Implicit and Parametric Differentiation (Further)

Most curves are not the graph of a function. A circle $x^2+y^2=25$x2+y2=25, a folium $x^3+y^3=6xy$x3+y3=6xy, or an ellipse traced by $x=a\cos t,\,y=b\sin t$x=acost,y=bsint all fail the vertical-line test, yet each has a well-defined gradient at almost every point. Implicit differentiation and parametric differentiation are the two techniques that extract that gradient, and both are nothing more than the chain rule applied with discipline. The single idea behind implicit differentiation — treat $y$y as an unknown function of $x$x and tag every $y$y-derivative with $\tfrac{dy}{dx}$dxdy​ — is one of the most powerful and re-usable moves in the whole A-Level.

This lesson builds both methods from the chain rule, extends them to second derivatives (where the subtleties live), and applies them to tangents and normals with full mark schemes. The techniques are prerequisites for related rates (the next lesson), for polar and parametric area/arc-length in further calculus, and for the classification of stationary points on implicit curves.

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Where this sits in AQA 7367

This is the further differentiation strand of the compulsory pure content examined on Paper 1 and Paper 2 (each 2 hours, 100 marks, 33⅓% of the A-Level). It promotes A-Level Maths implicit and parametric differentiation to second derivatives and to more demanding curves. Carrying out the differentiation is AO1; the "find the tangent / show that / classify the stationary point" framings carry the AO2 reasoning marks. It feeds directly into related rates of change (next lesson) and into further calculus (parametric and polar area and arc length).

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Core theory I: implicit differentiation

A curve given by an equation $F(x,y)=0$F(x,y)=0 that you cannot (or do not wish to) rearrange for $y$y is implicit. To differentiate, apply $\tfrac{d}{dx}$dxd​ to both sides, treating $y$y as a function of $x$x. The only new ingredient is the chain rule for a function of $y$y:

$\frac{d}{dx}\big[g(y)\big] = g'(y)\,\frac{dy}{dx}.$dxd​[g(y)]=g′(y)dxdy​.

So $\tfrac{d}{dx}(y^2) = 2y\,\tfrac{dy}{dx}$dxd​(y2)=2ydxdy​, $\tfrac{d}{dx}(y^3) = 3y^2\,\tfrac{dy}{dx}$dxd​(y3)=3y2dxdy​, $\tfrac{d}{dx}(\sin y) = \cos y\,\tfrac{dy}{dx}$dxd​(siny)=cosydxdy​. A product of $x$x and $y$y, such as $xy$xy, needs the product rule and the chain rule:

$\frac{d}{dx}(xy) = 1\cdot y + x\cdot\frac{dy}{dx} = y + x\frac{dy}{dx}.$dxd​(xy)=1⋅y+x⋅dxdy​=y+xdxdy​.

After differentiating every term, collect all $\tfrac{dy}{dx}$dxdy​ terms on one side, factorise, and divide. This last discipline — gather, factorise, divide — prevents the algebra from sprawling and is where method marks are banked.

To see why the discipline matters, consider a typical messy line after differentiating: $3x^2 + 5\tfrac{dy}{dx} + 2y + 7x\tfrac{dy}{dx} = 0$3x2+5dxdy​+2y+7xdxdy​=0. The temptation is to start dividing term by term, which scatters $\tfrac{dy}{dx}$dxdy​ across both sides and invites sign errors. The reliable move is to first move every non-$\tfrac{dy}{dx}$dxdy​ term to the right, then factorise the left: $(5 + 7x)\tfrac{dy}{dx} = -(3x^2 + 2y)$(5+7x)dxdy​=−(3x2+2y), giving $\tfrac{dy}{dx} = -\dfrac{3x^2 + 2y}{5 + 7x}$dxdy​=−5+7x3x2+2y​ in one clean step. Examiners explicitly reward this structure because it makes the gradient a single tidy fraction whose numerator and denominator each carry meaning (numerator zero $\Rightarrow$⇒ horizontal tangent; denominator zero $\Rightarrow$⇒ vertical tangent, as we will exploit below).

