AP Calculus AB · College Board CED Unit 3

Composite · Implicit
& Inverse Derivatives

The chain rule unlocks every nested function you've ever seen — sin(x²), √(1 + cos t), (3x+1)⁵. Implicit differentiation lets you find slopes on curves you can't even solve for y. Inverse derivatives finish the toolkit.

Press → to begin

38 SLIDES · 70 MIN · BUILT FROM CED SKILLS 3.1/3.2/3.3/3.4

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2 · Learning Objectives

By the end of this unit, you will be able to…

LO 1 State and apply the chain rule to differentiate composite functions in any form.
LO 2 Differentiate implicitly-defined functions and find slopes of tangent lines on closed-form curves (circles, ellipses, etc.).
LO 3 Compute derivatives of inverse functions using the inverse function theorem.
LO 4 Differentiate arcsin, arccos, arctan, arccot, arcsec, arccsc and combinations with the chain rule.
LO 5 Set up and solve related-rates problems using chain rule (foundation for Unit 4).
LO 6 Justify the chain rule using the limit definition and explain each step in a free response.

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3 · AP CED Mapping

Unit 3 → AP Topic reference

Unit 3 pulls skills across the 2024-25 CED Topic 3 (Differentiation) and Topic 4 (Contextual Applications).

Sub-topic	CED Skill ID	What it covers
3.1 Composite functions & chain rule	FUN-3.C	differentiate f∘g, f∘g∘h, table form
3.2 Implicit differentiation	FUN-3.D	tangent lines on closed curves, second derivatives
3.3 Inverse function theorem	FUN-3.E	computing (f⁻¹)'(b) from f'
3.4 Derivatives of inverse trig	FUN-3.F	arcsin, arccos, arctan, arccot, arcsec, arccsc
3.5 Differentiation strategy	FUN-3.A/B/C/D/E/F	picking the right rule in any combination

📌

Don't memorize these separately — they're all the same idea: derivative of a function whose argument is itself a function. The chain rule.

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4 · Key Terms

Key terms to know before slide 5

Composite function

(f∘g)(x) = f(g(x)) — apply g, then f to the result. Order matters.

Inner function

The "g(x)" inside f — the one you evaluate first.

Outer function

The "f" applied last — its derivative gets multiplied by the inner derivative.

Chain rule

(f∘g)'(x) = f'(g(x)) · g'(x). Multiply derivatives, evaluate.

Implicit relation

Equation defining y as a function of x without solving for y. e.g., x² + y² = 25.

Inverse function

f⁻¹ undoes f: f⁻¹(f(x)) = x. Reflection across y = x.

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5 · Key Terms · Notation

Notation for the same thing, multiple ways

Notation	Used for
d/dx [f(g(x))]	Chain rule in operational form
f'(g(x)) · g'(x)	Lagrange style — same thing, expanded
(f∘g)'(x)	When the composite has a name
dy/du · du/dx	Leibniz "cancelling" style — most intuitive for related rates
d/dx [y] = dy/dx	Implicit: differentiate both sides
(f⁻¹)'(b)	Derivative of inverse at point b; equals 1/f'(a) where b = f(a)

Most students prefer d/dx notation for the chain rule; Leibniz's chain (dy/du · du/dx) makes related rates much easier to set up. Pick one — and stay consistent.

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6 · Topic 3.1 · Chain Rule · Intuition

Why "outer times inner" works

Picture

y = f(g(x)). x changes by Δx. That makes g change by Δg ≈ g'(x)·Δx. That makes f change by Δy ≈ f'(g)·Δg. Multiplying the small ratios:

Δy / Δx ≈ (Δy / Δg) · (Δg / Δx) = f'(g(x)) · g'(x)

Take the limit as Δx → 0: (f∘g)'(x) = f'(g(x)) · g'(x). The intermediate variable "cancels" if you use Leibniz.

🎁

Meme break

I'd tell you a chain rule joke, but you'd have to unpack the outer joke first. And the funniness of the whole thing depends on the funniness of the inner joke times the funniness of the wrapper. dy/dx jokes = df/du · du/dx

🧅

Meme break

The chain rule walked into a gym. Trainer: "today we train the outside, then the inside." Chain rule: "so… every exercise, twice?" Trainer: "yes. And you multiply the pain." (f(g(x))′ = f′(g(x))·g′(x) — outer rate × inner rate.)

