Lesson 13: Implicit Differentiation

(a) Opening Question

You already know how to differentiate y = √(25 − x²). It's the chain rule:

y = (25 − x²)^(1/2)
y' = ½(25 − x²)^(−1/2)·(−2x) = −x/√(25 − x²)

But that equation is just the top half of the circle x² + y² = 25. The full circle isn't a function — for most x there are two y-values — so you can't solve it cleanly for a single y and differentiate the usual way.

Question: At the point (3, 4) on the circle x² + y² = 25, what is the slope of the tangent line? Try it two ways. First, use the top-half formula above and plug in x = 3. Then think: is there a way to get the slope directly from x² + y² = 25 without ever solving for y?

Hold onto your answer. The slope is a clean number, and by the end of section (b) you'll have a method that handles the entire circle — and curves far messier than a circle — in three lines.

(b) Core Concepts

The one new idea: y is secretly a function of x

Every technique in this lesson rests on a single shift in perspective. When an equation relates x and y — like x² + y² = 25 or x³ + y³ = 6xy — we treat y as an implicit function of x. We don't know a formula for it, but we assume one exists (locally), and we differentiate accordingly.

The consequence is the chain rule. Differentiating x² with respect to x gives 2x. But y is a function of x, so differentiating y² with respect to x requires the chain rule:

d/dx[y²] = 2y · dy/dx

The general pattern, for any power n:

d/dx[yⁿ] = n yⁿ⁻¹ · dy/dx

That extra factor of dy/dx is the entire game. Every time you differentiate a term containing y, a dy/dx tags along. Every time you differentiate a pure-x term, it doesn't. The simplest case is just d/dx[y] = dy/dx.

Implicit differentiation is the procedure: differentiate both sides of the equation with respect to x, attach a dy/dx to every y-term via the chain rule, then solve the resulting equation algebraically for dy/dx.

Worked derivation 1: the circle

Differentiate both sides of x² + y² = 25 with respect to x:

d/dx[x²] + d/dx[y²] = d/dx[25]
2x + 2y·(dy/dx) = 0

Now solve for dy/dx — it's just algebra:

2y·(dy/dx) = −2x
dy/dx = −x/y

That's the slope formula for the whole circle. At (3, 4):

dy/dx = −3/4

which matches your top-half answer from the Opening Question. But notice the formula −x/y also works at (3, −4) on the bottom half: there the slope is −3/(−4) = 3/4. One formula, both halves. The price of implicit differentiation is that the slope depends on both coordinates — you must supply a full point (x, y), not just an x.

Worked derivation 2: a curve with a product

The folium of Descartes is x³ + y³ = 6xy. The term 6xy is a product of x (a variable) and y (a function of x), so it needs the product rule — a classic trap. Differentiate term by term:

d/dx[x³] + d/dx[y³] = d/dx[6xy]
3x² + 3y²·(dy/dx) = 6·(x·dy/dx + y·1)
3x² + 3y²·(dy/dx) = 6x·(dy/dx) + 6y

Now collect every dy/dx on one side and everything else on the other:

3y²·(dy/dx) − 6x·(dy/dx) = 6y − 3x²
(3y² − 6x)·(dy/dx) = 6y − 3x²
dy/dx = (6y − 3x²)/(3y² − 6x) = (2y − x²)/(y² − 2x)

At the point (3, 3) (which lies on the curve, since 27 + 27 = 54 = 6·9):

dy/dx = (2·3 − 9)/(9 − 6) = (6 − 9)/3 = −3/3 = −1

The tangent line there has slope −1, passing through (3, 3):

y − 3 = −1·(x − 3)   →   y = −x + 6

That is the standard pattern for every tangent-line problem: get dy/dx, plug in the point to get a numerical slope m, then use point-slope form y − y₁ = m(x − x₁).

Horizontal and vertical tangents

Write dy/dx as a fraction N/D.

