Analyzing Implicit Relations

Not every curve is the graph of a function. Circles, tilted ellipses, and loops are described by equations in $x$ and $y$ together, like $x^2 + xy + y^2 = 12$. With implicit differentiation you can still find $\dfrac{dy}{dx}$, and with it you can analyze these curves the same way as functions: where the tangent is flat, where it’s vertical, where the curve is highest or lowest, and which way it bends.

Key ideas

Horizontal and vertical tangents

Implicit differentiation usually gives $\dfrac{dy}{dx}$ as a fraction involving both $x$ and $y$:

$$\frac{dy}{dx} = \frac{N(x, y)}{D(x, y)}$$

Tangent line	Condition	Plus…
Horizontal	numerator $N = 0$ and denominator $D \ne 0$	the point is on the curve
Vertical	denominator $D = 0$ and numerator $N \ne 0$	the point is on the curve
Can’t tell yet	$N = 0$ and $D = 0$	needs more work

The “on the curve” step is the one people forget. Setting $N = 0$ gives a relationship between $x$ and $y$ (like $y = -2x$), not a point. Substitute it into the original equation to find the actual points, and keep only real solutions.

Highest and lowest points

At a point with a horizontal tangent, the curve might have a local high point, a local low point, or neither, just like a critical point of a function. To decide, find $\dfrac{d^2y}{dx^2}$ by differentiating implicitly again, then evaluate it at the point:

• $\dfrac{d^2y}{dx^2} \gt 0$: concave up, so a local minimum of $y$ (a low point of the curve).
• $\dfrac{d^2y}{dx^2} \lt 0$: concave down, so a local maximum of $y$ (a high point).

This is the second derivative test applied to the part of the curve near the point.

Finding the second derivative

Two ways, and both work:

• Differentiate the expression for $\dfrac{dy}{dx}$ with the quotient rule, then substitute $\dfrac{dy}{dx}$ back in.
• Differentiate the equation you had after the first differentiation again, then solve for $\dfrac{d^2y}{dx^2}$.

A big time-saver: at a horizontal tangent, $\dfrac{dy}{dx} = 0$. Substitute $0$ for every $\dfrac{dy}{dx}$ before simplifying, and most terms disappear.

Worked examples

Example 1: Horizontal and vertical tangents on a tilted ellipse

Find all points on $x^2 + xy + y^2 = 12$ where the tangent line is horizontal or vertical.

Solution. Differentiate both sides with respect to $x$ (using the product rule on $xy$):

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

Horizontal: numerator $0$ means $y = -2x$. Substitute into the curve:

$$x^2 + x(-2x) + (-2x)^2 = 12 \quad\Rightarrow\quad 3x^2 = 12 \quad\Rightarrow\quad x = \pm 2$$

Points: $(2, -4)$ and $(-2, 4)$. Check the denominators: $2 + 2(-4) = -6 \ne 0$ and $-2 + 2(4) = 6 \ne 0$. ✓

Vertical: denominator $0$ means $x = -2y$. Substitute:

$$(-2y)^2 + (-2y)y + y^2 = 12 \quad\Rightarrow\quad 3y^2 = 12 \quad\Rightarrow\quad y = \pm 2$$

Points: $(-4, 2)$ and $(4, -2)$. Check the numerators: $2(-4) + 2 = -6 \ne 0$ and $2(4) - 2 = 6 \ne 0$. ✓

The horizontal tangents (orange) are at the highest and lowest points; the vertical tangents (green) are at the leftmost and rightmost points.

Example 2: Is it a high point or a low point?

For the same curve, use $\dfrac{d^2y}{dx^2}$ to decide whether $(2, -4)$ and $(-2, 4)$ are high or low points.

