Question

Apply the Chain Rule more than once to find the indicated derivative. \(\frac{d}{d x}\\{\sin [\cos (\sin 2 x)]\\}\)

Short answer

The derivative is \(-2 \cos [\cos (\sin 2x)] \cdot \sin [\sin (2x)] \cdot \cos (2x)\)."

Step-by-step solution

Identify Outer and Inner Functions

The outermost function is \( \sin [u] \) where \( u = \cos [v] \), and \( v = \sin [w] \) where \( w = 2x \).

Differentiate Outermost Function

Differentiate \( \sin [u] \) with respect to \( u \). This gives \( \cos [u] \cdot u' \).

Differentiate Second Function

Let \( u = \cos [v] \). Differentiate \( \cos [v] \) with respect to \( v \), which results in \(-\sin [v] \cdot v' \).

Differentiate Innermost Function

Let \( v = \sin [w] \). Differentiate \( \sin [w] \) with respect to \( w \), giving \( \cos [w] \cdot w' \).

Combine Derivatives

Combine each part's differentials: \( \cos [u] \cdot (-\sin [v]) \cdot \cos [w] \cdot w' \).

Substitute and Simplify

Substitute back \( u = \cos [\sin 2x] \), \( v = \sin 2x \), \( w = 2x \), and \( w' = 2 \). The expression becomes \( \cos [\cos (\sin 2x)] \cdot (-\sin [\sin 2x]) \cdot \cos (2x) \cdot 2 \).

Provide Final Derivative

The final derivative is: \(-2 \cos [\cos (\sin 2x)] \cdot \sin [\sin (2x)] \cdot \cos (2x)\).

Key concepts

Derivatives

Derivatives are a fundamental concept in calculus, allowing us to determine how a function changes with respect to changes in its input. In other words, derivatives provide the rate at which a quantity is changing. In the context of our problem, we are trying to find the derivative of the composite trigonometric expression \( \sin [\cos (\sin 2x)] \). To achieve this, we must apply the chain rule multiple times, as the expression involves several nested functions.
When differentiating, remember:

• The derivative \( \frac{d}{dx} f(x) \) measures how \( f(x) \) changes as \( x \) changes.
• The slope of the tangent line to the curve of \( f(x) \) at a given point can be determined using the derivative.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are essential in calculus, often appearing in problems involving periodic phenomena. These functions require special attention when differentiating.
In our problem, we encounter:

• \( \sin(x) \): the derivative of which is \( \cos(x) \)
• \( \cos(x) \): the derivative of which is \(-\sin(x) \)

Given the nature of the problem, understanding these derivative rules is crucial in solving it effectively. Our particular expression \( \sin [\cos (\sin 2x)] \) demonstrates how trigonometric functions can be compounded and requires careful application of derivative techniques.

Calculus Problem Solving

Calculus problem solving often involves applying established rules and techniques to work through complex expressions. The goal is to break down problems into simpler parts. For this exercise, we relied on the Chain Rule - a powerful method which allows us to differentiate composite functions.

• Step-by-step breakdown is crucial: start with the outermost function and work your way inward.
• Identify all inner functions and differentiate them separately, then combine their derivatives.

Using the Chain Rule requires practice, as it's important to keep track of each step and substitution mafe throughout the process. Proper organization and simplification during each step can significantly ease the solving process.

Composite Functions

Composite functions occur when one function is applied to the result of another function. They are common in calculus problems, requiring specific approaches when finding derivatives. For composite functions like \( \sin[\cos(\sin 2x)] \), identifying each layer of the function is essential:

• The outer function is \( \sin(u) \),
• the middle function is \( \cos(v) \),
• and the innermost function is \( \sin(2x) \).

To differentiate, apply the chain rule as many times as necessary. By carefully differentiating and back-substituting each level, we find the complete derivative efficiently. Practicing with different composite functions enhances one's skill set in navigating complex calculus problems.