IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~11 min read

Implicit Differentiation

Sometimes you can’t isolate y at all — curves like x² + y² = 25 (a circle) or x²y + xy² = 6 lock x and y together. Implicit differentiation lets you find dy/dx without ever solving for y. The whole technique is one chain-rule trick: treat y as a function of x, so d/dx[y²] = 2y · dy/dx. Then rearrange.

📘 What you need to know

The core chain-rule formula: d/dx[f(y)] = f′(y) · dy/dx. Every term containing y picks up a dy/dx when differentiated with respect to x.
Product rule for mixed terms: d/dx[x^myⁿ] = mx^m−1yⁿ + x^m · nyⁿ⁻¹ · dy/dx. Easy to get wrong under exam pressure.
Method in 3 lines: differentiate both sides w.r.t. x; collect all dy/dx terms on one side; factor out dy/dx and divide.
The result is in terms of x AND y — that’s expected. To find a numerical gradient, you need BOTH coordinates of the point.
Horizontal tangent: dy/dx = 0 → numerator = 0. Solve simultaneously with the curve equation.
Vertical tangent: dx/dy = 0 → denominator of dy/dx = 0. Solve simultaneously with the curve.
Tangent line at (x₀, y₀): y − y₀ = m(x − x₀) where m = dy/dx evaluated at (x₀, y₀). Normal: replace m with −1/m.
Don’t try to solve for y first — that’s the whole point. Even when you could, implicit differentiation is usually faster.

The technique — chain rule on every y

Implicit chain rule ddx [f(y)] = f′(y) · dydx

The intuition is “y is secretly a function of x“. So when you see y³, differentiating with respect to x gives 3y² (the inner derivative would be 3y² if y were x) times the missing dy/dx factor. That extra dy/dx is the whole story.

Pure x term

ddx[x³] = 3x²

no chain rule — standard power rule

Pure y term

ddx[y³] = 3y² · dydx

chain rule — extra dy/dx factor

Mixed term (product rule)

ddx[x²y] = 2xy + x² · dydx

product rule + chain rule on the y factor

Composite y term

ddx[e^y] = e^y · dydx

any function of y works the same way

🧭 Recipe — find dy/dx implicitly

Differentiate both sides of the equation with respect to x. Take it term-by-term — every y-containing term picks up a dy/dx via the chain rule.
Use the product rule for any term containing both x and y (like xy, x²y, x sin y, etc.) — differentiate each factor while holding the other; one factor will produce a dy/dx.
Collect all terms containing dy/dx on one side; everything else on the other.
Factor dy/dx from its terms, then divide to isolate it. The answer is an expression in x and y.
Apply: for a numerical gradient, substitute both coordinates of the point. For tangent line: y − y₀ = m(x − x₀). For normal: slope = −1/m. For horizontal/vertical tangents: set numerator/denominator of dy/dx to 0.

Horizontal and vertical tangents on implicit curves

Once you have dy/dx as a fraction in x and y, two special tangent cases drop out automatically.

Horizontal tangent

dy/dx = 0 ⇒ numerator = 0

solve along with the curve equation to find the point(s)

Vertical tangent

dx/dy = 0 ⇒ denominator = 0

again, solve with the curve equation to find the point(s)

For the circle x² + y² = 16, implicit differentiation gives dy/dx = −x/y. The numerator is zero at x = 0 (top and bottom points, horizontal tangents — green); the denominator is zero at y = 0 (left and right points, vertical tangents — orange).

Why the technique works: along the curve, every point (x, y) satisfies the equation. Differentiating that identity with respect to x gives a relation between dy/dx and the point’s coordinates — no need to ever isolate y.

Worked examples

WE 1

Basic implicit differentiation, then evaluation at a point

A curve has equation 5x² − 4y² = 16. (a) Find dydx in terms of x and y. (b) Hence find the gradient of the curve at the point (2, 1).

