Implicit Differentiation

A2 · P3 · 12 min

Many curves are given by an equation linking $x$ and $y$ that cannot conveniently be rearranged into $y = f(x)$ : the circle $x^2 + y^2 = 25$ , the syllabus example $x^2 + y^2 = xy + 7$ , or $x^3 + y^3 = 3xy$ . Such equations define $y$ implicitly. Implicit differentiation finds the gradient of these curves directly from the equation, without ever solving for $y$ . P3 regularly asks for the gradient at a point, the equation of a tangent or normal, or the points where the tangent is parallel to an axis.

The idea

On the curve $x^2 + y^2 = 25$ , near any point, $y$ behaves like a function of $x$ (the top half of the circle is $y = \sqrt{25 - x^2}$ , the bottom half $y = -\sqrt{25 - x^2}$ ). So it makes sense to differentiate the whole equation with respect to $x$ , both sides, treating $y$ as "some function of $x$ that we do not need to write down".

The only new fact you need is how to differentiate an expression in $y$ with respect to $x$ . By the chain rule,

\frac{d}{dx}\big(g(y)\big) = \frac{d}{dy}\big(g(y)\big)\times\frac{dy}{dx} = g'(y)\frac{dy}{dx}

So: differentiate with respect to $y$ as normal, then multiply by $\dfrac{dy}{dx}$ .

Differentiating terms in y with respect to x

Term	Derivative with respect to $x$
$y$	$\dfrac{dy}{dx}$
$y^n$	$ny^{n-1}\dfrac{dy}{dx}$
$e^y$	$e^y\dfrac{dy}{dx}$
$\ln y$	$\dfrac{1}{y}\dfrac{dy}{dx}$
$\sin y$	$\cos y\dfrac{dy}{dx}$
$xy$	$x\dfrac{dy}{dx} + y$ (product rule)
$x^2y^3$	$x^2\cdot 3y^2\dfrac{dy}{dx} + 2xy^3$ (product rule)
$e^{xy}$	$e^{xy}\left(x\dfrac{dy}{dx} + y\right)$ (chain rule, then product rule)
constant	$0$

The products are where marks are lost. A term like $xy$ involves two functions of $x$ , namely $x$ and $y$ , so it needs the product rule. Writing $\dfrac{d}{dx}(xy) = \dfrac{dy}{dx}$ or $= y$ are both wrong.

The method

Implicit differentiation

Differentiate every term on both sides with respect to $x$ . Each term in $y$ picks up a factor $\dfrac{dy}{dx}$ ; each product of $x$ and $y$ needs the product rule. Constants become $0$ .
Collect all terms containing $\dfrac{dy}{dx}$ on one side and everything else on the other.
Factorise out $\dfrac{dy}{dx}$ .
Divide to make $\dfrac{dy}{dx}$ the subject.
For a gradient at a point, substitute both $x$ and $y$ . (You may substitute before step 4 if the numbers are simple; it is often faster.)

The answer for $\dfrac{dy}{dx}$ is normally in terms of both $x$ and $y$ . That is expected and correct: a single $x$ can correspond to several points on the curve, each with its own gradient, so you need $y$ as well to say which point you mean.

A first example: the circle

Differentiate $x^2 + y^2 = 25$ with respect to $x$ :

2x + 2y\frac{dy}{dx} = 0 \quad\Rightarrow\quad \frac{dy}{dx} = -\frac{x}{y}

At $(3, 4)$ the gradient is $-\tfrac{3}{4}$ , and at $(3, -4)$ it is $+\tfrac{3}{4}$ : the same $x$ , two different points, two gradients.

Check with geometry: the radius to $(3, 4)$ has gradient $\tfrac{4}{3}$ , and the tangent is perpendicular to the radius, so it has gradient $-\tfrac{3}{4}$ . Implicit differentiation agrees with what you know from circles.

Tangents parallel to the axes

Once $\dfrac{dy}{dx}$ is written as a single fraction $\dfrac{N}{D}$ (numerator and denominator in $x$ and $y$ ):

the tangent is parallel to the $x$ -axis (horizontal, a stationary point) where $N = 0$ and $D \ne 0$ ;
the tangent is parallel to the $y$ -axis (vertical) where $D = 0$ and $N \ne 0$ .

In both cases the condition is an equation linking $x$ and $y$ , so solve it simultaneously with the equation of the curve to find the actual points. Usually you rearrange the condition to $y = kx$ (or $x = ky$ ) and substitute.

