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Differentiation: Composite, Implicit, and Inverse Functions

Hannah Hill

Hannah Hill

7 min read

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Study Guide Overview

This study guide covers differentiation for the AP Calculus AB/BC exam, focusing on the chain rule (for composite functions), implicit differentiation, and higher-order derivatives. It also connects these concepts to graphs, tables, and equations, and provides practice questions covering these key areas. It emphasizes understanding the meaning of derivatives, particularly the second derivative and its relationship to concavity.

AP Calculus AB/BC: Differentiation Deep Dive 🚀

Hey there, future AP Calculus master! Let's get you prepped and confident for the exam. This guide is designed to be your go-to resource, especially the night before the big day. We'll break down the trickiest concepts and make sure everything clicks. Let's dive in!

1. Chain Rule: Unlocking Composite Functions 🔗

What are Composite Functions?

Composite functions are like functions within functions. Think of it as a series of nested operations. The output of one function becomes the input of another. For example, in f(g(x))f(g(x)), g(x)g(x) is the inner function, and ff is the outer function. The chain rule is your key to differentiating these.

Key Concept

The chain rule states that the derivative of a composite function is the derivative of the outer function (evaluated at the inner function) multiplied by the derivative of the inner function. Mathematically, if y=f(u)y = f(u) and u=g(x)u = g(x), then dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.

Breaking It Down

Let's use an example: If y=sin(x2)y = \sin(x^2), then:

  1. Identify the inner and outer functions: Here, u=x2u = x^2 (inner) and y=sin(u)y = \sin(u) (outer).
  2. Differentiate each part: The derivative of yy with respect to uu is dydu=cos(u)\frac{dy}{du} = \cos(u), and the derivative of uu with respect to xx is dudx=2x\frac{du}{dx} = 2x.
  3. Apply the chain rule: dydx=dydududx=cos(u)2x=cos(x2)2x=2xcos(x2)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos(u) \cdot 2x = \cos(x^2) \cdot 2x = 2x\cos(x^2).
Memory Aid

Think of the chain rule like peeling an onion 🧅. You differentiate the outer layer, then move inward, multiplying by the derivative of each layer as you go.

Common Mistakes to Avoid

Common Mistake

Forgetting to differentiate the inner function is a very common error. Always remember to multiply by the derivative of the 'inside' function.

Another common pitfall is confusing t...

Question 1 of 9

What is the derivative of y=sin(2x)y = \sin(2x)? 🤔

2\cos(2x)

cos(2x)\cos(2x)

2cos(2x)-2\cos(2x)

2sin(2x)-2\sin(2x)