• A horizontal tangent occurs where the slope is 0 — that is, where the numerator N = 0 (and D ≠ 0).
• A vertical tangent occurs where the slope is undefined — where the denominator D = 0 (and N ≠ 0).

In each case you solve N = 0 (or D = 0) together with the original curve equation as a system, because the point must lie on the curve.

The inverse trig derivatives

Here is the table to know cold. Each one comes from implicit differentiation, and the chain-rule versions (composite argument u) are on the right.

d/dx[arcsin x]  =  1/√(1 − x²)           d/dx[arcsin u]  =  u'/√(1 − u²)
d/dx[arccos x]  = −1/√(1 − x²)           d/dx[arccos u]  = −u'/√(1 − u²)
d/dx[arctan x]  =  1/(1 + x²)            d/dx[arctan u]  =  u'/(1 + u²)
d/dx[arccot x]  = −1/(1 + x²)            d/dx[arccot u]  = −u'/(1 + u²)
d/dx[arcsec x]  =  1/(|x|√(x² − 1))      d/dx[arcsec u]  =  u'/(|u|√(u² − 1))
d/dx[arccsc x]  = −1/(|x|√(x² − 1))      d/dx[arccsc u]  = −u'/(|u|√(u² − 1))

Two patterns make these easy to remember. First, the "co-" functions just add a minus sign: arccos, arccot, arccsc are the negatives of arcsin, arctan, arcsec. Second, there are only three shapes: 1/√(1−x²), 1/(1+x²), and 1/(|x|√(x²−1)). (On AP Calculus AB you will overwhelmingly see arcsin, arccos, and arctan — the other three are rare, so anchor those first.)

Where they come from: derivation of d/dx[arctan x]

Let y = arctan x. Rewrite this without the inverse:

tan y = x

Now differentiate both sides with respect to x — implicit differentiation, because y is a function of x:

sec²y · (dy/dx) = 1
dy/dx = 1/sec²y

We need this in terms of x. Using the identity sec²y = 1 + tan²y, and recalling tan y = x:

dy/dx = 1/(1 + tan²y) = 1/(1 + x²)

The same trick gives arcsin: from sin y = x, differentiating gives cos y·(dy/dx) = 1, so dy/dx = 1/cos y. Since cos y = √(1 − sin²y) = √(1 − x²) (positive because arcsin outputs angles in [−π/2, π/2] where cosine ≥ 0), we get d/dx[arcsin x] = 1/√(1 − x²). Every inverse-trig derivative is built this exact way — implicit differentiation plus a Pythagorean identity.

(c) Worked Examples

Example 1 — Basic implicit dy/dx (no calculator)

Problem. Find dy/dx if x² + xy + y² = 12.

Strategy. Differentiate both sides with respect to x. The xy term needs the product rule; the y² term needs the chain rule.

Solution.

2x + (x·dy/dx + y·1) + 2y·(dy/dx) = 0
2x + y + x·(dy/dx) + 2y·(dy/dx) = 0
(x + 2y)·(dy/dx) = −(2x + y)
dy/dx = −(2x + y)/(x + 2y)

Justification. Each y-term carried a dy/dx (chain rule); the mixed term xy was differentiated as a product. Collecting the dy/dx terms and isolating gives the slope at any point (x, y) on the curve.

Example 2 — Tangent line on an implicit curve (no calculator)

Problem. Find the equation of the tangent line to x² + xy + y² = 12 at the point (2, 2).

Strategy. Use the dy/dx from Example 1, evaluate at the point for a numerical slope, then point-slope form. (First confirm the point is on the curve.)

Solution. Check: 4 + 4 + 4 = 12 ✓. Then:

dy/dx = −(2x + y)/(x + 2y)
At (2, 2):  dy/dx = −(4 + 2)/(2 + 4) = −6/6 = −1

Tangent line through (2, 2) with slope −1:

y − 2 = −1(x − 2)   →   y = −x + 4

Justification. The slope of the tangent equals dy/dx evaluated at the point of tangency; point-slope form then gives the line.

