Implicit Differentiation

The equation of a circle with radius 5 centred at the origin is

You want the slope of this circle at the point (3, 4). To use the tools you already have — differentiate a function, read off the slope — you would first need to solve for y:

Then differentiate: \frac{dy}{dx} = \frac{-2x}{2\sqrt{25 - x^2}} = \frac{-x}{\sqrt{25 - x^2}}. At x = 3: \frac{dy}{dx} = \frac{-3}{\sqrt{16}} = \frac{-3}{4}.

That worked, but only because you could solve for y. Try the same approach on

This is the folium of Descartes — a curve that loops through the origin and crosses itself. Solving this equation for y in terms of x requires the cubic formula, produces three branches, and makes finding the slope at any point a nightmare.

There is a better way. Instead of solving for y first, differentiate the equation as it stands — treating y as a function of x that you don't know explicitly. Every time you differentiate a term involving y, the chain rule produces a factor of dy/dx. Collect those factors, solve for dy/dx, and you have the slope — without ever solving for y.

This is implicit differentiation, and it works on any equation relating x and y, no matter how tangled.

Explicit versus implicit

Before stating the method, the terminology is worth understanding.

An explicit function is one where y is isolated on one side: y = x^2 - 3x + 1, or y = \sin x, or y = e^{2x}. You can compute y directly from x. Differentiating is straightforward: apply the rules you already know.

An implicit equation is one where x and y are tangled together: x^2 + y^2 = 25, or x^3 + y^3 = 6xy, or e^y = xy + 1. The variable y is not isolated — it is implicitly defined by the equation. Sometimes you can solve for y and make it explicit (as with the circle), but often you cannot (as with the folium), and even when you can, the explicit form may be ugly and hard to differentiate.

Implicit differentiation lets you find dy/dx from the implicit equation directly, skipping the step of solving for y entirely.

Here is the contrast at a glance:

The method

The technique has three steps, and they are always the same.

Step 1. Differentiate both sides with respect to x. Treat y as a function of x throughout. Every time you differentiate a term containing y, the chain rule adds a factor of \frac{dy}{dx}.

• The derivative of x^2 is 2x (no dy/dx — this is purely in x).
• The derivative of y^2 is 2y \cdot \frac{dy}{dx} (chain rule: outer derivative 2y, times inner derivative dy/dx).
• The derivative of xy is x\frac{dy}{dx} + y (product rule).
• The derivative of \sin y is \cos y \cdot \frac{dy}{dx} (chain rule).
• The derivative of a constant is 0.

Step 2. Collect all terms containing \frac{dy}{dx} on one side. Move everything else to the other side.

Step 3. Factor out \frac{dy}{dx} and divide. The result is \frac{dy}{dx} expressed in terms of both x and y.

That last point is important: the answer will typically involve both x and y. This is not a problem. When you want the slope at a specific point (x_0, y_0) on the curve, you plug in both coordinates. You need to know the point (both x and y) before you can compute the slope there — and that makes sense, because a curve like x^2 + y^2 = 25 has two different y-values at most x-values, and the slope is different at each.

The circle, done implicitly

Return to x^2 + y^2 = 25. Differentiate both sides with respect to x:

Solve for \frac{dy}{dx}:

At (3, 4): \frac{dy}{dx} = -\frac{3}{4}. Same answer as before, in two lines instead of six.

At (3, -4) — the point on the bottom half of the circle: \frac{dy}{dx} = -\frac{3}{-4} = \frac{3}{4}. The slope is positive there, which is correct — the bottom arc of the circle slopes upward as you move right past x = 3.

This is the power of the implicit approach: a single formula dy/dx = -x/y handles the entire circle — both halves, every point — without needing to split into y = \sqrt{25 - x^2} and y = -\sqrt{25 - x^2} and differentiate each separately.

Why the chain rule is the engine

The entire method rests on one idea: y depends on x, so when you differentiate any expression involving y, the chain rule fires.

Consider y^3. If y were just a constant, its derivative with respect to x would be zero. If y were x itself, the derivative would be 3x^2. But y is some function of x that you don't know — call it y(x). The chain rule gives:

The outer function u^3 has derivative 3u^2. The inner function is y(x), whose derivative is dy/dx. Multiply them: 3y^2 \cdot dy/dx.

This is why implicit differentiation is not a new rule — it is the chain rule applied in a setting where you cannot (or choose not to) write y explicitly in terms of x.

The folium of Descartes

Now tackle the curve from the opening: x^3 + y^3 = 6xy.

