Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Opening subject page...

Loading your content

  1. My Subjects
  2. AP Calculus BC
  3. Flashcards

AP Calculus BC Flashcards: Chain Rule

Study Chain Rule in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

← Back to flashcard decks

What this deck covers

This deck focuses on Chain Rule, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

AP Calculus BC Flashcards: Chain Rule

1

/ 30

0 reviewed

0% Complete

0 reviewing
QUESTION

Identify uuu in h(x)=(3x2+2x)5h(x) = (3x^2 + 2x)^5h(x)=(3x2+2x)5.

Tap or drag to reveal answer

ANSWER

u=3x2+2xu = 3x^2 + 2xu=3x2+2x. The expression being raised to the 5th power.

Swipe Right = I Know It! 🎉

Swipe Left = Still Learning

All flashcards

Flashcard 1: Identify uuu in h(x)=(3x2+2x)5h(x) = (3x^2 + 2x)^5h(x)=(3x2+2x)5.

Answer: u=3x2+2xu = 3x^2 + 2xu=3x2+2x. The expression being raised to the 5th power.

Flashcard 2: Find the derivative of sin2(x)\text{sin}^2(x)sin2(x).

Answer: 2sin(x)cos(x)2\text{sin}(x)\text{cos}(x)2sin(x)cos(x). Chain rule: 2sin⁡(x)×cos⁡(x)2\sin(x) \times \cos(x)2sin(x)×cos(x) or sin⁡(2x)\sin(2x)sin(2x).

Flashcard 3: Differentiate y=sin(ln(x2))y = \text{sin}(\text{ln}(x^2))y=sin(ln(x2)).

Answer: 2xcos(ln(x2))x2\frac{2x\text{cos}(\text{ln}(x^2))}{x^2}x22xcos(ln(x2))​. Chain rule twice: cos⁡(ln⁡(x2))×2xx2\cos(\ln(x^2)) \times \frac{2x}{x^2}cos(ln(x2))×x22x​.

Flashcard 4: Differentiate f(x)=ln(ex+1)f(x) = \text{ln}(\text{e}^x + 1)f(x)=ln(ex+1).

Answer: exex+1\frac{\text{e}^x}{\text{e}^x + 1}ex+1ex​. Chain rule: 1u×u′\frac{1}{u} \times u'u1​×u′ where u=ex+1u = e^x + 1u=ex+1.

Flashcard 5: Find dy/dxdy/dxdy/dx if y=ln(ex2)y = \text{ln}(\text{e}^{x^2})y=ln(ex2).

Answer: 2x2x2x. Simplifies to x2x^2x2 since ln⁡\lnln and eee cancel.

Flashcard 6: Identify uuu in y=(ln(x))4y = (\text{ln}(x))^4y=(ln(x))4.

Answer: u=ln(x)u = \text{ln}(x)u=ln(x). The natural logarithm function raised to power 4.

Flashcard 7: Differentiate y=etan(x)y = \text{e}^{\text{tan}(x)}y=etan(x).

Answer: sec2(x)etan(x)\text{sec}^2(x)\text{e}^{\text{tan}(x)}sec2(x)etan(x). Chain rule: eu×u′e^u \times u'eu×u′ where u=tan⁡(x)u = \tan(x)u=tan(x).

Flashcard 8: Which function is the inner function in f(g(x))=e3x+7f(g(x)) = \text{e}^{3x+7}f(g(x))=e3x+7?

Answer: g(x)=3x+7g(x) = 3x+7g(x)=3x+7. The linear expression in the exponent.

Flashcard 9: What is y′y'y′ if y=cos(5x)y = \text{cos}(5x)y=cos(5x)?

Answer: −5sin(5x)-5\text{sin}(5x)−5sin(5x). Chain rule: −sin⁡(u)×u′-\sin(u) \times u'−sin(u)×u′ where u=5xu = 5xu=5x.

Flashcard 10: Differentiate y=e3x+7y = \text{e}^{3x+7}y=e3x+7.

Answer: 3e3x+73\text{e}^{3x+7}3e3x+7. Chain rule: eu×u′e^u \times u'eu×u′ where u=3x+7u = 3x + 7u=3x+7.

Flashcard 11: Differentiate y=eln(x)y = \text{e}^{\text{ln}(x)}y=eln(x).

Answer: 1x\frac{1}{x}x1​. Simplifies to xxx since eln⁡(x)=xe^{\ln(x)} = xeln(x)=x.

Flashcard 12: Differentiate y=cos−1(5x)y = \text{cos}^{-1}(5x)y=cos−1(5x).

Answer: −5sqrt(1−25x2)-\frac{5}{\text{sqrt}(1 - 25x^2)}−sqrt(1−25x2)5​. Chain rule: −11−u2×u′-\frac{1}{\sqrt{1-u^2}} \times u'−1−u2​1​×u′ where u=5xu = 5xu=5x.

Flashcard 13: Find f′(x)f'(x)f′(x) if f(x)=cos(ex)f(x) = \text{cos}(\text{e}^x)f(x)=cos(ex).

Answer: −exsin(ex)-\text{e}^x\text{sin}(\text{e}^x)−exsin(ex). Chain rule: −sin⁡(u)×u′-\sin(u) \times u'−sin(u)×u′ where u=exu = e^xu=ex.

Flashcard 14: Find d/dxd/dxd/dx of y=cos(e2x)y = \text{cos}(\text{e}^{2x})y=cos(e2x).

