Home / A-Level Further Maths 9231 / Further differentiation / Second derivatives: implicit and parametric
Cambridge International AS & A Level · Topic 2.3
A-Level Further Maths 9231 / Topic 2.3 / AS & A Level

Differentiate the derivative, then divide by dx/dt.

You already find \(\dfrac{dy}{dx}\) for curves given implicitly or by a parameter. Now take the next step: find \(\dfrac{d^{2}y}{dx^{2}}\) by differentiating the first derivative again, evaluate it at a point, and use its sign to classify stationary points and describe the shape of the curve.

Topic 2.3All further differentiation lessons →
The key idea

Parametric: \(\dfrac{d^{2}y}{dx^{2}} = \dfrac{d}{dx}\!\left(\dfrac{dy}{dx}\right) = \dfrac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}\), NOT \(\dfrac{d^{2}y/dt^{2}}{d^{2}x/dt^{2}}\). Implicit: differentiate the whole equation twice with respect to \(x\), using the product rule on terms such as \(2y\,\dfrac{dy}{dx}\), then substitute the known value of \(\dfrac{dy}{dx}\). At a stationary point, \(\dfrac{d^{2}y}{dx^{2}} \gt 0\) means a minimum and \(\dfrac{d^{2}y}{dx^{2}} \lt 0\) a maximum.

x y O 1 2 4 2 1 −1 −2 t = −1: maximum (1, 2) d²y/dx² = −3/2 < 0 t = 1: minimum (1, −2) d²y/dx² = 3/2 > 0 (3, 0) x = t² y = t³ − 3t vertical tangent at O (t = 0)
Fig. 1: the curve x = t², y = t³ - 3t. Both stationary points lie on the line x = 1: t = -1 gives a maximum at (1, 2) and t = 1 gives a minimum at (1, -2), decided by the sign of d²y/dx². The curve crosses itself at (3, 0) and has a vertical tangent at O, where dx/dt = 0
Section 01

Parametric curves: divide by dx/dt again.

For \(x = f(t)\), \(y = g(t)\) you know \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\), which is a function of \(t\). The second derivative is the derivative of that with respect to \(x\), not \(t\). The chain rule handles the change of variable: \(\dfrac{d}{dx}(\text{anything}) = \dfrac{d}{dt}(\text{anything}) \div \dfrac{dx}{dt}\). Apply it to \(\dfrac{dy}{dx}\) itself.

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

Simplify \(\dfrac{dy}{dx}\) BEFORE differentiating it again: splitting a fraction into separate terms often saves a quotient rule.

Worked example A cubic in t
A curve has parametric equations \(x = t^{2}\), \(y = t^{3} - 3t\). Find \(\dfrac{d^{2}y}{dx^{2}}\) in terms of \(t\), and evaluate it at \(t = 2\).
  1. \(\dfrac{dx}{dt} = 2t\) and \(\dfrac{dy}{dt} = 3t^{2} - 3\), so \(\dfrac{dy}{dx} = \dfrac{3t^{2} - 3}{2t} = \tfrac{3}{2}\left(t - t^{-1}\right)\).
  2. Differentiate with respect to \(t\): \(\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right) = \tfrac{3}{2}\left(1 + t^{-2}\right) = \dfrac{3(t^{2} + 1)}{2t^{2}}\).
  3. Divide by \(\dfrac{dx}{dt} = 2t\):
  4. \[\frac{d^{2}y}{dx^{2}} = \frac{3(t^{2} + 1)}{2t^{2}} \times \frac{1}{2t} = \frac{3(t^{2} + 1)}{4t^{3}}.\]
  5. At \(t = 2\): \(\dfrac{3 \times 5}{4 \times 8} = \dfrac{15}{32}\).
Answer: \(\dfrac{d^{2}y}{dx^{2}} = \dfrac{3(t^{2} + 1)}{4t^{3}}\); at \(t = 2\) it is \(\dfrac{15}{32}\)
Worked example An ellipse, and the wrong shortcut
For the ellipse \(x = 2\cos t\), \(y = 3\sin t\), find \(\dfrac{d^{2}y}{dx^{2}}\) in terms of \(t\). Compare your answer with \(\dfrac{d^{2}y/dt^{2}}{d^{2}x/dt^{2}}\).
  1. \(\dfrac{dx}{dt} = -2\sin t\), \(\dfrac{dy}{dt} = 3\cos t\), so \(\dfrac{dy}{dx} = -\tfrac{3}{2}\cot t\).
  2. \(\dfrac{d}{dt}\!\left(-\tfrac{3}{2}\cot t\right) = \tfrac{3}{2}\operatorname{cosec}^{2} t\).
  3. \[\frac{d^{2}y}{dx^{2}} = \frac{\tfrac{3}{2}\operatorname{cosec}^{2} t}{-2\sin t} = -\tfrac{3}{4}\operatorname{cosec}^{3} t.\]
  4. The shortcut gives \(\dfrac{-3\sin t}{-2\cos t} = \tfrac{3}{2}\tan t\). At \(t = 0.8\) the true value is about \(-2.03\) and the shortcut gives about \(1.54\): not even the right sign. Second derivatives do not divide like first derivatives.
Answer: \(\dfrac{d^{2}y}{dx^{2}} = -\tfrac{3}{4}\operatorname{cosec}^{3} t\); the ratio \(\tfrac{3}{2}\tan t\) is wrong
Your turn
Given \(x = e^{t}\) and \(y = te^{t}\), find \(\dfrac{d^{2}y}{dx^{2}}\) in terms of \(t\).
Examiner trap

