5.14 Implicit differentiation, related rates and optimisation · Educerie Mathematics: analysis and approaches

This whole subtopic is higher level. Nothing in it is on an SL paper.

Educerie · IB Diploma · Mathematics: analysis and approaches

Topic 5 Calculus · 5.14 Implicit differentiation, related rates and optimisation

Level: HL only, the whole subtopic. If you are SL, none of this is on your papers.
Themes (key concepts): relationships, change, modelling. An implicit equation is a relationship between x and y that you differentiate as it stands; related rates track how one quantity's change drives another's; optimisation turns a real design problem into a model and finds its best value.
The question this unit answers: how do you differentiate a relationship you cannot solve for y, how fast does one quantity change when another linked to it is changing, and where exactly is the best value of a model, including when it sits at the edge of what is allowed?
Where it is examined: Paper 1 (no calculator): implicit differentiation to find a gradient, a tangent or a normal (5 to 7 marks), points where the tangent is horizontal or vertical, and related rates with exact answers such as 2/π. Paper 2: related rates and optimisation in context, with a GDC to check a maximum or solve the final equation (6 to 10 marks in Section B). Paper 3: an optimisation model built step by step, often with an end-point twist.

What you must be able to do

You must be able to	Level	What it looks like in the exam
Differentiate an implicit equation term by term, using the chain rule on terms in y and the product rule on terms in xy	HL only	"Find dy/dx in terms of x and y" (4 to 5 marks)
Find the gradient, tangent and normal at a point on an implicit curve	HL only	"Find the equation of the normal to the curve at P(2, 0)" (5 to 6 marks)
Find points where the tangent is horizontal or vertical	HL only	"Find the coordinates of the points where the tangent is parallel to the x-axis" (6 marks)
Set up and solve related rates problems, with units and a sign	HL only	"Find the rate at which the water level is rising when…" (5 to 7 marks)
Build an optimisation model from a description and find the optimum with calculus	HL only	"Find the value of x that minimises the cost, and justify that it is a minimum" (6 to 8 marks)
Recognise when the optimum is at an end point of the domain rather than at a stationary point	HL only	"Find the maximum possible total area" where the answer is at x = 0 (3 marks)

Before you start

You need the chain rule and the product rule (SL 5.6), and the SL work on stationary points and their nature, including the second derivative test (SL 5.7, 5.8). Tangents and normals are SL 5.4, and the equation of a line through a point is SL 2.1. Volume and surface-area formulas for spheres and cones are prior learning, and they are in the formula booklet. One example uses d/dx(tan x) = sec²x, which is AHL 5.15.

1The idea in one paragraph

Some relationships between x and y cannot be rearranged into y = f(x), or can only be rearranged with a mess. Implicit differentiation skips the rearranging: differentiate both sides with respect to x as they stand, remembering that y is a function of x, so every term in y picks up a factor dy/dx by the chain rule. Then solve for dy/dx. Related rates is the same idea with time: when two quantities are linked by an equation and both change with time, differentiate the equation with respect to t and the rates are linked too. Optimisation at HL means building the function to be maximised or minimised from a description, differentiating it (often with the chain rule or implicitly), and then checking the stationary points and the end points of the domain, because the best value is sometimes at the edge.

2Implicit differentiation

A circle of radius 5 centred at the origin is x² + y² = 25. It is not the graph of one function: each x between −5 and 5 has two y-values. You could split it into y = √(25 − x²) and y = −√(25 − x²) and differentiate each half, but there is a cleaner way.

Differentiate every term with respect to x. The x² term is ordinary: 2x. The y² term needs thought. y is itself a function of x, so y² is a composite function, and the chain rule gives

d/dx(y²) = 2y · dy/dx, and in general d/dx(g(y)) = g′(y) · dy/dx.

Differentiate "as if y were x", then multiply by dy/dx. That one habit is the whole of implicit differentiation.

Term	Derivative with respect to x	Rule used
y³	3y² dy/dx	chain rule
eʸ	eʸ dy/dx	chain rule
sin y	cos y · dy/dx	chain rule
ln y	(1/y) dy/dx	chain rule
xy	y + x dy/dx	product rule, then chain rule on y
x²y	2xy + x² dy/dx	product rule
a constant	0

Worked example 1. Paper 1. Find dy/dx for x² + y² = 25, and the gradient of the circle at (3, 4) and at (3, −4).

d/dx(x²) + d/dx(y²) = d/dx(25)

2x + 2y dy/dx = 0

dy/dx = −x/y

at (3, 4): dy/dx = −3/4

at (3, −4): dy/dx = 3/4

The answer contains y as well as x. That is normal for implicit differentiation, and it is why you need both coordinates of the point, not just x. It is also how one formula gives two different gradients at x = 3, one for each half of the circle. Figure 1 draws both tangents.