Points where the tangent is parallel to an axis

Find $\dfrac{dy}{dx} = \dfrac{N}{D}$ .
Horizontal tangent: set $N = 0$ . Vertical tangent: set $D = 0$ .
Rearrange the condition to express one variable in terms of the other.
Substitute into the equation of the curve and solve.
Find the other coordinate of each point. Check that the other part of the fraction is not also zero there.

Worked examples

Tangent to a circle

Find the equation of the tangent to $x^2 + y^2 = 25$ at the point $(3, 4)$ .

Solution

$2x + 2y\dfrac{dy}{dx} = 0$ , so $\dfrac{dy}{dx} = -\dfrac{x}{y} = -\dfrac{3}{4}$ at $(3, 4)$ .

Tangent: $y - 4 = -\tfrac{3}{4}(x - 3)$ . Multiply by $4$ : $4y - 16 = -3x + 9$ , so $3x + 4y = 25$ .

The syllabus curve

The curve $C$ has equation $x^2 + y^2 = xy + 7$ .

(a) Show that $\dfrac{dy}{dx} = \dfrac{y - 2x}{2y - x}$ .

(b) Find the equation of the tangent to $C$ at the point $(3, 1)$ .

Solution

(a) Differentiate both sides with respect to $x$ , using the product rule on $xy$ :

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

Collect the $\dfrac{dy}{dx}$ terms:

2y\frac{dy}{dx} - x\frac{dy}{dx} = y - 2x \quad\Rightarrow\quad \frac{dy}{dx}(2y - x) = y - 2x \quad\Rightarrow\quad \frac{dy}{dx} = \frac{y - 2x}{2y - x}

(b) Check $(3, 1)$ is on $C$ : $9 + 1 = 3 + 7$ . Yes.

Gradient: $\dfrac{1 - 6}{2 - 3} = \dfrac{-5}{-1} = 5$ .

Tangent: $y - 1 = 5(x - 3)$ , i.e. $y = 5x - 14$ .

An exponential term

The curve $e^xy + y^3 = 2$ passes through $(0, 1)$ . Find the equation of the normal at this point.

Solution

Differentiate, using the product rule on $e^xy$ :

e^xy + e^x\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0

Substitute $x = 0$ , $y = 1$ straight away:

1 + \frac{dy}{dx} + 3\frac{dy}{dx} = 0 \quad\Rightarrow\quad \frac{dy}{dx} = -\frac{1}{4}

The normal has gradient $4$ : $y - 1 = 4x$ , i.e. $y = 4x + 1$ .

Tangents parallel to both axes

The curve $x^2 + 2xy + 3y^2 = 6$ is an ellipse.

(a) Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .

(b) Find the coordinates of the points where the tangent is parallel to the $x$ -axis.

Solution

(a) $2x + 2x\dfrac{dy}{dx} + 2y + 6y\dfrac{dy}{dx} = 0$ , so $\dfrac{dy}{dx}(2x + 6y) = -(2x + 2y)$ and

\frac{dy}{dx} = -\frac{x + y}{x + 3y}

(b) Numerator zero: $y = -x$ . Substitute into the curve: $x^2 - 2x^2 + 3x^2 = 6$ , so $2x^2 = 6$ , $x = \pm\sqrt{3}$ .

Points: $(\sqrt{3}, -\sqrt{3})$ and $(-\sqrt{3}, \sqrt{3})$ . (The denominator $x + 3y = \mp 2\sqrt{3} \ne 0$ .)

Points: $(-3, 1)$ and $(3, -1)$ .

x^2 + 2x y + 3y^2 = 6 y = -sqrt(3) x = 3

The tilted ellipse $x^2 + 2xy + 3y^2 = 6$ . The dashed line $y = -\sqrt{3}$ is the horizontal tangent at the lowest point $(\sqrt{3}, -\sqrt{3})$ ; the dotted line $x = 3$ is the vertical tangent at $(3, -1)$ .

Exam-hard: the folium of Descartes

The curve $x^3 + y^3 = 3xy$ has a loop in the first quadrant.

(a) Show that $\dfrac{dy}{dx} = \dfrac{y - x^2}{y^2 - x}$ .

(b) Find the exact coordinates of the point on the loop, other than the origin, where the tangent is parallel to the $x$ -axis.