\(\dfrac{d^{2}y}{dx^{2}}\) is NOT \(\dfrac{d^{2}y/dt^{2}}{d^{2}x/dt^{2}}\), and it is not just \(\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)\) either. Differentiate \(\dfrac{dy}{dx}\) with respect to \(t\), then divide by \(\dfrac{dx}{dt}\) a second time. Forgetting that final division is the most common lost mark.

Section 02

Implicit relations: differentiate twice.

When \(x\) and \(y\) are tied together by an equation, differentiate every term with respect to \(x\), treating \(y\) as a function of \(x\). Do it once to get an equation in \(y'\), then differentiate THAT equation again. Terms such as \(2y\,y'\) or \(x\,y'\) are products, so they need the product rule: \(\dfrac{d}{dx}\big(2y\,y'\big) = 2(y')^{2} + 2y\,y''\). Finally substitute the known values of \(x\), \(y\) and \(y'\).

Worked example y = tan x without the quotient rule
Given \(y = \tan x\), show that \(\dfrac{dy}{dx} = 1 + y^{2}\). Hence find \(\dfrac{d^{2}y}{dx^{2}}\) in terms of \(y\), and its value at \(x = \tfrac{\pi}{4}\).
  1. \(\dfrac{dy}{dx} = \sec^{2} x = 1 + \tan^{2} x = 1 + y^{2}\).
  2. Differentiate \(y' = 1 + y^{2}\) with respect to \(x\); \(y^{2}\) is a function of a function: \(y'' = 2y\,y'\).
  3. Substitute \(y' = 1 + y^{2}\): \(y'' = 2y(1 + y^{2})\).
  4. At \(x = \tfrac{\pi}{4}\), \(y = 1\), so \(y'' = 2 \times 1 \times 2 = 4\). (Check: \(\dfrac{d^{2}}{dx^{2}}\tan x = 2\sec^{2} x\tan x = 2 \times 2 \times 1 = 4\).)
Answer: \(\dfrac{d^{2}y}{dx^{2}} = 2y(1 + y^{2})\); at \(x = \tfrac{\pi}{4}\) it equals \(4\)
Worked example A cubic curve at a point
The curve \(x^{3} + y^{3} = 9\) passes through \((1, 2)\). Find the values of \(\dfrac{dy}{dx}\) and \(\dfrac{d^{2}y}{dx^{2}}\) there.
  1. Differentiate once: \(3x^{2} + 3y^{2}y' = 0\), so \(x^{2} + y^{2}y' = 0\). At \((1, 2)\): \(1 + 4y' = 0\), \(y' = -\tfrac{1}{4}\).
  2. Differentiate \(x^{2} + y^{2}y' = 0\) again; \(y^{2}y'\) is a product of \(y^{2}\) and \(y'\):
  3. \[\begin{aligned} 2x + 2y\,y' \cdot y' + y^{2}y'' &= 0 \\ 2x + 2y(y')^{2} + y^{2}y'' &= 0. \end{aligned}\]
  4. Substitute \(x = 1\), \(y = 2\), \(y' = -\tfrac{1}{4}\): \(2 + 4 \times \tfrac{1}{16} + 4y'' = 0\), so \(4y'' = -\tfrac{9}{4}\).
  5. \(y'' = -\tfrac{9}{16}\).
Answer: \(\dfrac{dy}{dx} = -\tfrac{1}{4}\), \(\dfrac{d^{2}y}{dx^{2}} = -\tfrac{9}{16}\) at \((1, 2)\)
Your turn
Given \(y^{3} + y = x\), find \(\dfrac{d^{2}y}{dx^{2}}\) at the point \((2, 1)\).
Examiner trap

Differentiating \(y^{2}y'\) as \(2y\,y''\), or \(2y\,y'\) as \(2y''\), skips the product rule. Every term that contains both \(y\) (or \(x\)) and \(y'\) is a product and gives TWO terms. The missing \((y')^{2}\) term is where most marks go.

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For a curve given by \(x = f(t)\), \(y = g(t)\), which expression equals \(\dfrac{d^{2}y}{dx^{2}}\)?

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Given \(x = t^{2}\) and \(y = t^{3}\), find \(\dfrac{d^{2}y}{dx^{2}}\).

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What is \(\dfrac{d^{2}}{dx^{2}}\big(y^{2}\big)\), where \(y\) is a function of \(x\)?

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Given \(y = \tan x\), so that \(\dfrac{dy}{dx} = 1 + y^{2}\), express \(\dfrac{d^{2}y}{dx^{2}}\) in terms of \(y\).

Section 03

Stationary points and the shape of a curve.