Figure 1 · One formula, dy/dx = −x/y, gives both tangents at x = 3

A check: the radius to (3, 4) has gradient 4/3, and −3/4 is its negative reciprocal, so the tangent is perpendicular to the radius, as it must be.

Worked example 2. Paper 1. The curve C has equation x²y + y³ = 10. Show that (3, 1) lies on C, and find the equation of the tangent there.

check: 3²(1) + 1³ = 9 + 1 = 10 ✓

2xy + x² dy/dx + 3y² dy/dx = 0product rule on x²y, chain rule on y³

dy/dx (x² + 3y²) = −2xycollect the dy/dx terms on one side

dy/dx = −2xy / (x² + 3y²)

at (3, 1): dy/dx = −6 / (9 + 3) = −1/2

tangent: y − 1 = −(1/2)(x − 3), so y = −x/2 + 5/2

Worked example 3. Paper 1. The curve x³ + y³ = 6xy passes through (3, 3). Find the equations of the tangent and the normal there.

This curve (Figure 2) crosses itself at the origin and has a loop. There is no way to write it as y = f(x) with the tools of this course.

3x² + 3y² dy/dx = 6y + 6x dy/dxproduct rule on 6xy

dy/dx (3y² − 6x) = 6y − 3x²

dy/dx = (2y − x²) / (y² − 2x)divide top and bottom by 3

at (3, 3): dy/dx = (6 − 9)/(9 − 6) = −1

tangent: y − 3 = −(x − 3), so y = −x + 6

normal: gradient 1, so y − 3 = x − 3, y = x

Figure 2 · The tangent and normal to x³ + y³ = 6xy at (3, 3)

The normal is the line y = x, which is the curve's line of symmetry: swap x and y in the equation and it does not change.

Horizontal and vertical tangents. A tangent is horizontal where dy/dx = 0, which means the numerator of your dy/dx is zero (and the denominator is not). It is vertical where the denominator is zero (and the numerator is not). Either condition gives a relationship between x and y, which you then substitute back into the curve's equation.

Worked example 4. Paper 1. Find the points on x² + xy + y² = 27 where the tangent is horizontal, and where it is vertical.

2x + y + x dy/dx + 2y dy/dx = 0

dy/dx = −(2x + y)/(x + 2y)

horizontal: 2x + y = 0, so y = −2x

x² + x(−2x) + 4x² = 27 → 3x² = 27 → x = ±3substitute into the curve

points (3, −6) and (−3, 6)

vertical: x + 2y = 0, so x = −2y

4y² − 2y² + y² = 27 → y = ±3

points (−6, 3) and (6, −3)

In 5.15 the same method finds the derivatives of arcsin x, arccos x, arctan x and aˣ.

3Related rates of change

In a related rates problem, two or more quantities change with time and are linked by an equation. You know how fast one of them is changing and you want how fast another is changing, at one particular instant. The chain rule links them:

dA/dt = dA/dr × dr/dt

Or, when the link is an implicit equation, differentiate both sides with respect to t, exactly as in section 2 but with t in place of x.

The method, in five steps.

Draw a diagram and label every changing quantity with a letter. Do not put in the numbers for the particular instant yet; they are only true for a moment.
Write the rate you are given and the rate you want, with units, as derivatives with respect to t. A quantity that is decreasing has a negative rate.
Write an equation linking the quantities. If it has too many variables, use a fixed fact (similar triangles, a constant length) to remove one.
Differentiate the equation with respect to t.
Only now substitute the values at the instant, and solve for the rate you want.

Worked example 5. A stone dropped in a pond makes a circular ripple whose radius grows at 3 cm s⁻¹. How fast is the area inside the ripple increasing when the radius is 10 cm?

A = πr², dr/dt = 3

dA/dt = dA/dr × dr/dt = 2πr × 3 = 6πr

at r = 10: dA/dt = 60π ≈ 188 cm² s⁻¹

The rate of the radius is constant, but the rate of the area is not: the same 3 cm of new radius adds more area on a larger circle.

Worked example 6. Paper 1. A ladder 5 m long leans against a vertical wall. Its foot slides away from the wall at 0.4 m s⁻¹. How fast is the top sliding down the wall when the foot is 3 m from the wall? Figure 3 sets it up.