The mechanic in one line

The reason this works is that on the curve, $y$y genuinely is a (locally defined) function of $x$x, even if we never write it explicitly; so $F(x,y(x))=0$F(x,y(x))=0 is an identity in $x$x, and differentiating an identity is legitimate. The result $\tfrac{dy}{dx}$dxdy​ typically depends on both $x$x and $y$y — which is exactly right, because at a self-intersection a single $x$x can have several gradients, one per branch (per $y$y-value).

A catalogue of $y$y-derivatives

It pays to have the common chain-rule patterns at your fingertips, so the differentiation flows without hesitation. Each is just $\tfrac{d}{dy}(\cdot)$dyd​(⋅) times the chain factor $\tfrac{dy}{dx}$dxdy​:

Term in $y$y	Its $\tfrac{d}{dx}$dxd​
$y^n$yn	$n y^{n-1}\,\tfrac{dy}{dx}$nyn−1dxdy​
$\sin y$siny	$\cos y\,\tfrac{dy}{dx}$cosydxdy​
$e^{y}$ey	$e^{y}\,\tfrac{dy}{dx}$eydxdy​
$\ln y$lny	$\dfrac{1}{y}\,\tfrac{dy}{dx}$y1​dxdy​
$xy$xy (product)	$y + x\,\tfrac{dy}{dx}$y+xdxdy​
$x^2 y$x2y (product)	$2xy + x^2\,\tfrac{dy}{dx}$2xy+x2dxdy​

The only genuinely new skill beyond A-Level Maths is fluency — recognising instantly that a term such as $x^2 y^3$x2y3 needs the product rule on the $x$x-and-$y$y parts and the chain rule on the $y^3$y3: $\tfrac{d}{dx}(x^2 y^3) = 2x y^3 + x^2\cdot 3y^2\tfrac{dy}{dx}$dxd​(x2y3)=2xy3+x2⋅3y2dxdy​. Build that fluency on a few mixed terms and the rest is bookkeeping.

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Worked examples (with mark scheme)

Example 1 — a circle, and its second derivative

Find $\tfrac{dy}{dx}$dxdy​ and $\tfrac{d^2y}{dx^2}$dx2d2y​ for $x^2 + y^2 = 25$x2+y2=25.

Differentiate term by term:

$2x + 2y\frac{dy}{dx} = 0 \;\Rightarrow\; \frac{dy}{dx} = -\frac{x}{y}. \quad (\text{M1 chain rule on } y^2;\ \text{A1})$2x+2ydxdy​=0⇒dxdy​=−yx​.(M1 chain rule on y2; A1)

For the second derivative, differentiate $\tfrac{dy}{dx} = -\tfrac{x}{y}$dxdy​=−yx​ by the quotient rule, remembering $\tfrac{d}{dx}(y) = \tfrac{dy}{dx}$dxd​(y)=dxdy​:

$\frac{d^2y}{dx^2} = -\frac{(1)(y) - x\,\frac{dy}{dx}}{y^2} = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{y^2 + x^2}{y^3}. \quad (\text{M1 quotient; M1 substitute } \tfrac{dy}{dx})$dx2d2y​=−y2(1)(y)−xdxdy​​=−y2y−x(−yx​)​=−y2y+yx2​​=−y3y2+x2​.(M1 quotient; M1 substitute dxdy​)

Using the original equation $x^2+y^2=25$x2+y2=25:

$\frac{d^2y}{dx^2} = -\frac{25}{y^3}. \quad (\text{A1})$dx2d2y​=−y325​.(A1)

(M1 chain rule; A1 first derivative; M1 quotient rule; M1 for substituting $\tfrac{dy}{dx}=-x/y$dxdy​=−x/y back in — the step most often forgotten; A1 simplified using the curve's equation. The negative sign for $y>0$y>0 confirms the upper semicircle is concave down ✓.)