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7 · Topic 3.1 · Chain Rule · Formal

The chain rule, formally · FUN-3.C.1

If g is differentiable at x, and f is differentiable at g(x)

(f∘g)'(x) = f'(g(x)) · g'(x)

Or in Leibniz: dy/dx = (dy/du) · (du/dx), where y = f(u) and u = g(x).

Type	Pattern
Polynomial inside polynomial	d/dx [(3x² + 1)⁵] = 5(3x² + 1)⁴ · 6x
Trig inside polynomial	d/dx [sin(πx²)] = cos(πx²) · 2πx
Exp inside polynomial	d/dx [e^5x+1] = e^5x+1 · 5
Any nested combination	differentiate outer (treat inside as a variable) · multiply by derivative of inside

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8 · Chain Rule · Example 1

Basic chain rule

Step-by-step

Q. d/dx [(3x² + 1)⁵]

Step 1 Identify outer function: f(u) = u⁵. Inner: g(x) = 3x² + 1.

Step 2 f'(u) = 5u⁴. g'(x) = 6x.

Step 3 Multiply, evaluate inner: f'(g(x)) · g'(x) = 5(3x² + 1)⁴ · 6x = 30x(3x² + 1)⁴.

🪜

Common shortcut: Bring the original power down, multiply by the inner derivative, drop the outer power by 1. (Same thing, faster to write.)

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9 · Chain Rule · Trig

Chain rule with trig

Two examples

Q1. d/dx [sin(5x)]

→ cos(5x) · 5 = 5 cos(5x).

Q2. d/dx [sin(x²)]

→ cos(x²) · 2x = 2x cos(x²).

Q3. d/dx [cos(ln x)]

→ −sin(ln x) · (1/x) = −sin(ln x) / x.

Original comic · ap-study.com

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10 · Chain Rule · Multiple Nesting

Triple-nested functions

Worked

Q. d/dx [ √(sin(x²) + 1) ]

Step 1 Outer: √u → 1/(2√u). Middle: sin(w) → cos(w). Inner: w = x² → 2x.

Step 2 Multiply: [1/(2√(sin(x²) + 1))] · cos(x²) · 2x = (x · cos(x²)) / √(sin(x²) + 1).

🌳

Tree pattern: peel the outer layer, multiply by the inner derivative, peel that layer, multiply by the inner-inner derivative, and so on until you hit x.

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11 · Chain Rule · Table

Chain rule in a table (when there are 3+ layers)

FUN-3.C.2

Q. d/dx [ sin(tan(x²)) ]

Function	sin	tan	·	x²
Derivative rule	cos(u)	sec²(v)		2x
Substitute	cos(tan(x²))	sec²(x²)	·	2x
Multiply	2x · sec²(x²) · cos(tan(x²))

This table form is your friend — AP graders love it because it makes the chain rule transparent and reduces sign errors.

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12 · Chain Rule · Exponentials & Logs

Chain rule with e^x and ln x

Practice

Q1. d/dx [e^5x+1]

→ e^5x+1 · 5 = 5 e^5x+1.

Q2. d/dx [ln(3x² − 4)]

→ 1 / (3x² − 4) · 6x = 6x / (3x² − 4).

Q3. d/dx [e^{sin x}]

→ e^{sin x} · cos x.

Q4. d/dx [ln(sin x)]

→ (1 / sin x) · cos x = cos x / sin x = cot x.

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13 · Topic 3.2 · Implicit Differentiation · Idea

When you can't (or won't) solve for y

Idea

If F(x, y) = 0 defines y implicitly as a function of x, then we can find dy/dx without solving for y. Differentiate both sides with respect to x. Treat y as an unknown function of x; apply the chain rule any time you differentiate a term involving y.

Why bother?

Curves like x² + y² = 25 (a circle) or x³ + y³ = 6xy are very hard or impossible to solve for y. But we still want the slope of the tangent at, say, (3, 4).

🗣️

Meme break

Implicit differentiation is basically gossip: x and y refuse to define their relationship in public, so we differentiate the rumours and solve for dy/dx anyway.

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14 · Implicit Diff · Circle

Find the slope at a point on a circle

Worked

Q. Find dy/dx on x² + y² = 25. Then find the slope at (3, 4).

Step 1 Differentiate both sides w.r.t. x: 2x + 2y · dy/dx = 0.

Step 2 Solve for dy/dx: dy/dx = −x / y.

Step 3 At (3, 4): dy/dx = −3/4.