Figure 3 · The ladder: x and y change, the 5 m does not

x² + y² = 25Pythagoras; the ladder's length is fixed

2x dx/dt + 2y dy/dt = 0differentiate with respect to t

dy/dt = −(x/y) dx/dt

at the instant: x = 3, so y = 4; dx/dt = 0.4

dy/dt = −(3/4)(0.4) = −0.3

The top slides down at 0.3 m s⁻¹. The negative sign says y is decreasing. In the answer sentence, give the speed and the direction in words.

Worked example 7. Paper 1. A water tank is a cone with its vertex at the bottom, height 6 m and top radius 2 m. Water flows in at 2 m³ per minute. Find the rate at which the depth is rising when the water is 3 m deep.

The water forms a smaller cone, with depth h and surface radius r, both changing (Figure 4). The volume formula has two variables, so remove r first.

Figure 4 · The water is a cone similar to the tank

r/h = 2/6, so r = h/3similar triangles, true at every moment

V = (1/3)πr²h = (1/3)π(h/3)² h = πh³/27

dV/dt = (πh²/9) dh/dtdifferentiate with respect to t

2 = (π × 9/9) dh/dtat h = 3, dV/dt = 2

dh/dt = 2/π ≈ 0.637 m per minute

Substituting r = 1 (its value when h = 3) before differentiating would treat r as a constant and give the wrong answer. That is the reason for step 5.

4Optimisation, including end points

SL 5.8 taught you to find a maximum or minimum by setting the derivative to zero and checking its nature. At HL the problems are harder in two ways the guide names. The function often has to be built from a description, and differentiating it needs the chain rule or implicit differentiation. And the domain has limits, so the best value may be at an end point, not at a stationary point.

On a closed interval a ≤ x ≤ b, the greatest and least values are found either at a stationary point inside the interval or at an end point. Check all of them.

Worked example 8. Paper 2. A power cable must run from a station A on one bank of a straight river 600 m wide to a factory B on the other bank, 2000 m downstream. Cable under water costs €50 per metre and cable along the bank costs €30 per metre. The cable runs under water from A to a point P on the far bank, x m downstream of the point directly opposite A, then along the bank to B (Figure 5). Find the value of x that minimises the cost.

Figure 5 · The cable crosses to P, then follows the bank

C(x) = 50√(600² + x²) + 30(2000 − x), 0 ≤ x ≤ 2000

C′(x) = 50x/√(360000 + x²) − 30chain rule on the square root

C′(x) = 0: 50x = 30√(360000 + x²)

2500x² = 900(360000 + x²)square both sides (x ≥ 0)

1600x² = 324000000, x² = 202500, x = 450

C(450) = 50(750) + 30(1550) = 84000

end points: C(0) = 90000, C(2000) ≈ 104403

The least cost is €84,000, with x = 450 m. Checking the end points confirms that 450 gives the least value on the whole interval. The sign of C′ confirms the nature too: C′(0) = −30 < 0 and C′(2000) ≈ 17.9 > 0.

Now change one fact. Suppose the factory is only 300 m downstream. The cost function is C(x) = 50√(600² + x²) + 30(300 − x) with 0 ≤ x ≤ 300. The derivative is the same, and it is still zero only at x = 450, which is outside the domain. On the whole interval C′(x) < 0 (C′(300) ≈ −7.6), so C is decreasing all the way, and the minimum is at the end point x = 300: run the whole cable under water, straight to the factory.

C(300) = 50√450000 ≈ 33541all under water

C(0) = 30000 + 9000 = 39000straight across, then 300 m of bank

Figure 6 puts the two cost curves side by side. In the first the minimum is at the bottom of a valley. In the second there is no valley inside the domain, and the minimum is at the edge.

Figure 6 · The minimum at a stationary point, and at an end point

A stationary point outside the domain is not an answer. Solving C′(x) = 0, getting 450 and writing it down would score almost nothing in the second version.

Worked example 9. Paper 1. A cylinder is cut from a solid sphere of radius 3 cm, with its axis along a diameter (Figure 7). Find the greatest possible volume of the cylinder.

Figure 7 · A cylinder inside a sphere of radius 3

Let the cylinder have radius r and height h. The right-angled triangle in the figure links them.

r² + (h/2)² = 9, so r² = 9 − h²/4Pythagoras in the cross-section

V = πr²h = π(9 − h²/4)h = π(9h − h³/4), 0 < h < 6

dV/dh = π(9 − 3h²/4) = 0 → h² = 12, h = 2√3

d²V/dh² = −(3π/2)h < 0 at h = 2√3, so a maximum

r² = 9 − 3 = 6

V = π × 6 × 2√3 = 12√3π ≈ 65.3 cm³

You could instead keep both r and h and differentiate the constraint implicitly: 2r dr/dh + h/2 = 0. The guide asks for "appropriate use of the chain rule or implicit differentiation", and either route earns full marks. Substituting first, as above, is usually shorter.

