^3\). Find the equation of the tangent at the point where \(t = 2\).
At \(t = 2\): the point is \((12, 16)\) and the gradient is 2.
\(y - 16 = 2(x - 12)\).
Answer: \(y = 2x - 8\)
Worked example A trigonometric parameter
A curve has parametric equations \(x = 2\cos\theta\), \(y = 3\sin\theta\). Find the equation of the normal at the point where \(\theta = \dfrac{\pi}{6}\).
\(\dfrac{dx}{d\theta} = -2\sin\theta\), \(\dfrac{dy}{d\theta} = 3\cos\theta\), so \(\dfrac{dy}{dx} = -\dfrac{3\cos\theta}{2\sin\theta}\).
At \(\theta = \dfrac{\pi}{6}\): the point is \(\left(\sqrt{3}, \dfrac{3}{2}\right)\) and the gradient is \(-\dfrac{3(\sqrt{3}/2)}{2(1/2)} = -\dfrac{3\sqrt{3}}{2}\).
Normal gradient \(= \dfrac{2}{3\sqrt{3}} = \dfrac{2\sqrt{3}}{9}\).
The formula is \(\dfrac{dy/dt}{dx/dt}\), y-derivative on TOP. Dividing the other way gives \(\dfrac{dx}{dy}\), the reciprocal of the gradient.
Section 02
Implicit differentiation.
For an equation such as \(x^2 + y^2 = 25\) (Fig. 1), differentiate both sides with respect to \(x\). Terms in \(x\) only are done as usual. Every term containing \(y\) picks up a factor \(\dfrac{dy}{dx}\) by the chain rule, and a product such as \(xy\) needs the product rule.
Term
Derivative with respect to \(x\)
\(y^2\)
\(2y\dfrac{dy}{dx}\)
\(y^3\)
\(3y^2\dfrac{dy}{dx}\)
\(xy\)
\(x\dfrac{dy}{dx} + y\)
\(e^{y}\)
\(e^{y}\dfrac{dy}{dx}\)
\(\sin y\)
\(\cos y\dfrac{dy}{dx}\)
constant
\(0\)
Worked example A circle
Find the gradient of the circle \(x^2 + y^2 = 25\) at the point \(P(3, 4)\).
Differentiate: \(2x + 2y\dfrac{dy}{dx} = 0\).
\(\dfrac{dy}{dx} = -\dfrac{x}{y}\).
At \((3, 4)\): \(\dfrac{dy}{dx} = -\dfrac{3}{4}\).
Answer: \(-\dfrac{3}{4}\) (perpendicular to the radius, gradient \(\dfrac{4}{3}\))
Worked example With an xy term
The curve \(x^2 + 3xy + y^2 = 11\) passes through \((1, 2)\). Find the gradient there.
\(\dfrac{d}{dx}(3xy)\) is NOT \(3\dfrac{dy}{dx}\) or \(3y\). It is a product: \(3x\dfrac{dy}{dx} + 3y\). Missing one of the two terms is the most common error in implicit questions.
Stage 1 · Learn
Check your understanding
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For a curve with \(x = \text{f}(t)\) and \(y = \text{g}(t)\), which is \(\dfrac{dy}{dx}\)?
Quick check+10 XP
Differentiate \(y^2\) with respect to \(x\).
Quick check+10 XP
Differentiate \(xy\) with respect to \(x\).
Quick check+10 XP
What is the gradient of the circle \(x^2 + y^2 = 25\) at \((-4, 3)\)?
Section 03
Tangents and stationary points.
Stationary points (horizontal tangents) occur where the numerator of \(\dfrac{dy}{dx}\) is zero. That gives a relation between \(x\) and \(y\), which you substitute back into the curve's equation. Vertical tangents occur where the denominator is zero.
Worked example Horizontal tangents
Find the points on the curve \(x^2 - xy + y^2 = 12\) where the tangent is parallel to the \(x\)-axis.