Solution. Differentiate the first-derivative equation $2x + y + x\dfrac{dy}{dx} + 2y\dfrac{dy}{dx} = 0$ again. Write $y'$ for $\dfrac{dy}{dx}$ and $y''$ for $\dfrac{d^2y}{dx^2}$ to keep it short:

$$2 + y' + (y' + xy'') + \big(2y' \cdot y' + 2y\,y''\big) = 0$$

At a horizontal tangent, $y' = 0$, so almost everything vanishes:

$$2 + (x + 2y)\,y'' = 0 \quad\Rightarrow\quad y'' = -\frac{2}{x + 2y}$$

• At $(2, -4)$: $y'' = -\dfrac{2}{2 - 8} = \dfrac{1}{3} \gt 0$. Concave up, so $(2, -4)$ is a low point.
• At $(-2, 4)$: $y'' = -\dfrac{2}{-2 + 8} = -\dfrac{1}{3} \lt 0$. Concave down, so $(-2, 4)$ is a high point.

That matches the figure.

Example 3: Checking that the points are on the curve

Find the horizontal and vertical tangents of $y^2 = x^3 - 4x$, and classify the points with horizontal tangents.

Solution.

$$2y\frac{dy}{dx} = 3x^2 - 4 \quad\Rightarrow\quad \frac{dy}{dx} = \frac{3x^2 - 4}{2y}$$

Horizontal: $3x^2 - 4 = 0$ gives $x = \pm\dfrac{2}{\sqrt{3}}$. Now check the curve:

• $x = \dfrac{2}{\sqrt{3}}$: $y^2 = \dfrac{8}{3\sqrt{3}} - \dfrac{8}{\sqrt{3}} = -\dfrac{16}{3\sqrt{3}} \lt 0$. No real $y$, so no points here.
• $x = -\dfrac{2}{\sqrt{3}}$: $y^2 = \dfrac{16}{3\sqrt{3}} \approx 3.079$, so $y \approx \pm 1.755$.

Horizontal tangents at about $(-1.155, 1.755)$ and $(-1.155, -1.755)$.

Vertical: $2y = 0$ gives $y = 0$, so $x^3 - 4x = x(x - 2)(x + 2) = 0$: the points $(0, 0)$, $(2, 0)$, $(-2, 0)$. The numerator $3x^2 - 4$ is $-4$, $8$, $8$ there, never $0$. ✓

Classify: differentiate $2y\,y' = 3x^2 - 4$ again: $2(y')^2 + 2y\,y'' = 6x$. With $y' = 0$, $y'' = \dfrac{3x}{y}$.

• At $(-1.155, 1.755)$: $y'' \approx \dfrac{-3.464}{1.755} \lt 0$, so a high point.
• At $(-1.155, -1.755)$: $y'' \gt 0$, so a low point.

$y^2 = x^3 - 4x$: the horizontal tangents are only on the loop. The value $x = \frac{2}{\sqrt{3}}$ gave no points because it lies in a gap of the curve.

Example 4: Concavity of a circle

For the circle $x^2 + y^2 = 25$, show that $\dfrac{d^2y}{dx^2} = -\dfrac{25}{y^3}$, and explain what it says about the top and bottom halves.

Solution. First derivative: $2x + 2y\,y' = 0$, so $y' = -\dfrac{x}{y}$.

Differentiate with the quotient rule, then substitute $y' = -\dfrac{x}{y}$:

$$\begin{aligned} y'' &= -\frac{y \cdot 1 - x \cdot y'}{y^2} \\ &= -\frac{y + \frac{x^2}{y}}{y^2} \\ &= -\frac{y^2 + x^2}{y^3} && \text{multiply top and bottom by } y \\ &= -\frac{25}{y^3} && \text{since } x^2 + y^2 = 25 \end{aligned}$$

On the top half, $y \gt 0$, so $y'' \lt 0$: concave down. On the bottom half, $y \lt 0$, so $y'' \gt 0$: concave up. That’s exactly what a circle looks like.

Common mistakes

Stopping at a relationship instead of points. “Horizontal tangents where $y = -2x$” isn’t an answer. Substitute into the original equation to find the actual points.

Not checking the points are real and on the curve. In Example 3, $x = \frac{2}{\sqrt{3}}$ makes the numerator $0$, but there’s no point on the curve with that $x$. Always substitute back.

Ignoring the other part of the fraction. For a horizontal tangent, the denominator must be non-zero at the point. If both are $0$, you can’t conclude anything from $\frac{dy}{dx}$ alone.

Forgetting the chain rule on y terms. $\dfrac{d}{dx}\left(y^2\right) = 2y\dfrac{dy}{dx}$, not $2y$. When finding the second derivative, $\dfrac{d}{dx}\left(y\,y'\right) = (y')^2 + y\,y''$: product rule and chain rule.

Doing the second derivative the long way when you don’t need to. If you only need $\dfrac{d^2y}{dx^2}$ at a point where $\dfrac{dy}{dx} = 0$, set $y' = 0$ right away. The algebra gets much shorter.

Practice

1. (Warm-up) A curve has $\dfrac{dy}{dx} = \dfrac{x - 2}{y + 1}$. On what line do the points with a horizontal tangent lie? A vertical tangent?

Solution

Horizontal: numerator $0$, so on the line $x = 2$ (where $y \ne -1$).

Vertical: denominator $0$, so on the line $y = -1$ (where $x \ne 2$).

To get actual points, you’d substitute into the curve’s equation.

2. (Warm-up) Find the points on the circle $x^2 + y^2 = 25$ with horizontal and vertical tangents.

Solution

$\dfrac{dy}{dx} = -\dfrac{x}{y}$.

Horizontal: $x = 0$, so $y^2 = 25$: $(0, 5)$ and $(0, -5)$.

Vertical: $y = 0$, so $x^2 = 25$: $(5, 0)$ and $(-5, 0)$.

3. (Core) Find the points on $x^2 + y^2 - 6x + 4y = 12$ with horizontal and vertical tangents.

Solution

$$2x + 2y\frac{dy}{dx} - 6 + 4\frac{dy}{dx} = 0 \quad\Rightarrow\quad \frac{dy}{dx} = \frac{6 - 2x}{2y + 4} = \frac{3 - x}{y + 2}$$

Horizontal: $x = 3$. Then $9 + y^2 - 18 + 4y = 12$, so $y^2 + 4y - 21 = 0$, $(y + 7)(y - 3) = 0$. Points $(3, 3)$ and $(3, -7)$. (Denominators $5$ and $-5$, not $0$.)

Vertical: $y = -2$. Then $x^2 + 4 - 6x - 8 = 12$, so $x^2 - 6x - 16 = 0$, $(x - 8)(x + 2) = 0$. Points $(8, -2)$ and $(-2, -2)$. (Numerators $-5$ and $5$, not $0$.)

(This is a circle with centre $(3, -2)$ and radius $5$, so the answers make sense.)

4. (Core) Find the points on $x^2 - xy + y^2 = 3$ with horizontal and vertical tangents.

Solution

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

Horizontal: $y = 2x$. Then $x^2 - 2x^2 + 4x^2 = 3x^2 = 3$, so $x = \pm 1$. Points $(1, 2)$ and $(-1, -2)$. (Denominators $3$ and $-3$.)

Vertical: $x = 2y$. Then $4y^2 - 2y^2 + y^2 = 3y^2 = 3$, so $y = \pm 1$. Points $(2, 1)$ and $(-2, -1)$. (Numerators $-3$ and $3$.)

5. (Core) For the curve in question 4, find $\dfrac{d^2y}{dx^2}$ at $(1, 2)$. Is $(1, 2)$ a high point or a low point of the curve?

Solution

Differentiate $2x - y - x\,y' + 2y\,y' = 0$ again:

$$2 - y' - (y' + x\,y'') + \big(2(y')^2 + 2y\,y''\big) = 0$$

At $(1, 2)$, $y' = 0$: $2 - x\,y'' + 2y\,y'' = 0$, so $2 + (2y - x)\,y'' = 0$.

$$y'' = -\frac{2}{2(2) - 1} = -\frac{2}{3} \lt 0$$

Concave down, so $(1, 2)$ is a high point.

6. (Core) Show that the curve $y^3 - xy = 2$ has no horizontal tangents, and find the point where it has a vertical tangent.

Solution

$$3y^2\frac{dy}{dx} - \left(y + x\frac{dy}{dx}\right) = 0 \quad\Rightarrow\quad \frac{dy}{dx} = \frac{y}{3y^2 - x}$$

Horizontal would need $y = 0$, but then the equation gives $0 - 0 = 2$, which is false. So no point on the curve has $y = 0$, and there are no horizontal tangents.

