Implicit Differentiation (AI HL)
Most curves you differentiate are already written as \(y = \dots\), but some relations between \(x\) and \(y\) can't be rearranged that way - a circle, or a curve like \(x^2+xy+y^2=7\). Implicit differentiation lets you find \(\dfrac{dy}{dx}\) directly from the relation itself, and the same chain-rule idea underpins related rates of change questions, where two quantities change together over time. It's part of the broader Differentiation topic.
11 questions on this sub-topic.
The two techniques
Covered under IB syllabus reference AHL5.9, which extends the chain, product and quotient rules to related rates of change. Implicit differentiation isn't named separately in the syllabus wording, but it's the natural way to apply the chain rule when \(y\) can't be isolated first.
Differentiating an implicit relation
Differentiate each term with respect to \(x\). Any term in \(y\) picks up a \(\dfrac{dy}{dx}\) factor from the chain rule; a mixed term like \(xy\) needs the product rule too.
Not in the formula booklet - it's the chain rule applied to \(y\) as a function of \(x\), not a separate formula.
Related rates of change
Write a formula linking the two quantities (e.g. \(A=\pi r^2\)), differentiate both sides with respect to time, then substitute the known values.
Not in the formula booklet - applied method built from the chain rule and the shape formula the question gives you.
Need the surrounding derivative rules and formula-booklet reference table? See Differentiation.
Worked examples
A circular oil slick expands so its radius grows at 0.5 m/s.
Find the rate of increase of its area when \(r = 10\) m.
Worked solution
\(A = \pi r^2\), so \(\dfrac{dA}{dt}\) M1
\(= 2\pi r\dfrac{dr}{dt}.\) A1
\(= 2\pi(10)(0.5)\) M1 A1
\(= 10\pi \approx 31.4\) m²/s. A1
A curve is defined by \(x^2 + xy + y^2 = 7.\) Find \(\dfrac{dy}{dx}.\)
Worked solution
\(2x + (y + x\tfrac{dy}{dx}) + 2y\tfrac{dy}{dx}\) M1 \(= 0.\) A1
\(\tfrac{dy}{dx}\) terms: \(\tfrac{dy}{dx}(x + 2y)\) M1 \(= -(2x + y).\) A1
\(\dfrac{dy}{dx} = -\dfrac{2x + y}{x + 2y}.\) A1
Common mistakes
- Forgetting the \(\dfrac{dy}{dx}\) factor on \(y\) terms. Differentiating \(y^2\) gives \(2y\dfrac{dy}{dx}\), not \(2y\) - every term involving \(y\) needs the chain rule applied, since \(y\) is itself a function of \(x\).
- Missing the product rule on mixed terms. A term like \(xy\) differentiates to \(y + x\dfrac{dy}{dx}\), not just \(x\dfrac{dy}{dx}\) - dropping the \(y\) is a very common slip.
- Not matching units of time in related rates. If a radius grows in cm/s but the question asks for the rate in a different unit, or gives a diameter instead of a radius, convert before substituting - the chain rule step is often right but the numbers going in are wrong.
Ready to practise properly?
13 implicit differentiation and related rates questions, marked instantly like the real exam.
Quick answers
What is implicit differentiation?
It's a way of finding \(\dfrac{dy}{dx}\) from an equation relating \(x\) and \(y\) that hasn't been rearranged to give \(y\) on its own. Differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\) (so every \(y\) term picks up a \(\dfrac{dy}{dx}\) factor via the chain rule), then collect the \(\dfrac{dy}{dx}\) terms and solve.
How do related rates of change questions work?
Write down a formula linking the two quantities involved (such as area and radius), differentiate it with respect to time using the chain rule, then substitute the known rate and value to find the rate you're asked for. See the parent Differentiation page's GDC guidance for checking numerical substitutions on your calculator.