A stationary point needs \(\dfrac{dy}{dx} = 0\). For a parametric curve that means \(\dfrac{dy}{dt} = 0\) with \(\dfrac{dx}{dt} \ne 0\); for an implicit curve, set \(y' = 0\) in the first derivative equation and solve it together with the curve's equation. Then evaluate \(\dfrac{d^{2}y}{dx^{2}}\): positive means a minimum, negative a maximum (if it is zero, the test gives no answer).

A shortcut for implicit curves: at a stationary point \(y' = 0\), so every term containing \(y'\) in the second differentiation vanishes, leaving a simple equation for \(y''\).

Shape (a light extra): on an interval where \(\dfrac{d^{2}y}{dx^{2}} \gt 0\) the curve is convex (concave upwards, bending like a cup); where \(\dfrac{d^{2}y}{dx^{2}} \lt 0\) it is concave (concave downwards, like a cap). To show this, prove the sign of your expression for every value of the parameter in the interval.

Worked example Classify, parametrically
For the curve \(x = t^{2}\), \(y = t^{3} - 3t\) (Fig. 1), find the stationary points and determine their nature. You may use \(\dfrac{d^{2}y}{dx^{2}} = \dfrac{3(t^{2} + 1)}{4t^{3}}\) from section 1.
  1. \(\dfrac{dy}{dt} = 3t^{2} - 3 = 0\) gives \(t = \pm 1\); there \(\dfrac{dx}{dt} = \pm 2 \ne 0\), so both are genuine stationary points.
  2. \(t = 1\): the point is \((1, -2)\) and \(\dfrac{d^{2}y}{dx^{2}} = \dfrac{3 \times 2}{4} = \tfrac{3}{2} \gt 0\): a minimum.
  3. \(t = -1\): the point is \((1, 2)\) and \(\dfrac{d^{2}y}{dx^{2}} = \dfrac{3 \times 2}{-4} = -\tfrac{3}{2} \lt 0\): a maximum.
  4. Both points have \(x = 1\); the curve has two branches, and the sign of \(t^{3}\) decides which one each point is on. For \(t \gt 0\), \(\dfrac{d^{2}y}{dx^{2}} \gt 0\) throughout, so that branch is convex.
Answer: Minimum at \((1, -2)\) (\(t = 1\)); maximum at \((1, 2)\) (\(t = -1\))
Worked example Classify, implicitly
Find the stationary points of the curve \(x^{2} + xy + y^{2} = 12\) and determine their nature.
  1. Once: \(2x + y + xy' + 2y\,y' = 0\). Setting \(y' = 0\) gives \(2x + y = 0\), so \(y = -2x\).
  2. Substitute into the curve: \(x^{2} - 2x^{2} + 4x^{2} = 12\), so \(3x^{2} = 12\), \(x = \pm 2\). Points \((2, -4)\) and \((-2, 4)\).
  3. Again: \(2 + y' + y' + xy'' + 2(y')^{2} + 2y\,y'' = 0\). With \(y' = 0\) this is \(2 + (x + 2y)y'' = 0\), so \(y'' = -\dfrac{2}{x + 2y}\).
  4. At \((2, -4)\): \(y'' = -\dfrac{2}{-6} = \tfrac{1}{3} \gt 0\), a minimum. At \((-2, 4)\): \(y'' = -\dfrac{2}{6} = -\tfrac{1}{3} \lt 0\), a maximum.
Answer: Minimum at \((2, -4)\), maximum at \((-2, 4)\)
Your turn
Using \(\dfrac{d^{2}y}{dx^{2}} = -\tfrac{3}{4}\operatorname{cosec}^{3} t\) for the ellipse \(x = 2\cos t\), \(y = 3\sin t\), show that the upper half of the ellipse (\(0 \lt t \lt \pi\)) is concave.
Examiner trap

For a parametric curve, check \(\dfrac{dx}{dt} \ne 0\) where \(\dfrac{dy}{dt} = 0\). And for an implicit curve, solving \(y' = 0\) alone is not enough: the point must also lie ON the curve, so substitute back into the original equation to find the coordinates.

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A curve has parametric equations \(x = \sin t\), \(y = \cos 2t\). Find \(\dfrac{d^{2}y}{dx^{2}}\).

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The curve \(x^{2} + 4y^{2} = 8\) passes through \((2, 1)\). Find the value of \(\dfrac{d^{2}y}{dx^{2}}\) at this point.

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The curve \(x^{3} + y^{3} = 3xy\) has a stationary point at \(\big(2^{1/3}, 2^{2/3}\big)\). Find \(\dfrac{d^{2}y}{dx^{2}}\) there and state the nature of the point.

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A curve has parametric equations \(x = t^{3} + 3t\), \(y = t^{2} - 4t\). Find \(\dfrac{d^{2}y}{dx^{2}}\) at its stationary point and state its nature.

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Second derivatives found for implicit and parametric curves.

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Six original Cambridge-style questions
Parametric and implicit second derivatives: a quick parametric warm-up, successive differentiation of \(\tanh x\), an implicit curve at a point, the cycloid's concavity, implicit stationary points, and a hyperbola done both ways.