Lesson 13: Implicit Differentiation

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.

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.

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.

Common Mistakes

1. 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.

2. 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.

3. 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.

4. 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.

5. 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.

Practice Problems

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)

11. [NO CALC] (Short answer) Find dy/dx for sin y + x² = y. Show your work.

12. [NO CALC] (Short answer) Find the equation of the tangent line to x² − xy + y² = 7 at the point (−1, 2).

13. [NO CALC] (Short answer) Differentiate g(x) = arctan(eˣ).

14. [NO CALC] (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.

15. [NO CALC] (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.

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)