Step 1. Differentiate every term with respect to x:

The left side: \frac{d}{dx}[x^3] = 3x^2 and \frac{d}{dx}[y^3] = 3y^2 \frac{dy}{dx}.

The right side: \frac{d}{dx}[6xy] = 6\frac{d}{dx}[xy] = 6\left(1 \cdot y + x \cdot \frac{dy}{dx}\right) by the product rule.

Step 2. Expand and collect dy/dx terms:

Step 3. Factor and divide:

At the point (3, 3) on the folium: \frac{dy}{dx} = \frac{6 - 9}{9 - 6} = \frac{-3}{3} = -1. The tangent at (3, 3) has slope -1 — a line tilted at 45 degrees downward.

Notice that the formula \frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x} breaks down when y^2 - 2x = 0 — the denominator vanishes. At such points the tangent line is vertical (infinite slope). At the origin (0, 0), both numerator and denominator are zero, producing 0/0 — which reflects the fact that the curve crosses itself there, and two different tangent lines exist (one for each branch).

Finding the second derivative implicitly

You can differentiate again to find \frac{d^2y}{dx^2} — the second derivative — without ever solving for y explicitly.

Take the circle again: \frac{dy}{dx} = -\frac{x}{y}.

Differentiate both sides with respect to x:

Apply the quotient rule to x/y:

Substitute \frac{dy}{dx} = -\frac{x}{y}:

Since x^2 + y^2 = 25:

This tells you the concavity of the circle at every point. On the top half (y > 0), d^2y/dx^2 = -25/y^3 < 0 — the curve is concave down. On the bottom half (y < 0), d^2y/dx^2 = -25/y^3 > 0 — the curve is concave up. Both match the shape of a circle: the top curves downward, the bottom curves upward.

At the topmost point (0, 5): d^2y/dx^2 = -25/125 = -1/5. At (3, 4): d^2y/dx^2 = -25/64 \approx -0.39. The concavity is stronger (more negative) at (3, 4) than at (0, 5). This is because the circle is curving more sharply there — the top is the "flattest" part of the circle.

The key technique when finding the second derivative implicitly: substitute the first derivative back in wherever dy/dx appears, then simplify using the original equation.

A more involved example: second derivative of the ellipse

Take the ellipse \frac{x^2}{9} + \frac{y^2}{4} = 1. You will find dy/dx in the worked example below, and the result is \frac{dy}{dx} = -\frac{4x}{9y}.

Now differentiate again. Apply the quotient rule to -\frac{4x}{9y}:

Substitute \frac{dy}{dx} = -\frac{4x}{9y}:

From the original equation, \frac{x^2}{9} + \frac{y^2}{4} = 1 gives 4x^2 + 9y^2 = 36. Substitute:

On the top half of the ellipse (y > 0), d^2y/dx^2 < 0 — concave down. On the bottom half (y < 0), d^2y/dx^2 > 0 — concave up. The ellipse bends exactly as you would expect.

Computing one from start to finish

Common confusions

• "You can only use implicit differentiation when you can't solve for y." False — you can use it any time. For x^2 + y^2 = 25, you can solve for y and then differentiate, but implicit differentiation is faster. The technique is a choice, not a last resort.

• "The derivative dy/dx should be in terms of x only." Not when differentiating implicitly. The result dy/dx = -x/y involves both variables, and that is correct. To evaluate it at a point, you need both coordinates. This is unavoidable because the original equation defines a relation between x and y, not a single function — at a given x, there may be multiple values of y with different slopes.

• Forgetting the chain rule on y-terms. The derivative of y^3 with respect to x is 3y^2 \cdot dy/dx — not 3y^2. Every y-term picks up a dy/dx factor. Missing this factor is the single most common error.

• "The derivative of xy is just dy/dx." No — xy is a product of two functions of x: the function x and the function y(x). By the product rule: \frac{d}{dx}[xy] = 1 \cdot y + x \cdot \frac{dy}{dx} = y + x\frac{dy}{dx}.

• Confusing the second derivative computation. When finding d^2y/dx^2, you differentiate dy/dx with respect to x. Since dy/dx is itself a function of x and y, and y is a function of x, every y in the expression for dy/dx requires the chain rule again. You then substitute the known expression for dy/dx and simplify.

Going deeper

Why implicit differentiation works: the implicit function theorem

Implicit differentiation works because of a deep theorem in analysis called the implicit function theorem. Informally, it says: if F(x, y) = 0 defines a smooth curve, and the curve is not vertical at a point (meaning \frac{\partial F}{\partial y} \neq 0 there), then near that point y can be written as a function of x, and that function is differentiable.