Answer: −2e2xsin(e2x)-2\text{e}^{2x}\text{sin}(\text{e}^{2x})−2e2xsin(e2x). Chain rule: −sin⁡(u)×u′-\sin(u) \times u'−sin(u)×u′ where u=e2xu = e^{2x}u=e2x.

Flashcard 15: Differentiate y=tan2(3x)y = \text{tan}^2(3x)y=tan2(3x) using the Chain Rule.

Answer: 6tan(3x)sec2(3x)6\text{tan}(3x)\text{sec}^2(3x)6tan(3x)sec2(3x). Chain rule: 2u×u′2u \times u'2u×u′ where u=tan⁡(3x)u = \tan(3x)u=tan(3x).

Flashcard 16: Differentiate f(x)=ex3f(x) = \text{e}^{x^3}f(x)=ex3.

Answer: 3x2ex33x^2\text{e}^{x^3}3x2ex3. Chain rule: eu×u′e^u \times u'eu×u′ where u=x3u = x^3u=x3.

Flashcard 17: Differentiate f(x)=cos3(x)f(x) = \text{cos}^3(x)f(x)=cos3(x).

Answer: −3cos2(x)sin(x)-3\text{cos}^2(x)\text{sin}(x)−3cos2(x)sin(x). Chain rule: 3cos⁡2(x)×(−sin⁡(x))3\cos^2(x) \times (-\sin(x))3cos2(x)×(−sin(x)).

Flashcard 18: Which function is the outer function in f(g(x))=e3x+7f(g(x)) = \text{e}^{3x+7}f(g(x))=e3x+7?

Answer: f(u)=euf(u) = \text{e}^uf(u)=eu. The exponential function wraps the linear expression.

Flashcard 19: Evaluate ddx[ln(x2+1)]\frac{d}{dx}[\text{ln}(x^2 + 1)]dxd​[ln(x2+1)].

Answer: 2xx2+1\frac{2x}{x^2 + 1}x2+12x​. Chain rule: 1u×u′\frac{1}{u} \times u'u1​×u′ where u=x2+1u = x^2 + 1u=x2+1.

Flashcard 20: Identify the outer function in f(g(x))=tan⁡(x23)f(g(x)) = \tan(\frac{x^2}{3})f(g(x))=tan(3x2​).

Answer: f(u)=tan⁡(u)f(u) = \tan(u)f(u)=tan(u). The tangent function wraps the inner expression.

Flashcard 21: Differentiate y=(2x+5)6y = (2x + 5)^6y=(2x+5)6.

Answer: 12(2x+5)512(2x+5)^512(2x+5)5. Power rule with chain rule: 6u5×u′6u^5 \times u'6u5×u′.

Flashcard 22: Differentiate g(x)=ln(4x2+1)g(x) = \text{ln}(4x^2 + 1)g(x)=ln(4x2+1).

Answer: 8x4x2+1\frac{8x}{4x^2 + 1}4x2+18x​. Chain rule: 1u×u′\frac{1}{u} \times u'u1​×u′ where u=4x2+1u = 4x^2 + 1u=4x2+1.

Flashcard 23: Differentiate ex2e^{x^2}ex2 using the Chain Rule.

Answer: 2xex22xe^{x^2}2xex2. Chain rule: derivative of eue^ueu is eue^ueu times u′u'u′.

Flashcard 24: Identify the inner function in f(g(x))=tan⁡(x23)f(g(x)) = \tan(\frac{x^2}{3})f(g(x))=tan(3x2​).

Answer: g(x)=x23g(x) = \frac{x^2}{3}g(x)=3x2​. The expression inside the tangent function.

Flashcard 25: State the formula for the Chain Rule.

Answer: dydx=dydu×dudx\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}dxdy​=dudy​×dxdu​. Fundamental chain rule formula for composite functions.

Flashcard 26: Find y′y'y′ if y=tan(ln(x))y = \text{tan}(\text{ln}(x))y=tan(ln(x)).

Answer: 1xcos2(ln(x))\frac{1}{x\text{cos}^2(\text{ln}(x))}xcos2(ln(x))1​. Chain rule twice: sec⁡2(ln⁡(x))×1x\sec^2(\ln(x)) \times \frac{1}{x}sec2(ln(x))×x1​.

Flashcard 27: Differentiate y=(ln(x))4y = (\text{ln}(x))^4y=(ln(x))4 using the Chain Rule.

Answer: 4(ln(x))3x\frac{4(\text{ln}(x))^3}{x}x4(ln(x))3​. Power rule with chain rule: 4u3×u′4u^3 \times u'4u3×u′.

Flashcard 28: Find f′(x)f'(x)f′(x) if f(x)=sin(x3)f(x) = \text{sin}(x^3)f(x)=sin(x3).

Answer: 3x2cos(x3)3x^2 \text{cos}(x^3)3x2cos(x3). Chain rule: cos⁡(u)×u′\cos(u) \times u'cos(u)×u′ where u=x3u = x^3u=x3.

Flashcard 29: Differentiate f(x)=esin(x)f(x) = \text{e}^{\text{sin}(x)}f(x)=esin(x).

Answer: cos(x)esin(x)\text{cos}(x)\text{e}^{\text{sin}(x)}cos(x)esin(x). Chain rule: eu×u′e^u \times u'eu×u′ where u=sin⁡(x)u = \sin(x)u=sin(x).

Flashcard 30: What is the derivative of f(g(x))f(g(x))f(g(x)) using the Chain Rule?

Answer: f′(g(x))×g′(x)f'(g(x)) \times g'(x)f′(g(x))×g′(x). Chain rule applied to general composite function f(g(x))f(g(x))f(g(x)).