Example 3 — Horizontal and vertical tangents (no calculator)

Problem. For the curve x² + xy + y² = 12, find all points where the tangent line is horizontal and all points where it is vertical.

Strategy. With dy/dx = −(2x + y)/(x + 2y) = N/D: horizontal where N = 0, vertical where D = 0. Solve each condition with the curve equation.

Solution — horizontal (2x + y = 0, i.e. y = −2x):

x² + x(−2x) + (−2x)² = 12
x² − 2x² + 4x² = 12
3x² = 12  →  x = ±2
x = 2 → y = −4;   x = −2 → y = 4

Horizontal tangents at (2, −4) and (−2, 4). (Check D = x + 2y ≠ 0: at (2,−4), 2 − 8 = −6 ≠ 0 ✓.)

Solution — vertical (x + 2y = 0, i.e. x = −2y):

(−2y)² + (−2y)y + y² = 12
4y² − 2y² + y² = 12
3y² = 12  →  y = ±2
y = 2 → x = −4;   y = −2 → x = 4

Vertical tangents at (−4, 2) and (4, −2). (Check N = 2x + y ≠ 0: at (4,−2), 8 − 2 = 6 ≠ 0 ✓.)

Justification. Horizontal tangents require slope 0, forcing the numerator to vanish; vertical tangents require an undefined slope, forcing the denominator to vanish — in both cases at a point actually on the curve.

Example 4 — Inverse trig with the chain rule (no calculator)

Problem. Differentiate f(x) = arctan(x²) and find f '(1).

Strategy. This is arctan(u) with u = x², so u' = 2x. Use d/dx[arctan u] = u'/(1 + u²).

Solution.

f '(x) = (2x)/(1 + (x²)²) = 2x/(1 + x⁴)
f '(1) = 2/(1 + 1) = 2/2 = 1

Justification. The outer function is arctan, whose derivative is 1/(1 + u²); multiplying by the inner derivative u' = 2x (chain rule) gives the result. The slope of arctan(x²) at x = 1 is exactly 1.

(d) Common Mistakes

Forgetting the dy/dx factor on y-terms. Students write d/dx[y²] = 2y and stop. Why it's wrong: y is a function of x, so the chain rule demands d/dx[y²] = 2y·(dy/dx). Fix: every time you differentiate a term with a y in it, immediately write the trailing ·(dy/dx). No exceptions.

Not using the product rule on xy-terms. Students write d/dx[xy] = dy/dx or = y. Why it's wrong: xy is a product of two functions of x. Fix: d/dx[xy] = x·(dy/dx) + y·1 = x·(dy/dx) + y. Treat x and y as two separate functions and apply the product rule fully.

Sign errors on the inverse trig derivatives. Mixing up arcsin (+) with arccos (−), or arctan (+) with arccot (−). Why it's costly: a dropped minus sign turns a correct method into a wrong answer. Fix: memorize "the co-functions are negative." arccos, arccot, arccsc all carry the minus sign; their partners don't.

Algebra errors when solving for dy/dx. After differentiating, students factor incorrectly or move terms with sign mistakes. Why it's wrong: dy/dx appears in multiple terms; you must collect them all, factor dy/dx out, then divide. Fix: be mechanical — (i) get every dy/dx term on the left, (ii) everything else on the right, (iii) factor out dy/dx, (iv) divide.

Confusing the conditions for horizontal vs. vertical tangents. Setting the denominator to zero for a horizontal tangent. Fix: horizontal slope is 0 → numerator = 0; vertical (undefined) slope → denominator = 0.

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## (e) Practice Problems

(Short answer) Find dy/dx for sin y + x² = y. Show your work.

(Short answer) Find the equation of the tangent line to x² − xy + y² = 7 at the point (−1, 2).

(Short answer) Differentiate g(x) = arctan(eˣ).

(Justification) The point (1, 1) lies on the curve x³ + y³ = 2. A student claims the tangent line there is horizontal. Determine dy/dx at (1, 1) and justify, in one sentence, whether the student is correct.