At the ends of 0 < h < 6 the cylinder is flat or needle-thin and V → 0, so the stationary point is certainly the maximum.

Worked example 10. Find the point on the curve y = √x that is closest to A(4, 0).

The distance from A to a point (x, √x) is D = √((x − 4)² + x). Minimising a square root is awkward, but D is positive, so D is smallest exactly where D² is smallest. (By the chain rule, dD/dx = (1/(2D)) × d(D²)/dx, and the two are zero together.) So minimise D² instead.

D² = (x − 4)² + x = x² − 7x + 16

d(D²)/dx = 2x − 7 = 0 → x = 3.5

second derivative 2 > 0, so a minimum

D² = 3.75, D = √15/2 ≈ 1.94

closest point (3.5, √3.5) ≈ (3.5, 1.87)

The end point x = 0 gives D = 4, which is larger, so the stationary point is the answer.

5Where marks are lost

Forgetting dy/dx on a y-term. d/dx(y³) is 3y² dy/dx, not 3y². Every term in y gets the factor; every term in x does not.

Missing the product rule on xy. d/dx(xy) = y + x dy/dx. Writing just x dy/dx, or just 1 · dy/dx, loses the next three marks with it.

Differentiating the constant wrongly, or not at all. d/dx(25) = 0. Differentiating the right-hand side is still part of the method, even when it is a number.

Substituting the instant's values before differentiating. In a related rates problem, a quantity that is changing must stay a letter until after you have differentiated. Put in h = 3 first and h becomes a constant with derivative zero.

Losing the sign of a rate. A decreasing quantity has a negative rate. Put the sign into the calculation, and describe the direction in words in the answer.

Ignoring the domain in optimisation. A stationary point outside the allowed interval is not a candidate. And even when it is inside, the end points must be compared, or the answer is not justified.

No justification of the nature. "Find the maximum" needs a reason: the second derivative, a sign change of the first derivative, or a comparison with the end points.

Solving y = −2x and stopping. A horizontal-tangent condition is a line. Substitute it into the curve to find the actual points.

6Work it right

Implicit: differentiate every term, including the right-hand side, and write dy/dx after every term in y.
Collect the dy/dx terms on one side, factorise, and divide; give dy/dx in terms of x and y.
Check the given point lies on the curve before you use it.
For a normal, take the negative reciprocal of the tangent's gradient.
Related rates: diagram, letters, the given and wanted rates with units, an equation, differentiate with respect to t, and only then substitute.
Use similar triangles or a fixed length to remove an extra variable before differentiating.
Optimisation: state the function and its domain, differentiate, solve, check the nature, then compare with the end points.
Answer the question asked: the dimension, the least cost or the rate, with units and in context.

7Try it

Marks in brackets. Q1 to Q4 are Paper 1 style (no calculator). Q5 and Q6 are Paper 2 style.

Q1. A curve has equation x² + 3xy − y² = 3.

(a) Find dy/dx in terms of x and y. 4 marks

(b) Find the equation of the tangent to the curve at the point (1, 1). 2 marks

Q2. The curve C has equation xeʸ + y = 2. Find the equation of the normal to C at the point (2, 0). 5 marks

Q3. Find the coordinates of the points on the curve x² − xy + y² = 12 at which the tangent is parallel to the x-axis. 6 marks

Q4. A spherical balloon is inflated so that its volume increases at 50 cm³ s⁻¹. When the radius is 10 cm, find

(a) the rate at which the radius is increasing 3 marks

(b) the rate at which the surface area is increasing. 2 marks

Q5. A runner moves along a straight track at 8 m s⁻¹. A camera at C is 40 m from the track; O is the point on the track closest to C. When the runner is x m past O, the camera's line of sight makes an angle θ with CO, so that tan θ = x/40. Find the rate at which θ is changing, in radians per second, when x = 30. 5 marks

Q6. A wire 40 cm long is cut into two pieces. One piece, of length x cm, is bent into a square, and the other into a circle. Either piece may have zero length, so 0 ≤ x ≤ 40.

