Implicit Differentiation

The curve $x^2y + y^3 = 10$ passes through the point $(1, 2)$, since $1\cdot 2 + 2^3 = 10$. To find its slope there the way you have so far, you would first solve for $y$, and that means solving a cubic equation in $y$ for every $x$. Implicit differentiation skips that step. You differentiate the equation as it stands, treating $y$ as an unknown function of $x$, and the chain rule from topic 3.1 tells you what to do with every $y$ you meet.

One equation, several functions

A function is defined explicitly when $y$ is written alone on one side: $y = \sqrt{100 - x^2}$. It is defined implicitly when an equation ties $x$ and $y$ together without isolating $y$: $x^2 + y^2 = 100$.

An implicit equation can describe a curve that is not the graph of any single function. The circle $x^2 + y^2 = 100$ fails the vertical line test, since the line $x = 6$ meets it at $(6, 8)$ and $(6, -8)$. But near any one point other than $(10, 0)$ and $(-10, 0)$, where the tangent is vertical, a small piece of the circle is the graph of a function. The top half is $y = \sqrt{100 - x^2}$ and the bottom half is $y = -\sqrt{100 - x^2}$, and both satisfy the equation. When you differentiate implicitly, you are differentiating whichever of these functions passes through the point you care about, and the coordinates you plug in pick it out for you.

That is why implicit derivatives usually contain both $x$ and $y$. At $x = 6$ the circle has two different slopes, one at $(6, 8)$ and one at $(6, -8)$, so a formula that depended on $x$ alone could not give both.

Why every $y$ term picks up $\frac{dy}{dx}$

Since $y$ stands for a function of $x$, any expression in $y$ is a composition: the inside is $y(x)$ and the outside is whatever is done to it. So $y^4$ is $[y(x)]^4$, and by the chain rule

This is the row $[g(x)]^n$ from topic 3.1's table, with $g = y$. Compare $\frac{d}{dx}\left[x^4\right] = 4x^3$: there the inside is $x$ itself, whose derivative is 1, so no extra factor appears. The factor $\frac{dy}{dx}$ is the inside derivative, and you cannot write it as a number or a formula yet because you don't know $y(x)$. You carry it as an unknown and solve for it at the end.

The last three rows need more than the chain rule. A term such as $xy$ is a product of two functions of $x$, so it needs the product rule, and the factor $y$ inside it still brings its own $\frac{dy}{dx}$. In $e^{xy}$, the inside is the product $xy$, so the inside derivative is the whole product-rule result.

Try it on the circle. Differentiating both sides of $x^2 + y^2 = 100$ gives

At $(6, 8)$ the slope is $-\frac{6}{8} = -\frac{3}{4}$, and at $(6, -8)$ it is $\frac{3}{4}$, one formula giving both. You can confirm the first one explicitly: on the top half, $y = (100 - x^2)^{1/2}$, so $y' = \frac{-x}{\sqrt{100 - x^2}}$, which at $x = 6$ is $-\frac{6}{8}$. There is also a geometric check. The radius to $(6, 8)$ has slope $\frac{8}{6}$, and a tangent to a circle is perpendicular to the radius, so its slope is the negative reciprocal, $-\frac{6}{8}$.

Solving for $\frac{dy}{dx}$

After you differentiate, $\frac{dy}{dx}$ appears in one or more terms, and the equation is linear in it: it is never squared or inside a function. So you can always isolate it with algebra.

Back to the opening curve, $x^2y + y^3 = 10$. The first term is a product, so

Collecting and factoring,

At $(1, 2)$, the slope is $\frac{-4}{1 + 12} = -\frac{4}{13}$.

When you only need the slope at one point, you can substitute the coordinates right after step 1 and then solve. At $(1, 2)$ the differentiated equation becomes $4 + \frac{dy}{dx} + 12\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{4}{13}$ with almost no algebra. Substitute only after differentiating; the last trap below shows what goes wrong otherwise.

Reading horizontal and vertical tangents from the formula

Write the implicit derivative as a fraction, $\frac{dy}{dx} = \frac{N}{D}$, where $N$ and $D$ are expressions in $x$ and $y$.

