Implicit Differentiation

Some curves — like the circle \(x^2 + y^2 = 25\) — can't be conveniently solved for \(y\) as a single function of \(x\). Implicit differentiation finds \(\tfrac{dy}{dx}\) anyway: differentiate both sides, treating \(y\) as a hidden function of \(x\), then solve for the derivative.

By the end you'll be able to differentiate a relation like \(x^2 + y^2 = 25\) implicitly, solve for \(\tfrac{dy}{dx}\), and compute the tangent slope at any point on the curve.

Predict: where on the circle is the tangent line horizontal, and where is it vertical? Drag the point around the circle (or focus it and use the arrow keys) and check the slope readout against your prediction.

This is \(x^2 + y^2 = r^2\). Implicit differentiation says the tangent slope at any point is \(-x/y\) — the orange x over the green y, negated. The small dashes at the top, bottom, and sides mark where that formula goes to 0 (horizontal) or blows up (vertical). Resize with the radius slider: the formula never changes.

Point (x, y) = (3.00, 4.00) · slope dy/dx = −x/y = −(3.00)/(4.00) = −0.750

x² + y² = r² — drag the point; the dashed tangent has slope −x/y
circle tangent at the point x of the point y of the point horizontal / vertical tangent marks

Implicit differentiation differentiates both sides of a relation with respect to \(x\), treating \(y\) as a function of \(x\) via the chain rule, then solves for \(\tfrac{dy}{dx}\) — giving you slopes on curves that were never solved for \(y\).

Formal

The strategy has three steps. 1) Differentiate both sides of the equation with respect to \(x\). 2) Treat \(y\) as an unknown function \(y(x)\), so every \(y\)-term is a composite and picks up a factor of \(\tfrac{dy}{dx}\) by the chain rule: \(\frac{d}{dx}[y^2] = 2y\,\frac{dy}{dx}\), exactly like \(\frac{d}{dx}[(g(x))^2] = 2g(x)\,g'(x)\) — here \(y\) plays the role of the inner function. Terms in \(x\) alone differentiate normally. 3) Collect the \(\tfrac{dy}{dx}\) terms on one side and solve algebraically. The answer typically involves both \(x\) and \(y\) — that's normal for implicit results, since one \(x\) can sit above several \(y\) values.

Applied

The circle above is the picture to keep. \(x^2 + y^2 = r^2\) isn't the graph of any single function — a vertical line hits it twice — yet every point still has a perfectly good tangent line. Implicit differentiation delivers its slope, \(-\)\(x\)\(/\)\(y\), and the geometry checks out: at the top and bottom (where \(x = 0\)) the tangent is horizontal with slope 0, and at the far left and right (where \(y = 0\)) the formula divides by zero — a vertical tangent, something no formula of the form \(y = f(x)\) could ever produce.

Worked example

Find \(\tfrac{dy}{dx}\) for \(x^2 + y^2 = 25\), and the slope at \((3, 4)\). Differentiate both sides: \(2x + 2y\,\frac{dy}{dx} = 0\) — the \(y^2\) term carries the chain-rule factor \(\tfrac{dy}{dx}\). Solve: \(\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}\). At \((3, 4)\) the slope is \(-\tfrac{3}{4}\) — exactly what the demo above shows at its starting point.

Your turn

A different relation: find \(\tfrac{dy}{dx}\) for \(x^3 + y^3 = 9\) at the point \((1, 2)\). Differentiate both sides: \(3x^2 + 3y^2\,\frac{dy}{dx} = 0\). Now finish it: solve for \(\tfrac{dy}{dx}\) and evaluate at \((1, 2)\): \(\frac{dy}{dx} = \) ____, which at \((1, 2)\) equals ____.

Reveal the answer

Solving \(3y^2\,\frac{dy}{dx} = -3x^2\) gives \(\frac{dy}{dx} = -\frac{x^2}{y^2}\). At \((1, 2)\) that's \(-\frac{1^2}{2^2} = \) \(-\tfrac{1}{4}\). Note the pattern: the \(y^3\) term produced \(3y^2\,\frac{dy}{dx}\) — power rule on the outside, chain-rule factor for the inside.

More info — why the chain rule shows up at all

Recall the chain rule from the previous lesson: a composite \(f(g(x))\) differentiates to \(f'(g(x))\cdot g'(x)\). On a curve like \(x^2 + y^2 = 25\), \(y\) is some function of \(x\) near most points — you just don't have its formula. So \(y^2\) is secretly the composite \((y(x))^2\): outer function "square", inner function \(y(x)\). The chain rule gives \(2y(x)\cdot y'(x)\), i.e. \(2y\,\frac{dy}{dx}\). The same logic makes \(\frac{d}{dx}[\sin y] = \cos y\,\frac{dy}{dx}\). Implicit differentiation isn't a new rule — it's the chain rule applied to an inner function you can't write down. Two pitfalls to watch: mixed terms like \(xy\) need the product rule (\(\frac{d}{dx}[xy] = y + x\,\frac{dy}{dx}\)), and you're not done until \(\tfrac{dy}{dx}\) is isolated. The OpenStax link in Dive deeper below works through the full strategy step by step.

Check your understanding

Question 1 of 4

The curve \(x^2 + y^2 = 100\) passes through the point \((8, 6)\). Using implicit differentiation, what is the slope of the tangent line there?

Question 2 of 4

Differentiating \(x^2 + y^2 = 49\) with respect to \(x\), a student writes \(2x + 2y = 0\). What went wrong?

Question 3 of 4

During an implicit differentiation, you need \(\frac{d}{dx}[\sin y]\), where \(y\) is a function of \(x\). What is it?

Question 4 of 4

At which points on the circle \(x^2 + y^2 = r^2\) is the tangent line vertical?

Recap

  • To differentiate a relation not solved for \(y\): differentiate both sides with respect to \(x\), treat \(y\) as \(y(x)\), then solve for \(\tfrac{dy}{dx}\).
  • Every \(y\)-term picks up a chain-rule factor: \(\frac{d}{dx}[y^2] = 2y\,\frac{dy}{dx}\), \(\frac{d}{dx}[\sin y] = \cos y\,\frac{dy}{dx}\). Forgetting that factor is the classic mistake.
  • Circle: \(x^2 + y^2 = r^2\) gives \(\frac{dy}{dx} = -x/y\) — horizontal tangents at the top and bottom (\(x = 0\)), vertical tangents at the sides (\(y = 0\)).
  • Implicit answers usually contain both \(x\) and \(y\); mixed terms like \(xy\) need the product rule, and the job isn't finished until \(\tfrac{dy}{dx}\) is isolated.

Dive deeper

Sources

  • Implicit Differentiation