Example 2 — an implicit cubic

Find $\tfrac{dy}{dx}$dxdy​ for $x^3 + 3xy + y^3 = 1$x3+3xy+y3=1.

$3x^2 + 3\!\left(y + x\frac{dy}{dx}\right) + 3y^2\frac{dy}{dx} = 0. \quad (\text{M1 product rule on } 3xy;\ \text{M1 chain on } y^3)$3x2+3(y+xdxdy​)+3y2dxdy​=0.(M1 product rule on 3xy; M1 chain on y3)

Divide by 3 and gather the $\tfrac{dy}{dx}$dxdy​ terms:

$x^2 + y + (x + y^2)\frac{dy}{dx} = 0 \;\Rightarrow\; \frac{dy}{dx} = -\frac{x^2 + y}{x + y^2}. \quad (\text{M1 factorise; A1})$x2+y+(x+y2)dxdy​=0⇒dxdy​=−x+y2x2+y​.(M1 factorise; A1)

(M1 product rule; M1 chain rule; M1 gather/factorise; A1. The "gather then factorise" step is the examiner's method mark — never leave $\tfrac{dy}{dx}$dxdy​ scattered across the line.)

Example 3 — parametric first and second derivatives

For $x = t^2,\ y = t^3$x=t2, y=t3, find $\tfrac{dy}{dx}$dxdy​ and $\tfrac{d^2y}{dx^2}$dx2d2y​.

$\frac{dx}{dt} = 2t,\qquad \frac{dy}{dt} = 3t^2 \;\Rightarrow\; \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2t} = \frac{3t}{2}. \quad (\text{M1; A1})$dtdx​=2t,dtdy​=3t2⇒dxdy​=dx/dtdy/dt​=2t3t2​=23t​.(M1; A1)

For the second derivative, differentiate $\tfrac{dy}{dx}$dxdy​ with respect to $t$t, then divide by $\tfrac{dx}{dt}$dtdx​ — not by $\tfrac{d^2x}{dt^2}$dt2d2x​:

$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{\frac{d}{dt}\!\left(\frac{3t}{2}\right)}{2t} = \frac{3/2}{2t} = \frac{3}{4t}. \quad (\text{M1 correct structure; A1})$dx2d2y​=dtdx​dtd​(dxdy​)​=2tdtd​(23t​)​=2t3/2​=4t3​.(M1 correct structure; A1)

(M1 $\tfrac{dy/dt}{dx/dt}$dx/dtdy/dt​; A1 $\tfrac{3t}{2}$23t​; M1 the correct second-derivative structure (the single biggest parametric pitfall); A1 $\tfrac{3}{4t}$4t3​. Note $\tfrac{d^2y}{dx^2} \ne \tfrac{d^2y/dt^2}{d^2x/dt^2} = \tfrac{6t}{2} = 3t$dx2d2y​=d2x/dt2d2y/dt2​=26t​=3t — that "obvious" formula is wrong.)

Example 4 — a tangent to an implicit curve

Find the equation of the tangent to $x^2 + xy + y^2 = 7$x2+xy+y2=7 at $(1,2)$(1,2).

$2x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0. \quad (\text{M1 differentiate})$2x+(y+xdxdy​)+2ydxdy​=0.(M1 differentiate)

Substitute $(x,y)=(1,2)$(x,y)=(1,2) immediately (no need to find $\tfrac{dy}{dx}$dxdy​ in general):

$2 + \big(2 + 1\cdot\tfrac{dy}{dx}\big) + 4\tfrac{dy}{dx} = 0 \;\Rightarrow\; 4 + 5\frac{dy}{dx} = 0 \;\Rightarrow\; \frac{dy}{dx} = -\frac{4}{5}. \quad (\text{M1 substitute; A1})$2+(2+1⋅dxdy​)+4dxdy​=0⇒4+5dxdy​=0⇒dxdy​=−54​.(M1 substitute; A1)