Solution

(a) Differentiate, with the product rule on $3xy$ :

3x^2 + 3y^2\frac{dy}{dx} = 3x\frac{dy}{dx} + 3y

Divide by $3$ and collect: $\dfrac{dy}{dx}(y^2 - x) = y - x^2$ , so $\dfrac{dy}{dx} = \dfrac{y - x^2}{y^2 - x}$ .

(b) Horizontal tangent: $y = x^2$ . Substitute into the curve:

x^3 + x^6 = 3x\cdot x^2 = 3x^3 \quad\Rightarrow\quad x^6 = 2x^3 \quad\Rightarrow\quad x^3(x^3 - 2) = 0

$x = 0$ gives the origin, where both numerator and denominator are zero (the curve crosses itself there, so the gradient formula gives no information). The other solution is $x = 2^{1/3}$ , with $y = x^2 = 2^{2/3}$ .

The point is $\left(2^{1/3}, 2^{2/3}\right)$ , about $(1.26, 1.59)$ . The denominator there is $2^{4/3} - 2^{1/3} \ne 0$ , so the tangent really is horizontal.

x^3 + y^3 = 3x y y = 2^(2/3)

The folium of Descartes with its horizontal tangent at $y = 2^{2/3}$ , the top of the loop.

Exam-hard: an unknown constant

The curve $xy(x + y) = 2a^3$ , where $a$ is a positive constant, has exactly one point at which the tangent is parallel to the $x$ -axis. Find its coordinates in terms of $a$ .

Solution

Expand: $x^2y + xy^2 = 2a^3$ . Differentiate with the product rule on each term:

2xy + x^2\frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} = 0 \quad\Rightarrow\quad \frac{dy}{dx} = -\frac{y(2x + y)}{x(x + 2y)}

Horizontal tangent: $y(2x + y) = 0$ . Now $y = 0$ is impossible on the curve, since it would make the left side $0 \ne 2a^3$ . So $y = -2x$ .

Substitute: $x(-2x)(x - 2x) = 2x^3 = 2a^3$ , so $x = a$ (the only real root) and $y = -2a$ .

The point is $(a, -2a)$ . Check the denominator: $a(a - 4a) = -3a^2 \ne 0$ .

Common mistakes

Forgetting the $\dfrac{dy}{dx}$ on a $y$ term: $\dfrac{d}{dx}(y^2) = 2y\dfrac{dy}{dx}$ , not $2y$ .
Missing the product rule on $xy$ , $x^2y$ , $e^xy$ and similar.
Forgetting to differentiate the right-hand side, or differentiating a constant on the right to something other than $0$ .
Substituting only $x$ when finding a gradient at a point. You need both coordinates.
Finding the condition $y = -x$ and stopping. That is a line, not a point. Solve with the curve's equation.
Dividing by zero. At a point where both numerator and denominator vanish, the formula is undefined; say so rather than giving a gradient.

Exam tip

"Show that $\dfrac{dy}{dx} = \ldots$ " questions award a mark for the correct differentiation of each type of term, often: one for the $y^n$ terms, one for the product term, one for the correct rearrangement. Show the line with all terms differentiated before collecting.
If you only need a numerical gradient, substitute the point immediately after differentiating; the arithmetic is easier than rearranging in general.
For tangents parallel to the axes, the mark scheme expects: the condition (numerator or denominator $= 0$ ), substitution into the curve, then both coordinates of every point.
Check that a given point lies on the curve before using it. A quick substitution catches misreadings.

Summary

Differentiate both sides with respect to $x$ ; each $y$ -term gets a factor $\dfrac{dy}{dx}$ .
$\dfrac{d}{dx}(y^n) = ny^{n-1}\dfrac{dy}{dx}$ ; $\dfrac{d}{dx}(xy) = x\dfrac{dy}{dx} + y$ .
Collect $\dfrac{dy}{dx}$ terms, factorise, divide.
$\dfrac{dy}{dx}$ is usually in terms of both $x$ and $y$ ; substitute both to get a gradient.
Horizontal tangent: numerator $= 0$ ; vertical tangent: denominator $= 0$ . Solve each with the curve's equation.
Implicit differentiation proves $\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1+x^2}$ from $\tan y = x$ .