🌍

Sanity check: y = √(25 − x²) ⇒ dy/dx = −x / √(25 − x²) = −x / y. ✓ Same answer, the implicit version avoided the algebra.

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15 · Implicit Diff · Folium

A curve you cannot solve for y

Famous curve

Q. Folium of Descartes: x³ + y³ = 6xy. Find dy/dx.

Step 1 Differentiate: 3x² + 3y² · dy/dx = 6y + 6x · dy/dx.

Step 2 Collect dy/dx on one side: (3y² − 6x) dy/dx = 6y − 3x².

Step 3 Solve: dy/dx = (6y − 3x²) / (3y² − 6x) = (2y − x²) / (y² − 2x).

This is impossible to solve explicitly — but the implicit derivative still gives us the slope at every point the curve passes through. Bonus: it tells us where the tangent is horizontal (set numerator = 0) and vertical (set denominator = 0).

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16 · Implicit Diff · Second Derivative

Implicit second derivatives

AP-level

Q. Find d²y/dx² on x² + y² = 25 at the point (3, 4).

Step 1 We already found dy/dx = −x / y.

Step 2 Differentiate dy/dx using quotient rule, treating y = y(x): d²y/dx² = (−1 · y − (−x) · dy/dx) / y² = (−y + x · (−x/y)) / y².

Step 3 Simplify: d²y/dx² = (−y² − x²) / y³ = −25 / y³.

Step 4 At (3, 4): = −25 / 64.

🪞

Watch out: y is a function of x, even though we can't write it. Every time we differentiate a term involving y, we get a y' carried along by chain rule.

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17 · Topic 3.3 · Inverse Functions

What inverse means, geometrically

🔄

Meme break

f⁻¹ walked into a bar and ordered everything backwards. The bartender simply undid the order. (f∘f⁻¹)(x) = x — a perfectly balanced breakfast.

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18 · Topic 3.3 · Inverse Function Theorem · FUN-3.E

The inverse derivative formula

If f is differentiable at a with f'(a) ≠ 0, and f⁻¹ is the inverse

(f⁻¹)'(b) = 1 / f'(a) where b = f(a)

"The derivative of the inverse is the reciprocal of the derivative of the original, evaluated at the corresponding point."

If you want (f⁻¹)'(b), find the input a such that f(a) = b, then it's 1/f'(a).
Geometric: reciprocal slopes of perpendicular lines.

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19 · Inverse Function · Worked

Find (f⁻¹)'(b) without finding f⁻¹

Worked

Q. f(x) = x³ + 2x + 1. Find (f⁻¹)'(4).

Step 1 Find a with f(a) = 4. Try a = 1: 1 + 2 + 1 = 4. ✓ So a = 1.

Step 2 f'(x) = 3x² + 2. f'(1) = 5.

Step 3 (f⁻¹)'(4) = 1 / f'(1) = 1/5.

🪞

The beauty: We never wrote down the inverse (which is messy). We just used the chain rule and the fact that x = f⁻¹(f(x)) to derive the formula.

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20 · Inverses of xⁿ

x^1/n via the inverse theorem

From the inverse theorem

d/dx [x^1/n] = (1/n) x^{(1/n) − 1}

This is the power rule applied with exponent 1/n — but you can also derive it through the inverse function theorem.

Three cases

Q1. d/dx [√x] = (1/2) x^−1/2 = 1 / (2√x).

Q2. d/dx [∛x] = (1/3) x^−2/3.

Q3. d/dx [x^1/5] = (1/5) x^−4/5.

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21 · Topic 3.4 · Inverse Trig Functions

The six inverse trig functions

arcsin x

"the angle whose sine is x." Range [−π/2, π/2].

arccos x

Range [0, π].

arctan x

Range (−π/2, π/2). Defined for all real x.

arccot x

Range (0, π).

arcsec x

Range [0, π], y ≠ π/2. Defined for |x| ≥ 1.

arccsc x

Range [−π/2, π/2], y ≠ 0. Defined for |x| ≥ 1.

📐

AP exam focus: arcsin, arccos, arctan. Memorize those three. The other three occasionally appear on MCQ.