Tangent through $(1,2)$(1,2) with gradient $-\tfrac45$−54​:

$y - 2 = -\frac{4}{5}(x - 1) \;\Rightarrow\; 5y - 10 = -4x + 4 \;\Rightarrow\; 4x + 5y = 14. \quad (\text{A1})$y−2=−54​(x−1)⇒5y−10=−4x+4⇒4x+5y=14.(A1)

(M1 differentiate; M1 substitute the point; A1 gradient; A1 tangent equation. Substituting the point before rearranging is a real time-saver and avoids algebra slips.)

Example 5 — an implicit relation with exponentials and logs

Find $\tfrac{dy}{dx}$dxdy​ for $e^{2y} + x\ln y = 4$e2y+xlny=4 at the point where $y=0$y=0… — careful: $\ln y$lny is undefined at $y=0$y=0, so instead consider the well-posed curve $e^{y} + xy = 1$ey+xy=1 and find $\tfrac{dy}{dx}$dxdy​ at $(0,0)$(0,0).

Differentiate, using the chain rule on $e^y$ey and the product rule on $xy$xy:

$e^{y}\frac{dy}{dx} + \left(y + x\frac{dy}{dx}\right) = 0. \quad (\text{M1 chain on } e^y;\ \text{M1 product on } xy)$eydxdy​+(y+xdxdy​)=0.(M1 chain on ey; M1 product on xy)

Gather the $\tfrac{dy}{dx}$dxdy​ terms:

$(e^{y} + x)\frac{dy}{dx} = -y \;\Rightarrow\; \frac{dy}{dx} = -\frac{y}{e^{y} + x}. \quad (\text{A1})$(ey+x)dxdy​=−y⇒dxdy​=−ey+xy​.(A1)

At $(0,0)$(0,0): $\tfrac{dy}{dx} = -\dfrac{0}{e^{0}+0} = -\dfrac{0}{1} = 0$dxdy​=−e0+00​=−10​=0 — a horizontal tangent. (A1)

(M1 chain rule on the exponential; M1 product rule; A1 the general gradient; A1 the value at the point. The point $(0,0)$(0,0) lies on the curve since $e^0 + 0 = 1$e0+0=1 ✓. This example shows the method is unchanged for transcendental terms — only the derivatives $\tfrac{d}{dx}(e^y) = e^y\tfrac{dy}{dx}$dxd​(ey)=eydxdy​, $\tfrac{d}{dx}(\ln y) = \tfrac1y\tfrac{dy}{dx}$dxd​(lny)=y1​dxdy​ change.)

A subtlety worth absorbing: an implicit curve can have a vertical tangent where the denominator of $\tfrac{dy}{dx}$dxdy​ vanishes (here $e^y + x = 0$ey+x=0). At such a point $\tfrac{dy}{dx}$dxdy​ is undefined but $\tfrac{dx}{dy} = 0$dydx​=0; switching to $\tfrac{dx}{dy}$dydx​ is the clean way to locate vertical tangents on implicit curves, mirroring how the numerator vanishing gives horizontal tangents. This numerator/denominator duality — zero on top for horizontal, zero on the bottom for vertical — is a reliable analytic tool for sketching implicit curves and locating their extreme points.

Example 6 — locating horizontal tangents on an implicit curve

Find the points on $x^2 + xy + y^2 = 12$x2+xy+y2=12 where the tangent is horizontal.