(a) Step 1 — differentiate both sides with respect to x d/dx[5x² − 4y² − 16] = 0 10x − 8y · (dy/dx) − 0 = 0 Step 2 — isolate dy/dx 8y · (dy/dx) = 10x dy/dx = 10x/(8y) = 5x/(4y) (a) dy/dx = 5x/(4y) (b) substitute (x, y) = (2, 1) check point on curve: 5(4) − 4(1) = 20 − 4 = 16 ✓ dy/dx |_(2,1) = 5(2)/(4(1)) = 10/4 = 5/2 (b) gradient = 5/2 notice the −4y² term: chain rule says d/dx[−4y²] = −8y · dy/dx — the chain factor dy/dx is what makes this “implicit” rather than standard.

WE 2

Product rule applied to two mixed terms

A curve has equation x²y + xy² = 6. Find the gradient of the curve at the point (1, 2).

Step 1 — check the point is on the curve (1)²(2) + (1)(2)² = 2 + 4 = 6 ✓ Step 2 — differentiate term-by-term, using product rule on BOTH terms d/dx[x²y] = 2xy + x² · (dy/dx) ← product rule d/dx[xy²] = y² + x · 2y · (dy/dx) = y² + 2xy · (dy/dx) ← product rule + chain rule on y² Step 3 — combine and collect dy/dx terms 2xy + x²(dy/dx) + y² + 2xy(dy/dx) = 0 (x² + 2xy)(dy/dx) = −(2xy + y²) Step 4 — divide to isolate dy/dx dy/dx = −(2xy + y²)/(x² + 2xy) = −y(2x + y)/(x(x + 2y)) Step 5 — substitute (1, 2) dy/dx |_(1,2) = −(2(1)(2) + 4)/(1 + 2(1)(2)) = −(4 + 4)/(1 + 4) = −8/5 gradient = −8/5 two terms with both x and y, so the product rule fires twice. The dy/dx coefficient (x² + 2xy) drops out as a common factor after collecting.

WE 3

Equation of the tangent at a point on an implicit curve

A curve C has equation x² + y² + 4xy = 6. Find the equation of the tangent to C at the point (1, 1).

Step 1 — check the point is on the curve 1 + 1 + 4(1)(1) = 6 ✓ Step 2 — differentiate implicitly 2x + 2y · (dy/dx) + 4·(y + x · (dy/dx)) = 0 ← product rule on 4xy 2x + 4y + (2y + 4x)(dy/dx) = 0 Step 3 — isolate dy/dx dy/dx = −(2x + 4y)/(2y + 4x) = −(x + 2y)/(y + 2x) Step 4 — evaluate at (1, 1) dy/dx |_(1,1) = −(1 + 2)/(1 + 2) = −3/3 = −1 Step 5 — tangent through (1, 1) with slope −1 y − 1 = −1(x − 1) y = −x + 2 tangent: y = −x + 2 (or x + y = 2) the symmetry of the curve (swap x and y leaves it unchanged) means the slope at the symmetric point (1, 1) must be −1 — the curve’s diagonal symmetry forces this.

WE 4

Horizontal AND vertical tangents on a rotated ellipse

A curve has equation x² + xy + y² = 12. Find the coordinates of all points on the curve where the tangent is (i) parallel to the x-axis, and (ii) parallel to the y-axis.

Step 1 — implicit differentiation 2x + (y + x · (dy/dx)) + 2y · (dy/dx) = 0 ← product rule on xy 2x + y + (x + 2y)(dy/dx) = 0 dy/dx = −(2x + y)/(x + 2y) (i) HORIZONTAL tangent: dy/dx = 0 ⟹ numerator = 0 2x + y = 0 ⟹ y = −2x substitute into curve: x² + x(−2x) + 4x² = 12 x² − 2x² + 4x² = 3x² = 12 ⟹ x = ±2 x = 2 ⟹ y = −4; x = −2 ⟹ y = 4 (i) horizontal tangents at (2, −4) and (−2, 4) (ii) VERTICAL tangent: dx/dy = 0 ⟹ denominator = 0 x + 2y = 0 ⟹ x = −2y substitute into curve: 4y² + (−2y)(y) + y² = 12 4y² − 2y² + y² = 3y² = 12 ⟹ y = ±2 y = 2 ⟹ x = −4; y = −2 ⟹ x = 4 (ii) vertical tangents at (−4, 2) and (4, −2) the curve is a tilted ellipse symmetric about the origin. The 4 special points pair up: (±2, ∓4) are top/bottom; (±4, ∓2) are left/right.