Practice

Question

Find $\dfrac{dy}{dx}$ for the curve $x^3 + y^3 = 9$ .
Find the equation of the tangent to $x^2 + 3xy + y^2 = 11$ at the point $(1, 2)$ .
Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ given $ye^{2x} = x + y$ .
The curve $\ln y + xy = 2$ passes through $(2, 1)$ . Find the equation of the normal at this point.
Given $\tan y = 2x$ , use implicit differentiation to show that $\dfrac{dy}{dx} = \dfrac{2}{1 + 4x^2}$ .
Find the points on the circle $x^2 + y^2 - 4x + 6y = 12$ where the tangent is parallel to the $x$ -axis.
Find the coordinates of the points on $x^2 - 2xy + 4y^2 = 12$ at which the tangent is parallel to the $y$ -axis.
The curve $x^2 + xy + y^2 = 3$ has two tangents parallel to the $x$ -axis. Find their equations.
The curve $2x^2 + xy + y^2 = 14$ meets the line $y = x$ at two points. Show that the tangents to the curve at these points are parallel, and find their equations in exact form.

Answers

$3x^2 + 3y^2\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{x^2}{y^2}$ .
$2x + 3y + 3x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0$ . At $(1, 2)$ : $2 + 6 + 3\dfrac{dy}{dx} + 4\dfrac{dy}{dx} = 0$ , so $\dfrac{dy}{dx} = -\dfrac{8}{7}$ . Tangent: $y - 2 = -\tfrac{8}{7}(x - 1)$ , i.e. $8x + 7y = 22$ .
$e^{2x}\dfrac{dy}{dx} + 2ye^{2x} = 1 + \dfrac{dy}{dx}$ , so $\dfrac{dy}{dx}(e^{2x} - 1) = 1 - 2ye^{2x}$ and $\dfrac{dy}{dx} = \dfrac{1 - 2ye^{2x}}{e^{2x} - 1}$ .
$\dfrac{1}{y}\dfrac{dy}{dx} + x\dfrac{dy}{dx} + y = 0$ . At $(2, 1)$ : $\dfrac{dy}{dx} + 2\dfrac{dy}{dx} + 1 = 0$ , $\dfrac{dy}{dx} = -\tfrac{1}{3}$ . Normal gradient $3$ : $y - 1 = 3(x - 2)$ , i.e. $y = 3x - 5$ .
$\sec^2 y\dfrac{dy}{dx} = 2$ , so $\dfrac{dy}{dx} = \dfrac{2}{\sec^2 y} = \dfrac{2}{1 + \tan^2 y} = \dfrac{2}{1 + 4x^2}$ .
$2x + 2y\dfrac{dy}{dx} - 4 + 6\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = \dfrac{2 - x}{y + 3}$ . Horizontal: $x = 2$ . Then $4 + y^2 - 8 + 6y = 12$ , $y^2 + 6y - 16 = 0$ , $(y + 8)(y - 2) = 0$ . Points $(2, 2)$ and $(2, -8)$ , the top and bottom of the circle with centre $(2, -3)$ and radius $5$ .
$2x - 2y - 2x\dfrac{dy}{dx} + 8y\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = \dfrac{y - x}{4y - x}$ . Vertical: $x = 4y$ . Then $16y^2 - 8y^2 + 4y^2 = 12$ , so $y^2 = 1$ . Points $(4, 1)$ and $(-4, -1)$ .
$\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}$ . Horizontal: $y = -2x$ . Then $x^2 - 2x^2 + 4x^2 = 3$ , $x = \pm 1$ , giving $(1, -2)$ and $(-1, 2)$ . Tangents: $y = -2$ and $y = 2$ .
With $y = x$ : $2x^2 + x^2 + x^2 = 14$ , $x^2 = \tfrac{7}{2}$ , so the points are $P\left(\sqrt{7/2}, \sqrt{7/2}\right)$ and $Q\left(-\sqrt{7/2}, -\sqrt{7/2}\right)$ . Differentiating: $4x + y + x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0$ , so $\dfrac{dy}{dx} = -\dfrac{4x + y}{x + 2y}$ . On $y = x$ this is $-\dfrac{5x}{3x} = -\dfrac{5}{3}$ at both points, so the tangents are parallel. Write $k = \sqrt{7/2} = \tfrac{\sqrt{14}}{2}$ . Tangent at $P$ : $y - k = -\tfrac{5}{3}(x - k)$ , i.e. $5x + 3y = 8k = 4\sqrt{14}$ . Tangent at $Q$ : $5x + 3y = -4\sqrt{14}$ .

The idea

The method

A first example: the circle

Tangents parallel to the axes

Worked examples

Practice

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