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AP Calculus BC 3.1 The Chain Rule Study Notes

AP Calculus BC 3.1 The Chain Rule Study Notes - New Syllabus

AP Calculus BC 3.1 The Chain Rule Study Notes- New syllabus

AP Calculus BC 3.1 The Chain Rule Study Notes – AP Calculus BC-  per latest AP Calculus BC Syllabus.

LEARNING OBJECTIVE

  • Calculate derivatives of compositions of differentiable functions.

Key Concepts: 

  • The Chain Rule

AP Calculus BC-Concise Summary Notes- All Topics

The Chain Rule: Calculating Derivatives of Compositions of Functions

The Chain Rule: Calculating Derivatives of Compositions of Functions

 If \( y = f(g(x)) \), where both \( f \) and \( g \) are differentiable, then the derivative is:

\( \dfrac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

Key Idea: Differentiate the outer function, keep the inside unchanged, then multiply by the derivative of the inside function.

Notation: If \( y = f(u) \) and \( u = g(x) \), then:

\( \dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} \).

Example:

Find \( \dfrac{d}{dx} \big( (3x^2 + 5)^4 \big) \).

▶️ Answer/Explanation

Let \( u = 3x^2 + 5 \), then \( y = u^4 \).

\( \dfrac{dy}{du} = 4u^3 \), \( \dfrac{du}{dx} = 6x \).

Apply the chain rule:

\( \dfrac{dy}{dx} = 4u^3 \cdot 6x = 24x(3x^2 + 5)^3 \).

Answer: \( 24x(3x^2 + 5)^3 \).

Example:

Find \( \dfrac{d}{dx} \big( \sin(x^3) \big) \).

▶️ Answer/Explanation

Outer function: \( \sin(u) \), inner function: \( u = x^3 \).

\( \dfrac{d}{du}(\sin u) = \cos u \), \( \dfrac{du}{dx} = 3x^2 \).

Chain rule:

\( \dfrac{d}{dx}[\sin(x^3)] = \cos(x^3) \cdot 3x^2 \).

Answer: \( 3x^2 \cos(x^3) \).

Example:

Differentiate \( y = \ln(5x^2 + 4) \).

▶️ Answer/Explanation

Outer function: \( \ln(u) \), inner: \( u = 5x^2 + 4 \).

\( \dfrac{d}{du}[\ln u] = \dfrac{1}{u} \), \( \dfrac{du}{dx} = 10x \).

Apply chain rule:

\( \dfrac{dy}{dx} = \dfrac{1}{5x^2 + 4} \cdot 10x = \dfrac{10x}{5x^2 + 4} \).

Answer: \( \dfrac{10x}{5x^2 + 4} \).

Example:

Find \( \dfrac{d}{dx} \big( e^{\sin x} \big) \).

▶️ Answer/Explanation

Outer function: \( e^u \), inner: \( u = \sin x \).

\( \dfrac{d}{du}[e^u] = e^u \), \( \dfrac{du}{dx} = \cos x \).

Chain rule:

\( \dfrac{d}{dx}[e^{\sin x}] = e^{\sin x} \cdot \cos x \).

Answer: \( e^{\sin x} \cos x \).

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