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AP - Calculus AB/BC:: Differentiation - Composite, Implicit, and Inverse Functions

Created by Chat Robotics Community  ·  Updated 2026-08-30

Curriculum Overview

Topics & Key Concepts

Ap Calculus

Sample Flashcard Questions & Answers

Showing 8 of 50 cards
Question #1 Active Recall

The Chain Rule for \(\dfrac{d}{dx}\left[f(g(x))\right]\) states:

- **A)** \(f'(g(x)) + g'(x)\)
- **B)** \(f'(x) \cdot g'(x)\)
- **C)** \(f(g'(x))\)
- **D)** \(f'(g(x)) \cdot g'(x)\)

Answer & Explanation:
**Answer: D)**

The Chain Rule differentiates the 'outer' function evaluated at the inner function, then multiplies by the derivative of the 'inner' function.
Question #2 Active Recall

Use the Chain Rule to find \(\dfrac{d}{dx}\left[(3x+1)^5\right]\).

- **A)** \(5(3x+1)^4 \cdot x\)
- **B)** \(15x^4\)
- **C)** \(15(3x+1)^4\)
- **D)** \(5(3x+1)^4\)

Answer & Explanation:
**Answer: C)**

Outer function (power rule): \(5(3x+1)^4\); inner derivative: \(3\). Multiply: \(5(3x+1)^4 \cdot 3 = 15(3x+1)^4\).
Question #3 Active Recall

Use the Chain Rule to find \(\dfrac{d}{dx}\left[\sin(x^2)\right]\).

- **A)** \(x^2\cos(x^2)\)
- **B)** \(2x\cos(x^2)\)
- **C)** \(2x\cos(x)\)
- **D)** \(\cos(x^2)\)

Answer & Explanation:
**Answer: B)**

Outer derivative \(\cos(x^2)\) times inner derivative \(2x\) gives \(2x\cos(x^2)\).
Question #4 Active Recall

Use the Chain Rule to find \(\dfrac{d}{dx}\left[e^{3x}\right]\).

- **A)** \(e^{3}\)
- **B)** \(3e^{3x}\)
- **C)** \(3xe^{3x-1}\)
- **D)** \(e^{3x}\)

Answer & Explanation:
**Answer: B)**

The outer derivative of \(e^u\) is \(e^u\); multiplying by the inner derivative \(3\) gives \(3e^{3x}\).
Question #5 Active Recall

Use the Chain Rule to find \(\dfrac{d}{dx}\left[\ln(5x)\right]\).

- **A)** \(5\ln x\)
- **B)** \(\dfrac{1}{5x}\)
- **C)** \(\dfrac{1}{x}\)
- **D)** \(\dfrac{5}{x}\)

Answer & Explanation:
**Answer: C)**

Outer derivative \(\dfrac{1}{5x}\) times inner derivative \(5\) gives \(\dfrac{5}{5x} = \dfrac{1}{x}\).
Question #6 Active Recall

Use the Chain Rule to find \(\dfrac{d}{dx}\left[\cos(4x)\right]\).

- **A)** \(-4\sin(4x)\)
- **B)** \(4\sin(4x)\)
- **C)** \(-4\cos(4x)\)
- **D)** \(-\sin(4x)\)

Answer & Explanation:
**Answer: A)**

Outer derivative \(-\sin(4x)\) times inner derivative \(4\) gives \(-4\sin(4x)\).
Question #7 Active Recall

Use the Product Rule together with the Chain Rule to find \(\dfrac{d}{dx}\left[x^2 \sin(3x)\right]\).

- **A)** \(2x\cos(3x)\)
- **B)** \(2x\sin(3x) + 3x^2\cos(3x)\)
- **C)** \(2x\sin(3x) + x^2\cos(3x)\)
- **D)** \(6x\cos(3x)\)

Answer & Explanation:
**Answer: B)**

Product Rule: \(f'g+fg'\) with \(f=x^2, g=\sin(3x)\). Since \(g'=3\cos(3x)\) (Chain Rule), we get \(2x\sin(3x) + x^2 \cdot 3\cos(3x) = 2x\sin(3x) + 3x^2\cos(3x)\).
Question #8 Active Recall

Use the Quotient Rule together with the Chain Rule to find \(\dfrac{d}{dx}\left[\dfrac{\sin(2x)}{x}\right]\).

- **A)** \(\dfrac{2\cos(2x)}{1}\)
- **B)** \(\dfrac{2x\cos(2x) - \sin(2x)}{x^2}\)
- **C)** \(\dfrac{2x\cos(2x) + \sin(2x)}{x^2}\)
- **D)** \(\dfrac{\cos(2x)}{x}\)

Answer & Explanation:
**Answer: B)**

With \(f=\sin(2x)\) (so \(f'=2\cos(2x)\) by Chain Rule) and \(g=x\): \(\dfrac{2\cos(2x)\cdot x - \sin(2x)\cdot 1}{x^2} = \dfrac{2x\cos(2x)-\sin(2x)}{x^2}\).

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