(Justification) For the curve x² + y² = 25, find every point at which the tangent line is vertical, and justify your reasoning using the structure of dy/dx.

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## (f) AP Exam Focus

> Free-Response Question (No Calculator) — 9 points

>

> Consider the curve defined by x² + xy + y² = 12.

>

> (a) Show that dy/dx = −(2x + y)/(x + 2y). (2 points)

>

> (b) Write an equation for the line tangent to the curve at the point (2, 2). (2 points)

>

> (c) Find the coordinates of all points on the curve at which the tangent line is horizontal, or explain why no such points exist. (3 points)

>

> (d) Is there a point on the curve at which the tangent line is vertical? If so, give its coordinates; if not, justify why not. (2 points)

### Model Solution

(a) Differentiate both sides of x² + xy + y² = 12 with respect to x, treating y as a function of x. The xy term requires the product rule and the y² term the chain rule:

2x + (x·dy/dx + y) + 2y·(dy/dx) = 0

Collect the dy/dx terms:

(x + 2y)·(dy/dx) = −(2x + y)
dy/dx = −(2x + y)/(x + 2y)

∴ dy/dx = −(2x + y)/(x + 2y), as required.

(b) The point (2, 2) lies on the curve since 4 + 4 + 4 = 12. Evaluating:

dy/dx |₍₂,₂₎ = −(2·2 + 2)/(2 + 2·2) = −6/6 = −1

An equation for the tangent line is

y − 2 = −1(x − 2),   or equivalently   y = −x + 4.

(c) A horizontal tangent requires dy/dx = 0, which occurs where the numerator is zero and the denominator is nonzero:

2x + y = 0  →  y = −2x

Substitute into the curve equation:

x² + x(−2x) + (−2x)² = 12
x² − 2x² + 4x² = 12  →  3x² = 12  →  x = ±2

For x = 2, y = −4; for x = −2, y = 4. At each point the denominator x + 2y ≠ 0 (e.g., at (2, −4), x + 2y = −6 ≠ 0), so the slope is genuinely 0.

∴ The tangent line is horizontal at (2, −4) and (−2, 4).

(d) A vertical tangent requires dy/dx to be undefined, i.e., the denominator is zero while the numerator is nonzero:

x + 2y = 0  →  x = −2y
(−2y)² + (−2y)y + y² = 12  →  3y² = 12  →  y = ±2

For y = 2, x = −4; for y = −2, x = 4. At each, the numerator 2x + y ≠ 0 (e.g., at (4, −2), 2x + y = 6 ≠ 0), so the tangent is vertical.

∴ Yes — vertical tangents at (4, −2) and (−4, 2).

Scoring Commentary

• (a) 2 pts: 1 pt for correct implicit differentiation (product rule on xy, chain rule on y² — the most common point lost is dropping dy/dx on y² or mishandling xy); 1 pt for correctly solving for dy/dx.
• (b) 2 pts: 1 pt for the slope −1 at (2, 2); 1 pt for a correct tangent-line equation. A common loss: stating the slope but not producing an equation of a line.
• (c) 3 pts: 1 pt for setting the numerator equal to zero (2x + y = 0); 1 pt for substituting into the curve to solve; 1 pt for both correct points. Students who set the denominator to zero earn 0 of these 3.
• (d) 2 pts: 1 pt for setting the denominator to zero; 1 pt for both correct coordinate pairs. Stating "no vertical tangent exists" earns 0.
• Justification traps: On the AP exam, full credit on (c)/(d) requires identifying which part of the fraction must vanish. Vague statements like "the slope is zero so I plug in" without isolating the numerator typically lose the reasoning point.

dy/dx = −2x/(cos y − 1) = 2x/(1 − cos y)

y' = (y − 2x)/(2y − x)
At (−1, 2):  y' = (2 − (−2))/(4 − (−1)) = 4/5

g'(x) = eˣ/(1 + e²ˣ)