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22 · Topic 3.4 · Derivatives · FUN-3.F

Six inverse trig derivatives · memorize four

Function	Derivative	Domain restriction
arcsin x	1 / √(1 − x²)	−1 < x < 1
arccos x	−1 / √(1 − x²)	−1 < x < 1
arctan x	1 / (1 + x²)	all real x
arccot x	−1 / (1 + x²)	all real x
arcsec x	1 / (\|x\| √(x² − 1))	\|x\| > 1
arccsc x	−1 / (\|x\| √(x² − 1))	\|x\| > 1

Memory pattern: the "co" inverse trig derivatives are the negatives of the originals. arccos' = −arcsin', arccot' = −arctan', arccsc' = −arcsec'.

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23 · Deriving arcsin' from scratch

Derive d/dx [arcsin x]

Y = arcsin x means sin Y = x

Show every step.

Step 1 Write y = arcsin x. Then sin(y) = x.

Step 2 Differentiate both sides w.r.t. x (chain rule on the left): cos(y) · dy/dx = 1.

Step 3 dy/dx = 1 / cos(y) = 1 / √(1 − sin²y) = 1 / √(1 − x²).

Note We picked the +√ branch because cos(y) > 0 on the range of arcsin.

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24 · Inverse Trig + Chain Rule

Chain rule with inverse trig

Worked

Q1. d/dx [ arcsin(2x) ]

→ 1/√(1 − 4x²) · 2 = 2 / √(1 − 4x²).

Q2. d/dx [ arctan(e^x) ]

→ 1 / (1 + e^2x) · e^x = e^x / (1 + e^2x).

Q3. d/dx [ arctan(ln x) ]

→ 1 / (1 + (ln x)²) · 1/x = 1 / (x (1 + (ln x)²)).

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25 · Bonus · Integrals of Inverse Trig (Unit 6 preview)

Why inverse trig matters in integration

These integrals will be on Unit 6

∫ dx / √(a² − x²) = arcsin(x/a) + C

∫ dx / (a² + x²) = (1/a) arctan(x/a) + C

You'll memorize these in Unit 6. For now: it's the inverse of what we just learned.

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26 · Topic 3 Bonus · Related Rates Preview

Where the chain rule meets real life

Pattern

Two quantities are linked by an equation. Both change with time. The chain rule gives you a relationship between their rates. Solve for the unknown rate.

Preview · full lesson in Unit 4

Q. A spherical balloon grows so V = (4/3)πr³. If dV/dt = 10 in³/s, find dr/dt when r = 5 in.

Step 1 Differentiate both sides w.r.t. t: dV/dt = 4πr² · dr/dt.

Step 2 Solve: dr/dt = (dV/dt) / (4πr²) = 10 / (4π · 25) = 1 / (10π) in/s.

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27 · Worked Example 1 · Chain Rule Mastery

Chain rule practice run

Four at once

Q1. y = sin(5x² − 3x + 1) ⇒ y' = ?

→ cos(5x² − 3x + 1) · (10x − 3).

Q2. y = e^{cos 4x} ⇒ y' = ?

→ e^{cos 4x} · (−sin 4x) · 4 = −4 sin 4x · e^{cos 4x}.

Q3. y = (x³ + 1)²(x² − 5)³ ⇒ y' = ?

→ Product + chain: 2(x³ + 1)(3x²)(x² − 5)³ + (x³ + 1)² · 3(x² − 5)² · 2x.

Q4. y = √(1 + x²) ⇒ y' = ?

→ (1/(2√(1 + x²))) · 2x = x / √(1 + x²).

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28 · Worked Example 2 · Implicit Diff

Implicit diff practice run

Three scenarios

Q1. Find dy/dx for x²y + y³ = 7.

→ 2xy + x²y' + 3y²y' = 0 → y'(x² + 3y²) = −2xy → y' = −2xy / (x² + 3y²).

Q2. sin(y) + x = y. Find y'.

→ cos(y) · y' + 1 = y' → y'(cos(y) − 1) = 1 → y' = 1 / (cos(y) − 1).

Q3. Find the tangent line to xy² + 3x = 8 at the point (1, ?) where y > 0.

→ At x = 1: y² + 3 = 8 → y = √5. dy/dx = −2xy/(x² + 3y²) = −2√5/(1 + 15) = −√5/8. Tangent: y = √5 + (−√5/8)(x − 1).

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29 · Worked Example 3 · Inverse Derivatives

Inverse function derivative practice

Worked

Q1. f(x) = 5x − x². Find (f⁻¹)'(4) if you know f(1) = 4.

→ f'(x) = 5 − 2x. f'(1) = 3. (f⁻¹)'(4) = 1/3.

Q2. y = arcsin(x³). Find dy/dx.

