Some curves can’t be written as \(y=f(x)\) — a circle, an ellipse, the looping folium from
the unit cover. Implicit differentiation finds their slopes anyway, using the chain rule you just learned as its engine.
When an equation mixes \(x\) and \(y\) — like \(x^2+y^2=25\) — you often can’t (or don’t want to) solve for \(y\). Implicit differentiation gets \(\frac{dy}{dx}\) without solving: treat \(y\) as a function of \(x\), differentiate both sides, and apply the chain rule to every \(y\)-term — so \(y^2\) becomes \(2y\frac{dy}{dx}\). Then solve for \(\frac{dy}{dx}\). The whole method is the chain rule, used in reverse-gear.
A curve that isn’t a function
These three curves all fail the vertical-line test — no single \(y=f(x)\). Yet each has a
definite tangent at every point. Switch between them and drag along the curve: implicit differentiation reads
\(\frac{dy}{dx}\) straight off the equation, and the spots where the tangent turns
vertical or horizontal
light up — places you could never reach by solving for \(y\).
curve
implicit slope \(\frac{dy}{dx}=\)
at point \((x,y)\)
slope here
Every \(y\) is secretly a function of \(x\)
The whole method rests on one move. Because \(y\) depends on \(x\), differentiating any \(y\)-term needs the
chain rule — it picks up a factor of \(\frac{dy}{dx}\). An \(x\)-term doesn’t. Practice spotting which is which.
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Differentiate this term with respect to \(x\):
The process, in four moves
Differentiate both sides with respect to \(x\); every \(y\)-term gains a \(\frac{dy}{dx}\) from the chain rule while \(x\)-terms behave normally; gather the \(\frac{dy}{dx}\) terms on one side; then factor and solve. The two identities below are the only new thing — everything else is algebra.
Worked example: the circle
Find \(\frac{dy}{dx}\) for \(x^2+y^2=25\). Press step to advance.
press step
A curve with a product term
The folium from the unit cover, \(x^3+y^3=6xy\), brings in a twist: the \(xy\) on the right needs the
product rule — and the \(y\)-factor still picks up its \(\frac{dy}{dx}\).
The whole idea on one card
Differentiate both sides with respect to \(x\), leaving \(y\) as \(y(x)\)
Each \(y\)-term gains a \(\frac{dy}{dx}\) from the chain rule; \(x\)-terms differentiate normally
Collect every \(\frac{dy}{dx}\) term on one side, everything else on the other
Factor out \(\frac{dy}{dx}\) and divide — the answer usually involves both \(x\) and \(y\)
The ideas everything else builds on
Don’t solve for \(y\)
You differentiate the equation as it stands — no need to isolate \(y\) first.
Every \(y\) gets a \(dy/dx\)
It’s the chain rule: \(y\) is a function of \(x\).
\(x\)-terms are normal
No \(\frac{dy}{dx}\) on a pure \(x\)-term; mixed \(xy\) needs the product rule.
Then just solve
Collect, factor, divide — the rest is algebra.
Your turn
Quick quiz
Eight fast checks across the whole lesson.
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Common mistakes & exam tips
Forgetting the \(dy/dx\) on \(y\)-terms
The defining error. \(\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\), not \(2y\). Without that factor the whole method collapses.
Treating \(y\) as a constant
\(y\) is a function of \(x\), so it changes as \(x\) changes. That’s exactly why its derivative isn’t zero.
Skipping the product rule on \(xy\)
\(\frac{d}{dx}(xy)=y+x\frac{dy}{dx}\). The term is a product of two functions of \(x\), so the product rule applies.
Trying to solve for \(y\) first
For most implicit curves that’s impossible or ugly. Differentiate the equation as written instead.
Putting \(dy/dx\) on \(x\)-terms
A pure \(x\)-term differentiates the ordinary way — no chain-rule factor.
On the AP exam
Every time you differentiate a \(y\), tack on \(\frac{dy}{dx}\). Then collect those terms and solve.
For a tangent line at a point, find \(\frac{dy}{dx}\) first, then substitute the point’s \(x\) and \(y\).
A slope that comes out as \(\frac{0}{\text{something}}\) is a horizontal tangent; a zero in the denominator means a vertical tangent.
Mixed \(xy\) (or \(x^2y\), etc.) terms need the product rule — and the \(y\)-factor still carries its \(\frac{dy}{dx}\).
You can now differentiate any relation, solved for \(y\) or not. Next: a special, powerful case of exactly this idea —
differentiating a function’s inverse by reflecting across \(y=x\), which hands you the derivatives of \(\arcsin\), \(\arctan\), and the rest.
Inverse & Inverse-Trig Derivatives is next — or head back to the
Unit 3 Guide.