Differentiating (as in Example 4): $2x + y + (x + 2y)\tfrac{dy}{dx} = 0$2x+y+(x+2y)dxdy​=0, so

$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}. \quad (\text{M1})$dxdy​=−x+2y2x+y​.(M1)

A horizontal tangent needs the numerator zero (and denominator nonzero): $2x + y = 0$2x+y=0, i.e. $y = -2x$y=−2x. (M1) Substitute into the curve:

$x^2 + x(-2x) + (-2x)^2 = 12 \;\Rightarrow\; x^2 - 2x^2 + 4x^2 = 3x^2 = 12 \;\Rightarrow\; x = \pm 2. \quad (\text{A1})$x2+x(−2x)+(−2x)2=12⇒x2−2x2+4x2=3x2=12⇒x=±2.(A1)

So the horizontal tangents are at $(2, -4)$(2,−4) and $(-2, 4)$(−2,4) (using $y=-2x$y=−2x). (A1) At each, the denominator $x + 2y = 2 + 2(-4) = -6 \ne 0$x+2y=2+2(−4)=−6=0 ✓, so these are genuine horizontal tangents.

(M1 find $\tfrac{dy}{dx}$dxdy​; M1 set numerator to zero; A1 solve for $x$x; A1 give both points and confirm the denominator is nonzero. This "numerator = 0" technique is the standard route to the highest and lowest points of a closed implicit curve such as this tilted ellipse.)

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Core theory II: the parametric and tangent toolkit

For a curve $x=f(t),\,y=g(t)$x=f(t),y=g(t), the gradient is the ratio of the two $t$t-derivatives:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}, \qquad \frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}.$dxdy​=dx/dtdy/dt​=f′(t)g′(t)​,dx2d2y​=dtdx​dtd​(dxdy​)​.

This is just the chain rule $\tfrac{dy}{dx} = \tfrac{dy}{dt}\cdot\tfrac{dt}{dx}$dxdy​=dtdy​⋅dxdt​ with $\tfrac{dt}{dx} = 1/(dx/dt)$dxdt​=1/(dx/dt). The second-derivative formula deserves a moment's reflection, because it is the source of more lost marks than any other single result in the topic. It is not enough to differentiate $y$y twice with respect to $t$t and $x$x twice with respect to $t$t and divide. The reason is that $\tfrac{d^2y}{dx^2}$dx2d2y​ means "the rate of change of the gradient $\tfrac{dy}{dx}$dxdy​ with respect to $x$x", and the gradient is itself a function of $t$t; so we differentiate $\tfrac{dy}{dx}$dxdy​ with respect to $t$t and then convert that $t$t-rate into an $x$x-rate by dividing by $\tfrac{dx}{dt}$dtdx​. The structure is therefore $\tfrac{d}{dx}\big(\tfrac{dy}{dx}\big) = \tfrac{d}{dt}\big(\tfrac{dy}{dx}\big)\div\tfrac{dx}{dt}$dxd​(dxdy​)=dtd​(dxdy​)÷dtdx​ — the chain rule applied to the first derivative, treated as a new quantity. Internalise that sentence and the parametric second derivative will never trip you up.

Once a gradient $m=\tfrac{dy}{dx}$m=dxdy​ is known at a point $(x_0,y_0)$(x0​,y0​):

• Tangent: $y - y_0 = m(x - x_0)$y−y0​=m(x−x0​).
• Normal: $y - y_0 = -\dfrac{1}{m}(x - x_0)$y−y0​=−m1​(x−x0​) (perpendicular gradient).

A worked parametric tangent illustrates the second-derivative trap once more: for $x=a\cos t,\ y=a\sin t$x=acost, y=asint, $\tfrac{dx}{dt} = -a\sin t$dtdx​=−asint, $\tfrac{dy}{dt} = a\cos t$dtdy​=acost, so $\tfrac{dy}{dx} = \dfrac{a\cos t}{-a\sin t} = -\cot t$dxdy​=−asintacost​=−cott. The gradient depends on $t$t alone — and at $t=\tfrac{\pi}{2}$t=2π​ (the top of the circle) $\tfrac{dy}{dx} = -\cot\tfrac{\pi}{2} = 0$dxdy​=−cot2π​=0, a horizontal tangent, exactly as geometry demands.