You do not need to actually find the function y(x) — the theorem guarantees it exists, and that is enough to justify treating y as a differentiable function of x and applying the chain rule.

When \frac{\partial F}{\partial y} = 0 — that is, when y^2 - 2x = 0 in the folium example, or when y = 0 in the circle example — the implicit function theorem does not apply, and the tangent line is vertical (or the curve has a singularity). These are exactly the points where the formula for dy/dx has a zero denominator, confirming that the algebra and the theory agree.

For F(x, y) = x^2 + y^2 - 25, the formula from the theorem gives:

which is exactly what implicit differentiation produces. The technique is a shortcut for the general formula dy/dx = -F_x/F_y.

Application: tangent lines to conic sections

Every conic section — circle, ellipse, parabola, hyperbola — can be written as a second-degree equation in x and y, and implicit differentiation gives the tangent slope at any point in one quick computation.

For the general conic ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0, differentiating implicitly gives:

Solving:

This single formula handles the tangent to any conic at any point. For the circle x^2 + y^2 = 25: a = b = 1, h = 0, g = f = 0, c = -25, giving dy/dx = -x/y — exactly what you had before.

The equation of the tangent line at a point (x_1, y_1) on a conic has a beautiful form: replace x^2 with xx_1, y^2 with yy_1, xy with (xy_1 + yx_1)/2, x with (x + x_1)/2, and y with (y + y_1)/2. This "T = 0" substitution rule, often taught as a trick in JEE preparation, is actually a direct consequence of implicit differentiation applied to the general conic equation.

Deriving the derivative of \sin^{-1} x using implicit differentiation

Implicit differentiation gives the cleanest proofs of the derivatives of inverse trigonometric functions. Here is the derivative of \sin^{-1} x as an example.

Let y = \sin^{-1} x, so \sin y = x with -\pi/2 \le y \le \pi/2. Differentiate both sides with respect to x:

Now you need \cos y in terms of x. Since \sin y = x and \cos^2 y = 1 - \sin^2 y = 1 - x^2, and \cos y \ge 0 for y \in [-\pi/2, \pi/2]:

The derivative of \sin^{-1} x is \frac{1}{\sqrt{1 - x^2}}. No limit computation, no difference quotient — just implicit differentiation and a trigonometric identity. This is the standard method for all six inverse trig derivatives.

Related rates

Implicit differentiation extends naturally to problems where x and y both depend on a third variable — usually time t. If a balloon is being inflated and its radius r is growing at 2 cm/s, how fast is the volume V = \frac{4}{3}\pi r^3 increasing when r = 10 cm?

Differentiate implicitly with respect to t: \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}. At r = 10, dr/dt = 2: \frac{dV}{dt} = 4\pi(100)(2) = 800\pi \approx 2513 cm³/s.

This is exactly the same chain rule mechanism — V depends on r, r depends on t, so differentiating V with respect to t requires multiplying by dr/dt — just applied to a physical problem rather than a geometric curve.

Implicit differentiation and orthogonal trajectories

Two families of curves are orthogonal trajectories if every curve of one family intersects every curve of the other family at right angles. For example, the family of circles x^2 + y^2 = r^2 (centered at the origin) and the family of lines y = mx (through the origin) are orthogonal: at every intersection, the circle and the line meet at 90 degrees.

You can prove this with implicit differentiation. For the circle, dy/dx = -x/y. For the line y = mx, dy/dx = m = y/x. The product of the slopes is (-x/y)(y/x) = -1, which is the condition for perpendicularity. The two families are orthogonal at every point of intersection.

Finding orthogonal trajectories is a standard application in physics — in electrostatics, equipotential curves and electric field lines are orthogonal trajectories. The technique starts by finding dy/dx for the given family (usually by implicit differentiation), then replacing dy/dx with -dx/dy to get the differential equation of the orthogonal family.

Where this leads next

Implicit differentiation is a technique that appears everywhere derivatives are used. The most direct continuations:

• Tangent and Normal — using dy/dx (whether found explicitly or implicitly) to write the equations of tangent and normal lines.
• Logarithmic Differentiation — another technique that differentiates both sides of an equation, using \ln to simplify the right side before differentiating.
• Derivatives of Inverse Trigonometric Functions — many of these derivatives are most cleanly derived by implicit differentiation.
• Concavity and Points of Inflection — the second derivative, which you now know how to find implicitly, determines whether a curve bends up or down.
• Parabola: Tangent and Normal — a direct application of implicit differentiation to the standard parabola y^2 = 4ax.