Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 33-40, apply the Chain Rule more than once to find the indicated derivative. $$ \frac{d}{d x}\\{\sin [\cos (\sin 2 x)]\\} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Composition of Functions

The given function is \( \sin[\cos(\sin(2x))] \). It is a composition of three functions: \( f(u) = \sin(u) \), \( g(v) = \cos(v) \), and \( h(x) = \sin(2x) \). Then we have \( f(g(h(x))) \). To find the derivative, we need to apply the Chain Rule more than once.

Step 2: Apply the Chain Rule on the Outer Function

The outer function is \( f(u) = \sin(u) \). Its derivative is \( f'(u) = \cos(u) \). So, applying the Chain Rule, we get \( \frac{d}{dx} \sin[v] = \cos[v] \cdot \frac{d}{dx}[v] \), where \( v = \cos(\sin(2x)) \).

Step 3: Apply the Chain Rule on the Middle Function

For \( \frac{d}{dx}[v] \), the middle function \( v = \cos(w) \), where \( w = \sin(2x) \), needs its derivative calculated. The derivative \( \frac{d}{dv}[\cos(w)] = -\sin(w) \) leads to \( -\sin(w) \cdot \frac{d}{dx}[w] \).

Step 4: Apply the Chain Rule on the Innermost Function

In \( \frac{d}{dx}[w] \), we have \( w = \sin(2x) \), which involves \( u = 2x \). Differentiate to get \( \cos(2x) \cdot \frac{d}{dx}[2x] = \cos(2x) \cdot 2 = 2\cos(2x) \).

Step 5: Combine All Derivatives

Combine derivatives using the product from each chain step: \( \cos[\cos(\sin(2x))] \cdot (-\sin(\sin(2x))) \cdot (2\cos(2x)) \) or \( -2\cos[\cos(\sin(2x))] \cdot \sin(\sin(2x)) \cdot \cos(2x) \).

Key Concepts

Derivatives

Derivatives are a foundational concept in calculus that measure how a function changes as its input changes. They are commonly represented by \( \frac{d}{dx} \) and provide the rate of change or the slope of a function. Understanding derivatives allows us to predict how a quantity will change in response to a change in another quantity.

• In the exercise, derivatives are calculated for a composite function, which involves multiple layers of functions within each other.
• The chain rule is used to find the derivative of composite functions. It allows us to differentiate a function composed of multiple functions by considering each function's derivative step by step.
• This requires understanding not only the basic rules of differentiation, like the power rule, product rule, and trigonometric derivatives but also mastering how to nest these rules through the chain rule.

In simpler terms, the derivative gives us a mathematical way to see how sensitive a particular value of a function is to slight changes in its input.

Composite Functions

Composite functions are functions made by combining two or more simpler functions. The function in the original exercise, \( \sin[\cos(\sin(2x))] \), is a composition where one function is applied to the result of another function. Understanding composite functions requires us to:

• Identify the inner and outer functions, as well as any functions between them.
• Understand how each layer or function impacts the overall output when a value is fed into the entire function.
• For example, in the exercise, the outer function is a sine function, then a cosine function, and finally, another sine function at the innermost level. Each of these must be considered in sequence when differentiating.

To differentiate composite functions using the chain rule, it is important to differentiate each function with respect to its variable and then multiply these derivatives accordingly. This ensures that the entire structure of the composite function is accounted for.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, play a crucial role in calculus, especially when dealing with oscillatory behaviors. In the exercise provided:

• The trigonometric functions sine and cosine are central to solving the derivative of the composite function.
• Each trigonometric function has specific derivatives: the derivative of \( \sin(x) \) is \( \cos(x) \), while the derivative of \( \cos(x) \) is \( -\sin(x) \).
• Knowing these derivatives is essential when applying the chain rule, as demonstrated by the multiple steps necessary in the solution.
• Furthermore, understanding the behavior of these functions helps in anticipating their impact on the changes in the function's input values.

When applying the chain rule in the context of trigonometric functions, careful attention to their inherent properties ensures accurate and effective differentiation.