Implicit Differentiation (AA HL)
Most curves you differentiate are given as \(y = f(x)\), but plenty of HL curves - circles, ellipses, and equations that mix \(x\) and \(y\) together - can't easily be rearranged that way. Implicit differentiation lets you find \(\dfrac{dy}{dx}\) directly from the equation as it stands, no rearranging required. It's part of the broader Differentiation topic.
11 questions on this sub-topic.
The method
Covered under IB syllabus reference AHL5.14. There's no separate formula to learn here - implicit differentiation is a technique built entirely from the chain rule, so it isn't listed in the formula booklet.
Implicit differentiation
Differentiate both sides of an equation in \(x\) and \(y\) with respect to \(x\). Because \(y\) is itself a function of \(x\), every \(y\)-term picks up a \(\dfrac{dy}{dx}\) factor from the chain rule. Once every term is differentiated, rearrange to isolate \(\dfrac{dy}{dx}\).
Not in the formula booklet - techniqueMixed terms need the product rule too
A term like \(xy\) contains both variables, so differentiate it as a product: \(\dfrac{d}{dx}(xy) = y + x\dfrac{dy}{dx}\). Then collect every \(\dfrac{dy}{dx}\) term on one side and factorise before dividing through.
Useful for tangent/normal problems on circles, ellipses, and other implicitly-defined curves.
Need the full syllabus wording and formula-booklet reference table? See Differentiation.
Worked examples
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
Use logarithmic differentiation to find \(\dfrac{dy}{dx}\) for \(y = x^x\), \(x > 0.\)
Worked solution
\(\ln y = x\ln x.\) M1 A1
\(\dfrac{1}{y}\dfrac{dy}{dx} = \ln x + 1.\) M1 A1
\(\dfrac{dy}{dx} = x^x(\ln x + 1).\) M1 A1
Common mistakes
- Dropping \(\dfrac{dy}{dx}\) on a y-term. Every term containing \(y\) needs a \(\dfrac{dy}{dx}\) factor from the chain rule - forgetting it on even one term gives a completely wrong equation to rearrange.
- Treating \(xy\) as a simple product of constants. A mixed term like \(xy\) needs the product rule, not just "differentiate \(x\), differentiate \(y\)" - the correct result is \(y + x\dfrac{dy}{dx}\), not \(1 \cdot \dfrac{dy}{dx}\).
- Stopping before isolating \(\dfrac{dy}{dx}\). After differentiating, the \(\dfrac{dy}{dx}\) terms are usually scattered across the equation - you must collect them, factorise, and divide before the answer is complete.
Ready to practise properly?
11 implicit-differentiation questions, marked instantly like the real exam.
Quick answers
What is implicit differentiation?
A way of differentiating an equation in \(x\) and \(y\) that isn't rearranged into \(y = f(x)\). You differentiate both sides with respect to \(x\), applying the chain rule so every \(y\)-term gains a \(\dfrac{dy}{dx}\) factor, then rearrange to isolate \(\dfrac{dy}{dx}\).
Why does differentiating a y-term give a dy/dx?
Because \(y\) is itself a function of \(x\), so differentiating any expression in \(y\) with respect to \(x\) needs the chain rule: \(\dfrac{d}{dx}(y^n) = ny^{n-1}\dfrac{dy}{dx}\).
Need a calculator refresher? See Using your GDC on the full Differentiation page.