(a) Show that the total area enclosed is A(x) = x²/16 + (40 − x)²/(4π). 2 marks

(b) Find the value of x that minimises A, and the minimum area. 3 marks

8In one breath

To differentiate an equation in x and y without solving for y, differentiate every term with respect to x and multiply every y-derivative by dy/dx (chain rule), using the product rule on terms like xy; then collect and solve for dy/dx, which will contain both x and y, so use both coordinates. Horizontal tangents: numerator zero; vertical: denominator zero; substitute the condition back into the curve. Related rates: draw, label with letters, write the given and wanted rates with units and signs, link the quantities with one equation (similar triangles to remove extras), differentiate with respect to t, and substitute the instant's values last. Optimisation: build the function and its domain, differentiate (chain rule or implicitly), justify the nature, and on a closed interval compare the stationary points with the end points, because the best value can sit at the edge, and a stationary point outside the domain does not count.

Answers

Q1. (a) 2x + 3y + 3x dy/dx − 2y dy/dx = 0, so dy/dx(3x − 2y) = −(2x + 3y) and dy/dx = −(2x + 3y)/(3x − 2y). M1 for implicit differentiation attempted, A1 for 3y + 3x dy/dx from the product rule, A1 for −2y dy/dx, A1 for the rearranged result.

(b) At (1, 1): dy/dx = −5/1 = −5. Tangent: y − 1 = −5(x − 1), so y = −5x + 6. M1 for substituting (1, 1) into their dy/dx, A1 for the equation.

Q2. Check: 2e⁰ + 0 = 2 ✓. Differentiate: eʸ + xeʸ dy/dx + dy/dx = 0. At (2, 0): 1 + 2 dy/dx + dy/dx = 0, so dy/dx = −1/3. The normal has gradient 3: y − 0 = 3(x − 2), so y = 3x − 6. M1 for implicit differentiation, A1 for eʸ + xeʸ dy/dx from the product rule, A1 for dy/dx = −1/3, M1 for the negative reciprocal, A1 for the equation.

Q3. 2x − y − x dy/dx + 2y dy/dx = 0, so dy/dx = (y − 2x)/(2y − x). Horizontal tangent: y − 2x = 0, so y = 2x. Substituting: x² − 2x² + 4x² = 12, so 3x² = 12 and x = ±2. The points are (2, 4) and (−2, −4), and at both the denominator 2y − x = ±6 ≠ 0. M1 for implicit differentiation, A1 for a correct dy/dx, M1 for setting the numerator to zero, M1 for substituting y = 2x into the curve, A1 for x = ±2, A1 for both points.

Q4. (a) V = (4/3)πr³, so dV/dt = 4πr² dr/dt. At r = 10: 50 = 400π dr/dt, so dr/dt = 1/(8π) ≈ 0.0398 cm s⁻¹. M1 for the chain rule link, A1 for substituting r = 10 and dV/dt = 50, A1 for 1/(8π).

(b) S = 4πr², so dS/dt = 8πr dr/dt = 8π(10) × 1/(8π) = 10 cm² s⁻¹. M1 for dS/dt = 8πr dr/dt, A1 for 10.

Q5. x = 40 tan θ, so dx/dt = 40 sec²θ dθ/dt. When x = 30, tan θ = 3/4, so sec²θ = 1 + 9/16 = 25/16. Then 8 = 40 × (25/16) × dθ/dt, so dθ/dt = 0.128 rad s⁻¹. M1 for differentiating with respect to t, A1 for 40 sec²θ dθ/dt, M1 for finding sec²θ at x = 30, A1 for 25/16, A1 for 0.128. Working in degrees scores the method marks only.

Q6. (a) The square has side x/4, so area x²/16. The circle has circumference 40 − x, so radius (40 − x)/(2π) and area π((40 − x)/(2π))² = (40 − x)²/(4π). Adding gives the result. A1 for each area.

(b) A′(x) = x/8 − (40 − x)/(2π) = 0 gives πx = 4(40 − x), so x = 160/(π + 4) ≈ 22.4 cm. A″(x) = 1/8 + 1/(2π) > 0, so it is a minimum; A ≈ 56.0 cm². M1 for A′(x) = 0, A1 for x ≈ 22.4, A1 for 56.0 with the nature justified.

(c) The only stationary point is a minimum, so the maximum is at an end point. A(0) = 1600/(4π) ≈ 127 cm² (all circle) and A(40) = 100 cm² (all square), so the maximum is 127 cm², when the whole wire forms a circle. R1 for the maximum being at an end point because the stationary point is a minimum, A1 for both end values, A1 for 127 cm² with x = 0.

Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 5.14 Implicit differentiation, related rates and optimisation. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

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