→ 1/√(1 − x⁶) · 3x² = 3x² / √(1 − x⁶).

Q3. y = arctan(2x). Find dy/dx.

→ 1/(1 + 4x²) · 2 = 2/(1 + 4x²).

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30 · Quick Check

Quick check — chain rule

d/dx [ ln(cos x) ] = ?

A1 / cos x

B−tan x

C−tan x

D−1 / cos²x

✅ Answer: C (−tan x). 1/cos x · (−sin x) = −sin x / cos x = −tan x.

Click an option to check.

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31 · AP MCQ · Implicit Diff

AP MCQ · implicit differentiation

If x² + y² = 25, what is dy/dx at the point (3, −4)?

A−3/4

B3/4

C−4/3

DUndefined

✅ Answer: B (3/4). From slide 14, dy/dx = −x/y = −3 / (−4) = 3/4.

Click to check.

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32 · AP FRQ

AP FRQ · inverse function

Practice problem

Let f(x) = 2 + e^x² for x ≥ 0.

(a) Find f⁻¹(x). (Hint: it won't be pretty — solve for x.)

(b) Find (f⁻¹)'(3) using the inverse function theorem (no need to simplify f⁻¹).

Solution: (a) x = f⁻¹(y) means y = 2 + e^x² ⇒ x² = ln(y − 2) ⇒ x = √(ln(y − 2)). (b) f(a) = 3 ⇒ 2 + e^a² = 3 ⇒ a² = ln 1 = 0 ⇒ a = 0. f'(x) = 2x · e^x². f'(0) = 0. So (f⁻¹)'(3) = 1/f'(0) = undefined (vertical tangent at the inverse).

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33 · AP FRQ · Implicit

AP FRQ · implicit second derivative

AP-style

Consider the curve defined by x² + xy + y² = 7.

(a) Find dy/dx at the point (1, 2). (Implicit diff.)

(b) Is the curve concave up or concave down at (1, 2)?

Solution: (a) 2x + y + xy' + 2yy' = 0 ⇒ y'(x + 2y) = −(2x + y) ⇒ y' = −(2x + y)/(x + 2y). At (1, 2): y' = −4/5. (b) Differentiate y' again (with quotient rule, treating y = y(x)), simplify, evaluate at (1, 2). You'll find y'' = −18/25 → concave down.

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34 · Common Mistakes

Top 5 mistakes in Unit 3

① Dropping the inner derivative

"d/dx [sin(x²)] = cos(x²)" — missing the ·2x. Always multiply by the derivative of the inside.

② Forgetting to apply the chain rule to y when implicit-diffing

If your equation has y, y², sin(y), etc., you must bring dy/dx along with the chain rule. Example: differentiating y² gives 2y · dy/dx, not just 2y.

③ Wrong sign on arcsin' / arctan'

arcsin' is positive. arccos' is NEGATIVE. arctan' is positive. arccot' is negative. Don't memorize just two of them.

④ Using x = a vs x = (f⁻¹)(a) carelessly

The inverse function theorem mixes the two arguments. Be sure: (f⁻¹)'(b) = 1 / f'(a) where a = f⁻¹(b) (i.e., f(a) = b).

⑤ Differentiating arctan as 1/x instead of 1/(1+x²)

1/x is d/dx [ln x]. 1/(1+x²) is d/dx [arctan x]. Different formulas — different functions.

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35 · Real-World Applications

Chain rule, implicit diff, and inverses in the wild

Arctan in machine learning: The activation function arctan(x) is differentiable everywhere (unlike ReLU), so its gradient never vanishes. d/dx arctan(x) = 1/(1+x²) appears in millions of neural-network updates per second.
Implicit curves in design: Bézier curves (used in every font, logo, UI animation) are defined implicitly. Designers use calculus-of-implicit-curves to control lighting, shadows, and 3D motion.
Inverse function theorem in physics: The Jacobian of an inverse transformation (e.g., coordinate changes in fluid dynamics) gives you the inverse derivative framework. Critical for general relativity.
Drug concentration: If concentration C(t) depends on time, and you want to know how dose rate affects peak concentration, the chain rule through the inverse function lets you solve for t as a function of C.
Stock options pricing: The Black-Scholes formula uses arctan-like terms and implicit relationships between strike price, time, and volatility.

🤯

Most-Google'd AP topic: The chain rule is what students search for most. Master it now and the rest of calculus (Units 4-8) opens up.

