Problem 33 - Calculus Early Transcendentals

Question

Find \(d y / d x\) $$ y=\cos ^{3}(\sin 2 x) $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -6\cos^2(\sin(2x))\sin(\sin(2x))\cos(2x) \).

Step 1: Apply Chain Rule

To find the derivative \( \frac{dy}{dx} \), we first recognize that \( y = (\cos(\sin(2x)))^3 \). This is a composite function, so we will use the chain rule. Let \( u = \cos(\sin(2x)) \). Then \( y = u^3 \). The chain rule tells us \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

Step 2: Differentiate Outer Function

Differentiate the outer function \( y = u^3 \) with respect to \( u \). We have \( \frac{dy}{du} = 3u^2 \). Substitute \( u = \cos(\sin(2x)) \) back: \( \frac{dy}{du} = 3(\cos(\sin(2x)))^2 \).

Step 3: Differentiate Inner Function using Chain Rule Again

Now we find \( \frac{du}{dx} \) where \( u = \cos(\sin(2x)) \). This requires two applications of the chain rule: - First, differentiate \( \cos \left(\sin(2x)\right) \) with respect to \( \sin(2x) \), getting \( \frac{d}{d\sin(2x)}\cos(\sin(2x)) = -\sin(\sin(2x)) \).- Then differentiate \( \sin(2x) \) with respect to \( x \), yielding \( \frac{d}{dx}\sin(2x) = 2\cos(2x) \). So, \( \frac{du}{dx} = -\sin(\sin(2x)) \cdot 2\cos(2x) \).

Step 4: Multiply Derivatives

According to the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). Substitute the derivatives found: \[ \frac{dy}{dx} = 3(\cos(\sin(2x)))^2 \times -\sin(\sin(2x)) \times 2\cos(2x) \] Simplify to get:\[ \frac{dy}{dx} = -6(\cos(\sin(2x)))^2 \sin(\sin(2x))\cos(2x) \]

Step 5: Write Final Expression

Combine the terms to write the derivative in a single expression:\[ \frac{dy}{dx} = -6\cos^2(\sin(2x))\sin(\sin(2x))\cos(2x) \].

Key Concepts

Chain Rule

The Chain Rule is an essential tool in calculus for finding the derivative of composite functions. It allows us to differentiate a complex function by breaking it down into simpler functions. Imagine you have a function made up of several layers, like an onion. The Chain Rule lets you peel back each layer one at a time.

• Use the Chain Rule when you have a function inside another function, known as a composite function.
• Here’s the general idea: if you have a function \( y = f(g(x)) \), the derivative \( \frac{dy}{dx} \) is \( f'(g(x)) \cdot g'(x) \).
• In this context, you derive the outer layer first and then multiply it by the derivative of the inner function.

This approach is beneficial for our given function since it involves multiple layers of trigonometric and polynomial expressions.

Composite Functions

Composite functions are functions within functions. They are like nested dolls—each layer holds another function inside it. For example, in our exercise, we are given \( y = (\cos(\sin(2x)))^3 \). Here, we are dealing with a composition of a power function with trigonometric functions.

• The outer function is \( y = u^3 \) where \( u = \cos(\sin(2x)) \).
• The inner function is \( u = \cos(\sin(2x)) \).

Recognizing and properly separating these functions are crucial for correctly applying the Chain Rule. Each layer must be differentiated separately and then combined to find the derivative of the whole expression.

Trigonometric Functions

Trigonometric Functions play an important role in calculus. The common functions include \( \sin(x), \cos(x), \tan(x), \) and their reciprocal functions. These functions describe ratios of sides in right triangles and are periodic, making them useful in modeling cyclical phenomena.

• In the original problem, \( \sin(2x) \) and \( \cos(\sin(2x)) \) are examples of trigonometric functions.
• They add layers of complexity to the differentiation process because they change the way angles relate to lengths in their associated triangles.
• These functions have unique derivatives that must be memorized or understood conceptually for application in problems.

Understanding the properties of these functions helps us manage complex derivative problems more efficiently.

Derivatives of Trigonometric Functions

The derivatives of trigonometric functions are fundamental in calculus. They capture how the trigonometric function changes as the variable x changes.

• The derivative of \( \sin(x) \) is \( \cos(x) \). Essentially, this tells us that the rate of change of the sine function is the cosine function.
• Meanwhile, the derivative of \( \cos(x) \) is \(-\sin(x) \). This negative sign indicates a downward slope in the rate of change of the cosine function.
• Understanding these derivatives is crucial for breaking down and analyzing complex functions, especially when combined with other functions such as in chain and composite functions.

In our exercise, we use these derivatives in conjunction with the Chain Rule to find the rate of change of a composite function involving multiple trigonometric expressions.