WE 5

Implicit differentiation involving an exponential term

A curve has equation e^y + xy = e. Find dydx at the point (0, 1).

Step 1 — check the point is on the curve e¹ + (0)(1) = e + 0 = e ✓ Step 2 — differentiate. Chain rule on e^y, product rule on xy e^y · (dy/dx) + (y + x · (dy/dx)) = 0 Step 3 — collect dy/dx terms (e^y + x)(dy/dx) = −y dy/dx = −y/(e^y + x) Step 4 — evaluate at (0, 1) dy/dx |_(0,1) = −1/(e¹ + 0) = −1/e dy/dx = −1/e (≈ −0.368) e^y is just another function of y, so it picks up a dy/dx factor like any other y-term. The chain rule doesn’t care whether the y is inside a polynomial, an exponential, or a trig function.

WE 6

Implicit differentiation involving a logarithm — find tangent line

A curve has equation ln(y) + xy = 4. Find the equation of the tangent to the curve at the point (4, 1).

Step 1 — check the point is on the curve ln(1) + (4)(1) = 0 + 4 = 4 ✓ Step 2 — differentiate. Chain rule on ln(y), product rule on xy (1/y) · (dy/dx) + (y + x · (dy/dx)) = 0 Step 3 — collect dy/dx terms (1/y + x)(dy/dx) = −y dy/dx = −y/(1/y + x) = −y² / (1 + xy) ← multiply num and den by y Step 4 — evaluate at (4, 1) dy/dx |_(4,1) = −1²/(1 + 4·1) = −1/5 Step 5 — tangent through (4, 1) with slope −1/5 y − 1 = −(1/5)(x − 4) 5y − 5 = −x + 4 x + 5y = 9 tangent: x + 5y = 9 (or y = (9 − x)/5) d/dx[ln y] = (1/y) · dy/dx — the chain rule on natural log. Multiplying numerator and denominator by y clears the inner fraction.

💡 Top tips

The dy/dx factor on every y-term is non-negotiable — it’s the single most-marked error in implicit differentiation problems.
Mixed terms always trigger the product rule — xy, x²y, x sin y, etc. Differentiate one factor at a time, keep the other.
Don’t substitute numbers too early — finish isolating dy/dx as an expression in x, y, THEN substitute the point.
Always verify the point lies on the curve as the first step. A worked example built around a point that doesn’t satisfy the equation will give a meaningless gradient.
For horizontal/vertical tangents: numerator = 0 gives horizontal, denominator = 0 gives vertical. Always solve simultaneously with the curve equation — you need both the slope condition AND the curve.

⚠ Common mistakes

Differentiating y² as 2y (without the dy/dx factor). It’s 2y · dy/dx. The chain rule applies because y = y(x).
Forgetting the product rule on xy — d/dx[xy] = y + x · dy/dx, not just x · dy/dx.
Algebraic errors when collecting terms — keep dy/dx terms together, factor cleanly. A sign error here ripples through everything.
Trying to solve the implicit equation for y first — defeats the purpose, and for most HL implicit equations, you simply can’t.
Confusing horizontal-tangent and vertical-tangent conditions — horizontal is dy/dx = 0 (numerator zero), vertical is dx/dy = 0 (denominator zero). Swapping them gives the WRONG points.

Up next: Related Rates of Change. Same implicit-differentiation idea, but now BOTH x and y are functions of time t. Differentiating a relation like V = πr²h with respect to t gives dV/dt in terms of dr/dt and dh/dt — connecting how fast volume changes to how fast radius and height change. Real-world rate problems all reduce to this technique.

Need help with Calculus?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →