How do I know whether a tangent is horizontal or vertical?

Write dy/dx as a single fraction N/D. A horizontal tangent occurs where the numerator N = 0 but the denominator D ≠ 0 (slope zero). A vertical tangent occurs where D = 0 but N ≠ 0 (slope undefined). Always verify the point actually lies on the original curve.

Why does my second derivative still have dy/dx in it?

That is expected after the first round of differentiation. You must substitute your first-derivative expression dy/dx = N/D back into the result so that d^2y/dx^2 is written only in terms of x and y. If a problem gives a specific point, you can plug in the numeric value of dy/dx instead of simplifying symbolically.

What happens when both the numerator and denominator of dy/dx are zero at a point?

That point is indeterminate — the simple horizontal/vertical rule does not apply. The curve may have a cusp, a self-intersection, or cross itself. On the AP exam these cases are rare; if you meet one, additional analysis (such as examining nearby behavior or the original equation) is needed, and the problem will usually guide you.

Do I really need to check that the denominator is nonzero for a horizontal tangent?

Yes. Setting the numerator to zero only shows the slope could be zero. If the denominator is also zero there, the slope is not actually zero — it is indeterminate. AP graders often require this nonzero-denominator statement as part of a complete justification, so include it.

AP-CALCBC-5.12

U5.12 Exploring Behaviors of Implicit Relations

Master AP Calculus BC 5.12: use implicit differentiation to find horizontal and vertical tangents and analyze concavity of implicitly defined curves.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U5.12 Exploring Behaviors of Implicit Relations, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

Not every curve can be solved for

y

in terms of

x

. Circles, ellipses, and tangled relations like

x^3 + y^3 = 6xy

define

y

implicitly, yet they still have tangent lines and bends you can analyze. In this lesson you will take the same first- and second-derivative tools you used for explicit functions and apply them to implicit relations.

The payoff: you can locate where a curve has a horizontal tangent (a possible high or low point), where it turns vertical (where

\frac{dy}{dx}

blows up), and where it is concave up or down. These are classic AP free-response and multiple-choice targets, so we will focus on the algebra that trips students up and the reasoning graders reward.

Setting Up the Implicit Derivative

An implicit relation ties

x

and

y

together in one equation, such as

x^2 + xy + y^2 = 7

. To differentiate, apply

\frac{d}{dx}

to every term, treating

y

as a function of

x

so that each

y

-derivative carries a factor of

\frac{dy}{dx}

(the chain rule). Products of

x

and

y

require the product rule.

For

x^2 + xy + y^2 = 7

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

Collect the

\frac{dy}{dx}

terms on one side and factor:

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

Notice the result usually depends on both

x

and

y

. That is the key feature of implicit differentiation and the source of most of the analysis in this topic. A very common error is forgetting the

\frac{dy}{dx}

factor on

y

-terms or mishandling the product rule on the

xy

term. Always write

\frac{dy}{dx}

explicitly rather than in your head, and factor cleanly so you can read off a numerator and a denominator — those two pieces drive the entire tangent analysis that follows.

Horizontal and Vertical Tangents

Once

\frac{dy}{dx}

is written as a fraction

\frac{N}{D}

, the tangent behavior comes straight from the numerator and denominator.

Feature	Condition	Meaning
Horizontal tangent	$N = 0$ and $D \ne 0$	slope is $0$
Vertical tangent	$D = 0$ and $N \ne 0$	slope undefined
Indeterminate point	$N = 0$ and $D = 0$	needs further analysis (cusp, node, or crossing)

Critically, setting

N=0

D=0

alone is not enough. Those equations describe conditions on

x

and

y

, but the point must also lie on the original curve. So you solve a system: the numerator (or denominator) equation together with the original relation.

For

\frac{dy}{dx} = \frac{-(2x+y)}{x+2y}

, horizontal tangents need

2x + y = 0

, i.e.

y = -2x

, substituted back into

x^2 + xy + y^2 = 7

. Vertical tangents need

x + 2y = 0

combined with the same relation. The AP exam frequently asks you to justify that a point gives a horizontal tangent by verifying both that the numerator is zero and the denominator is nonzero there. Do not skip that nonzero check — it is a common place points are lost.

The Second Derivative and Concavity

To find

\frac{d^2y}{dx^2}

, differentiate

\frac{dy}{dx}

again with respect to

x

, using the quotient rule and remembering that any remaining

y

differentiates to

\frac{dy}{dx}

. This produces an expression containing

\frac{dy}{dx}

, which you then replace using your first-derivative formula. The result is written purely in terms of

x

and

y

.

For example, from

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

on the circle

x^2 + y^2 = r^2

\frac{d^2y}{dx^2} = \frac{-\left(y - x\frac{dy}{dx}\right)}{y^2} = \frac{-\left(y - x\cdot\frac{-x}{y}\right)}{y^2} = \frac{-(y^2 + x^2)}{y^3} = \frac{-r^2}{y^3}.

Concavity follows the usual rule: the curve is concave up where

\frac{d^2y}{dx^2} > 0

and concave down where it is negative. On the upper half of the circle (

y>0

) the expression is negative, so the curve is concave down — exactly what a circle looks like on top. A frequent misconception is trying to determine concavity from the sign of

\frac{dy}{dx}

; concavity comes only from the second derivative. Always substitute the first-derivative formula before evaluating, or you will leave a stray

\frac{dy}{dx}

in your answer.

How the Exam Tests This

On multiple-choice questions you are often given a point and asked for the slope, the equation of the tangent line, or the value of

\frac{d^2y}{dx^2}

at that point. These reward clean algebra and correct chain-rule bookkeeping. On free-response, you typically differentiate implicitly, then are asked to (a) find all points with a horizontal tangent, (b) show a tangent is vertical, or (c) determine concavity or classify a point.

Graders look for specific evidence. For a horizontal tangent, show the numerator equals zero and confirm the denominator does not. For concavity at a point, produce a numerical value of

\frac{d^2y}{dx^2}

and state its sign with a conclusion ("since

\frac{d^2y}{dx^2} = -\tfrac{1}{4} < 0

, the curve is concave down there").

Time-savers: keep

\frac{dy}{dx}

as a single fraction, and when a problem gives you a specific point, plug in numbers early once you have the first derivative — you rarely need the fully simplified symbolic second derivative. Watch units of justification: a bare answer without the sign statement or the nonzero-denominator check often costs a point even when the arithmetic is perfect.

Key terms

Implicit relation.: An equation relating $x$ and $y$ that is not solved for $y$ , such as $x^2 + y^2 = 25$ , yet still defines one or more curves.
Implicit differentiation.: Differentiating both sides of a relation with respect to $x$ , treating $y$ as a function of $x$ so every $y$ -term gains a factor of $\frac{dy}{dx}$ .
Horizontal tangent.: A point where $\frac{dy}{dx} = 0$ , which for a fraction $\frac{N}{D}$ means the numerator is zero while the denominator is nonzero.
Vertical tangent.: A point where $\frac{dy}{dx}$ is undefined because the denominator is zero while the numerator is nonzero; the tangent line is vertical.
Second derivative.: $\frac{d^2y}{dx^2}$ , found by differentiating $\frac{dy}{dx}$ again and substituting the first-derivative expression to remove any remaining $\frac{dy}{dx}$ .
Concavity.: The direction a curve bends: concave up where $\frac{d^2y}{dx^2} > 0$ and concave down where $\frac{d^2y}{dx^2} < 0$ .

Worked example

Consider the curve

x^2 - xy + y^2 = 3

. Find all points where the curve has a horizontal tangent, and determine whether the curve is concave up or down at the point

(1,2)

Differentiate implicitly. Using the product rule on

xy

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

Group the

\frac{dy}{dx}

terms:

\frac{dy}{dx}(2y - x) = y - 2x \;\Rightarrow\; \frac{dy}{dx} = \frac{y - 2x}{2y - x}.

Horizontal tangents require the numerator

y - 2x = 0

, so

y = 2x

, with denominator

2y - x \ne 0

. Substitute

y = 2x

into the original curve:

x^2 - x(2x) + (2x)^2 = x^2 - 2x^2 + 4x^2 = 3x^2 = 3 \;\Rightarrow\; x = \pm 1.

This gives the points

(1, 2)

and

(-1, -2)

. Check the denominator at

(1,2)

2(2) - 1 = 3 \ne 0

, and at

(-1,-2)

-4 + 1 = -3 \ne 0

. Both are genuine horizontal tangents.

Now concavity at

(1,2)

. First confirm

\frac{dy}{dx} = 0

there. Differentiate

\frac{dy}{dx} = \frac{y - 2x}{2y - x}

using the quotient rule:

\frac{d^2y}{dx^2} = \frac{\left(\frac{dy}{dx} - 2\right)(2y - x) - (y - 2x)\left(2\frac{dy}{dx} - 1\right)}{(2y - x)^2}.

(1,2)

\frac{dy}{dx} = 0

and

y - 2x = 0

, so the second term vanishes:

\frac{d^2y}{dx^2} = \frac{(0 - 2)(2\cdot 2 - 1)}{(2\cdot 2 - 1)^2} = \frac{(-2)(3)}{9} = -\frac{2}{3}.

Since

\frac{d^2y}{dx^2} = -\frac{2}{3} < 0

, the curve is concave down at

(1,2)

, consistent with this point being a local maximum on that branch.

Practice questions

For the curve

x^2 + y^2 = 25

, at which point does the curve have a vertical tangent line?

$(0, 5)$
$(5, 0)$
$(3, 4)$
$(0, -5)$

Answer: $(5, 0)$

Implicit differentiation gives

2x + 2y\frac{dy}{dx} = 0

, so

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

. A vertical tangent occurs where the denominator is zero and the numerator is nonzero:

y = 0

with

x \ne 0

. On the circle,

y = 0

gives

x = \pm 5

. Among the choices,

(5,0)

satisfies this. The points

(0,5)

and

(0,-5)

give

\frac{dy}{dx}=0

(horizontal tangents), and at

(3,4)

the slope is

-\frac{3}{4}

The curve defined by

y^2 = x^3 - 3x + 3

passes through the point

(1, 1)

. Find

\frac{dy}{dx}

at this point, and determine whether the curve is concave up or concave down there. Justify your answer.

Answer: $\frac{dy}{dx} = 0$ at $(1,1)$ , and the curve is concave up there.

Differentiate:

2y\frac{dy}{dx} = 3x^2 - 3

, so

\frac{dy}{dx} = \frac{3x^2 - 3}{2y}

. At

(1,1)

\frac{3 - 3}{2} = 0

, a horizontal tangent. Differentiate again with the quotient rule:

\frac{d^2y}{dx^2} = \frac{6x(2y) - (3x^2-3)(2\frac{dy}{dx})}{4y^2}

. At

(1,1)

the second term vanishes because

3x^2 - 3 = 0

, leaving

\frac{d^2y}{dx^2} = \frac{6(1)(2)}{4(1)} = \frac{12}{4} = 3 > 0

, meaning concave up. Because the first derivative is zero and the second derivative is positive,

(1,1)

is a local minimum on that branch and the curve is concave up there.

For the relation

xy + y^2 = 6

, find all

x

-values (with corresponding points) where the curve has a horizontal tangent line.

Answer: There are no points with a horizontal tangent.

Differentiating:

y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0

, so

\frac{dy}{dx}(x + 2y) = -y

, giving

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

. A horizontal tangent needs the numerator

-y = 0

, i.e.

y = 0

. But substituting

y = 0

into

xy + y^2 = 6

gives

0 = 6

, which is impossible. Since no point on the curve has

y = 0

, there are no horizontal tangents. This illustrates why you must always check the numerator condition against the original equation.

FAQ

How do I know whether a tangent is horizontal or vertical?: Write $\frac{dy}{dx}$ as a single fraction $\frac{N}{D}$ . A horizontal tangent occurs where the numerator $N = 0$ but the denominator $D \ne 0$ (slope zero). A vertical tangent occurs where $D = 0$ but $N \ne 0$ (slope undefined). Always verify the point actually lies on the original curve.
Why does my second derivative still have dy/dx in it?: That is expected after the first round of differentiation. You must substitute your first-derivative expression $\frac{dy}{dx} = \frac{N}{D}$ back into the result so that $\frac{d^2y}{dx^2}$ is written only in terms of $x$ and $y$ . If a problem gives a specific point, you can plug in the numeric value of $\frac{dy}{dx}$ instead of simplifying symbolically.
What happens when both the numerator and denominator of dy/dx are zero at a point?: That point is indeterminate — the simple horizontal/vertical rule does not apply. The curve may have a cusp, a self-intersection, or cross itself. On the AP exam these cases are rare; if you meet one, additional analysis (such as examining nearby behavior or the original equation) is needed, and the problem will usually guide you.
Do I really need to check that the denominator is nonzero for a horizontal tangent?: Yes. Setting the numerator to zero only shows the slope could be zero. If the denominator is also zero there, the slope is not actually zero — it is indeterminate. AP graders often require this nonzero-denominator statement as part of a complete justification, so include it.

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