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36 · YouTube Lessons

Best YouTube for this unit

Chain rule The Chain Rule — Visual Proof + Examples 📺 3Blue1Brown⏱ 16 min Implicit Implicit Differentiation — Step by Step 📺 The Organic Chemistry Tutor⏱ 26 min Inverse trig Inverse Trig Derivatives · Memorize in 18 min 📺 Professor Leonard⏱ 18 min

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37 · Unit 3 in 60 Seconds

Unit 3 recap

Chain rule: (f∘g)' = f'(g(x)) · g'(x). When in doubt, peel layers.
Implicit diff: Differentiate both sides w.r.t. x; bring dy/dx along via chain rule; solve.
Inverse function theorem: (f⁻¹)'(b) = 1 / f'(a) where b = f(a).
Inverse trig derivatives: arcsin' = 1/√(1−x²), arccos' = −1/√(1−x²), arctan' = 1/(1+x²), arccot' = −1/(1+x²).
Composite inverse derivative = 1 / (composite of original derivative evaluated at the inverse).

🔗

Next: Unit 4 · Contextual Applications of Differentiation — uses the chain rule to model rates of change in physics, biology, and finance.

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38 · Knowledge Coverage Checklist

Unit 3 · everything covered ✓

Mapped 1:1 against the 2024-25 College Board CED (across Topic 3 sub-skills).

3.1 / FUN-3.C Composite functions & chain rule (formal statement).
3.1 / FUN-3.C Chain rule in tables (3+ layer chain).
3.1 / FUN-3.C Combining chain rule with product, quotient, sum rules.
3.2 / FUN-3.D Implicit differentiation — first derivative.
3.2 / FUN-3.D Implicit differentiation — second derivatives.
3.3 / FUN-3.E Inverse function theorem.
3.3 / FUN-3.E Derivatives of inverse functions of xⁿ.
3.4 / FUN-3.F Derivatives of arcsin, arccos, arctan, arccot, arcsec, arccsc.
3.5 / FUN-3.* Inverse trig + chain rule combined.
3 bonus Related rates preview (full lesson in Unit 4).
Worked examples · 4 unlimited + 3 quiz · 5 practice problems embedded.
Common errors · 5 pitfalls explicitly called out.
3 curated YouTube lessons (chain / implicit / inverse trig).
Real-world connections · 5 applications across ML, design, physics, medicine, finance.

🎓

Done with Unit 3! Move to Unit 4 — chain rule in motion: position, velocity, acceleration, related rates, and linear approximation.

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Unit 3 1 / 38 Composite · Implicit · Inverse

Composite · Implicit& Inverse Derivatives

By the end of this unit, you will be able to…

Unit 3 → AP Topic reference

Key terms to know before slide 5

Composite function

Inner function

Outer function

Chain rule

Implicit relation

Inverse function

Notation for the same thing, multiple ways

Why "outer times inner" works

The chain rule, formally · FUN-3.C.1

Basic chain rule

Chain rule with trig

Triple-nested functions

Chain rule in a table (when there are 3+ layers)

Chain rule with ex and ln x

When you can't (or won't) solve for y

Find the slope at a point on a circle

A curve you cannot solve for y

Implicit second derivatives

What inverse means, geometrically

The inverse derivative formula

Find (f⁻¹)'(b) without finding f⁻¹

x1/n via the inverse theorem

The six inverse trig functions

arcsin x

arccos x

arctan x

arccot x

arcsec x

arccsc x

Six inverse trig derivatives · memorize four

Derive d/dx [arcsin x]

Chain rule with inverse trig

Why inverse trig matters in integration

Where the chain rule meets real life

Chain rule practice run

Implicit diff practice run

Inverse function derivative practice

Quick check — chain rule

AP MCQ · implicit differentiation

AP FRQ · inverse function

AP FRQ · implicit second derivative

Top 5 mistakes in Unit 3

① Dropping the inner derivative

② Forgetting to apply the chain rule to y when implicit-diffing

③ Wrong sign on arcsin' / arctan'

④ Using x = a vs x = (f⁻¹)(a) carelessly

⑤ Differentiating arctan as 1/x instead of 1/(1+x²)

Chain rule, implicit diff, and inverses in the wild

Best YouTube for this unit

Unit 3 recap

Unit 3 · everything covered ✓

Composite · Implicit
& Inverse Derivatives

Chain rule with e^x and ln x

x^1/n via the inverse theorem