Vertical: $x = 3y^2$. Substitute: $y^3 - 3y^3 = -2y^3 = 2$, so $y = -1$ and $x = 3$. The numerator is $-1 \ne 0$. Vertical tangent at $(3, -1)$.

7. (Core) Find all points on $x^2 - xy + y^2 = 3$ where the tangent line is parallel to the line $y = x$.

Solution

From question 4, $\dfrac{dy}{dx} = \dfrac{y - 2x}{2y - x}$. Set it equal to $1$:

$$y - 2x = 2y - x \quad\Rightarrow\quad y = -x$$

Substitute: $x^2 + x^2 + x^2 = 3x^2 = 3$, so $x = \pm 1$. Points $(1, -1)$ and $(-1, 1)$.

Check $(1, -1)$: $\dfrac{-1 - 2}{-2 - 1} = 1$. ✓ Check $(-1, 1)$: $\dfrac{1 + 2}{2 + 1} = 1$. ✓

8. (Challenge) Consider the curve $y^3 + 3y = x^3 - 3x + 6$.

• (a) Find $\dfrac{dy}{dx}$ and explain why the curve has no vertical tangents.
• (b) Find the point with a horizontal tangent where $x = 1$, and decide whether it’s a high or low point.
• (c) There’s also a horizontal tangent where $x = -1$. Use a calculator to find $y$ to three decimal places, and classify the point.

Solution

(a) $3y^2\,y' + 3y' = 3x^2 - 3$, so

$$\frac{dy}{dx} = \frac{x^2 - 1}{y^2 + 1}$$

The denominator $y^2 + 1 \ge 1$ is never $0$, so there are no vertical tangents.

(b) $y' = 0$ when $x = \pm 1$. At $x = 1$: $y^3 + 3y = 1 - 3 + 6 = 4$. $y = 1$ works, and it’s the only real solution because $y^3 + 3y$ is increasing. So the point is $(1, 1)$.

Differentiate $(y^2 + 1)\,y' = x^2 - 1$ again: $2y(y')^2 + (y^2 + 1)\,y'' = 2x$. With $y' = 0$:

$$y'' = \frac{2x}{y^2 + 1}$$

At $(1, 1)$: $y'' = \dfrac{2}{2} = 1 \gt 0$, so it’s a low point.

(c) At $x = -1$: $y^3 + 3y = -1 + 3 + 6 = 8$. Solving $y^3 + 3y - 8 = 0$ on a calculator gives $y \approx 1.513$. Then $y'' = \dfrac{-2}{y^2 + 1} \lt 0$, so $(-1, 1.513)$ is a high point.

9. (Challenge) The curve $x^3 + y^3 = 6xy$ is called a folium. Find $\dfrac{dy}{dx}$, find the point (other than the origin) with a horizontal tangent, and explain why the origin needs more work.

Solution

$$3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx} \quad\Rightarrow\quad \frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}$$

Horizontal: $2y = x^2$, so $y = \dfrac{x^2}{2}$. Substitute:

$$x^3 + \frac{x^6}{8} = 6x \cdot \frac{x^2}{2} = 3x^3 \quad\Rightarrow\quad \frac{x^6}{8} = 2x^3 \quad\Rightarrow\quad x^3(x^3 - 16) = 0$$

So $x = 0$ or $x = \sqrt[3]{16} = 2\sqrt[3]{2}$. For the second, $y = \dfrac{x^2}{2} = \dfrac{\sqrt[3]{256}}{2} = 2\sqrt[3]{4}$.

Check the denominator there: $y^2 - 2x = 4\sqrt[3]{16} - 4\sqrt[3]{2} = 8\sqrt[3]{2} - 4\sqrt[3]{2} = 4\sqrt[3]{2} \ne 0$. ✓ So there’s a horizontal tangent at $\left(2\sqrt[3]{2},\ 2\sqrt[3]{4}\right) \approx (2.520, 3.175)$.

At the origin, both the numerator and the denominator are $0$, so $\dfrac{dy}{dx}$ has the form $\frac{0}{0}$ and tells you nothing. (The curve actually crosses itself there, with two different tangent lines.)