• The tangent is horizontal at a point of the curve where $N = 0$ and $D \neq 0$, since the slope is 0 there.
• The tangent is vertical at a point of the curve where $D = 0$ and $N \neq 0$. The slope is undefined there, and the curve, viewed with $x$ as a function of $y$, has $\frac{dx}{dy} = \frac{D}{N} = 0$.
• Where $N$ and $D$ are both 0, the formula gives $\frac{0}{0}$ and decides nothing. The curve might cross itself there or have a cusp, and you would need another method.

Setting $N = 0$ gives one equation in two unknowns, so it describes many points, almost all of them off the curve. You find the actual points by solving it together with the original equation. On the circle, $N = -x$, so a horizontal tangent needs $x = 0$; the circle then gives $y^2 = 100$, so the horizontal tangents are at $(0, 10)$ and $(0, -10)$, where $D = y \neq 0$.

Differentiating an implicit equation a second time gives $\frac{d^2y}{dx^2}$ as a relation in $x$, $y$ and $\frac{dy}{dx}$. Topic 3.6 shows the computation, and topic 5.12 uses it to find and classify critical points of implicit curves.

Worked examples

Worked exampleExample 3: an exam-style curve question (no calculator)

Consider the curve given by $x^2 - 2xy + 3y^2 = 12$.

(a) Show that $\dfrac{dy}{dx} = \dfrac{y - x}{3y - x}$.

(b) Find all points on the curve at which the tangent line is horizontal.

(c) Find all points on the curve at which the tangent line is vertical.

(d) There are two points on the curve with $x$-coordinate 4. Find them, and find the slope of the curve at each.

Show solution

(a) Differentiate both sides with respect to $x$. The middle term is a product, so it needs the product rule:

$$2x - 2\left(y + x\frac{dy}{dx}\right) + 6y\frac{dy}{dx} = 0.$$

Collect the $\frac{dy}{dx}$ terms on the left and the rest on the right:

$$(6y - 2x)\frac{dy}{dx} = 2y - 2x, \qquad \frac{dy}{dx} = \frac{2(y - x)}{2(3y - x)} = \frac{y - x}{3y - x}.$$

A "show that" part is earned by the differentiation line and the algebra that reaches the given expression, so both belong on the page.

(b) A horizontal tangent needs the numerator to be 0, so $y = x$. Points with $y = x$ must also lie on the curve, so substitute into the original equation:

$$x^2 - 2x^2 + 3x^2 = 2x^2 = 12, \qquad x = \pm\sqrt{6}.$$

The points are $\left(\sqrt{6}, \sqrt{6}\right)$ and $\left(-\sqrt{6}, -\sqrt{6}\right)$. At each, the denominator is $3y - x = \pm 2\sqrt{6} \neq 0$, so the slope really is 0.

(c) A vertical tangent needs the denominator to be 0, so $x = 3y$. Substitute into the curve:

$$9y^2 - 6y^2 + 3y^2 = 6y^2 = 12, \qquad y = \pm\sqrt{2}.$$

The points are $\left(3\sqrt{2}, \sqrt{2}\right)$ and $\left(-3\sqrt{2}, -\sqrt{2}\right)$. At each, the numerator is $y - x = \mp 2\sqrt{2} \neq 0$, so these are vertical tangents, not $\frac{0}{0}$ points.

(d) Put $x = 4$ into the curve: $16 - 8y + 3y^2 = 12$, so $3y^2 - 8y + 4 = 0$, which factors as $(3y - 2)(y - 2) = 0$. The points are $(4, 2)$ and $\left(4, \frac{2}{3}\right)$.

$$\left.\frac{dy}{dx}\right|_{(4,\,2)} = \frac{2 - 4}{6 - 4} = -1, \qquad \left.\frac{dy}{dx}\right|_{(4,\,2/3)} = \frac{\frac{2}{3} - 4}{2 - 4} = \frac{5}{3}.$$

One $x$-value, two slopes: the curve is the tilted ellipse below, and the line $x = 4$ crosses it twice, once on the upper arc and once on the lower.

Where students lose points

On the exam

A free-response question on an implicit curve may open with "show that $\frac{dy}{dx} = \ldots$". The given expression lets you do the later parts even if your own differentiation goes wrong, so use it. The later parts can ask for a tangent line at a point, points with horizontal or vertical tangents, or the second derivative (topics 3.6 and 5.12). An answer may be left in terms of $x$ and $y$ unless a point is given. A multiple-choice question can ask for the slope at a given point, where substituting the point right after differentiating is the fastest route, or for the points where the tangent is horizontal or vertical, which means solving the slope condition together with the curve's equation.