Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-18, find \(D_{x} y\). $$ y=\sin x \cos x $$

Step-by-Step Solution

Answer

The derivative \( D_{x}y \) is \( \cos 2x \).

Step 1: Identify the Problem Type

The problem asks us to find the derivative of a function, specifically the function \( y = \sin x \cos x \) with respect to \( x \). This involves using derivative rules.

Step 2: Recognize Using Product Rule

Since the function is a product of two separate functions (\( \sin x \) and \( \cos x \)), we need to apply the product rule. The product rule states that \( (uv)' = u'v + uv' \), where \( u \) and \( v \) are functions of \( x \).

Step 3: Assign Functions and Derivatives

Let \( u = \sin x \) and \( v = \cos x \). Now find their individual derivatives: \( u' = \cos x \) and \( v' = -\sin x \).

Step 4: Apply the Product Rule

Substitute \( u \), \( v \), \( u' \), and \( v' \) into the product rule. We get: \[D_{x}y = u'v + uv' = (\cos x)(\cos x) + (\sin x)(-\sin x).\]

Step 5: Simplify the Expression

Simplify the resulting expression: \[D_{x}y = \cos^2 x - \sin^2 x.\]

Step 6: Recognize Trigonometric Identity

Notice that \( \cos^2 x - \sin^2 x \) can be expressed using the double angle formula: \( \cos 2x = \cos^2 x - \sin^2 x \).

Step 7: Write the Final Derivative

Using the identity, the derivative simplifies to: \[D_{x}y = \cos 2x.\]

Key Concepts

Product Rule

In calculus, one of the key techniques for differentiation is the product rule. This rule is essential when you have to find the derivative of a function that is the product of two or more functions. For instance, consider the problem where you need to differentiate the function \( y = \sin x \cos x \). Here, the function \( y \) is the product of \( \sin x \) and \( \cos x \). The product rule states:

• If \( u(x) \) and \( v(x) \) are both differentiable functions of \( x \), then the derivative of their product \( uv \) is \( (uv)' = u'v + uv' \).

In this problem, we assign \( u = \sin x \) and \( v = \cos x \), so we can use the product rule to find the derivative of the function as follows:

• Find \( u' = \frac{d}{dx} \sin x = \cos x \)
• Find \( v' = \frac{d}{dx} \cos x = -\sin x \)

Plug these derivatives into the product rule:

• \( D_{x}y = u'v + uv' = \cos x \cdot \cos x + \sin x \cdot (-\sin x) \)
• Simplified, it results in \( \cos^2 x - \sin^2 x \).

Trigonometric Identities

In mathematics, trigonometric identities are equations that involve trigonometric functions and are true for every value of the occurring variables. These identities play a vital role in simplifying trigonometric expressions and are very handy in calculus, especially when simplifying derivatives or integrals. In the given exercise, after applying the product rule, we were left with an expression:

• \( D_{x}y = \cos^2 x - \sin^2 x \)

This expression can be further simplified using a well-known trigonometric identity called the double angle identity for cosine. The identity states:

• \( \cos 2x = \cos^2 x - \sin^2 x \)

By recognizing this pattern, we can easily simplify the derivative to:

• \( D_{x}y = \cos 2x \)

Using these identities makes it easier to understand and work with trigonometric functions in calculus.

Derivative Calculation

Derivative calculation is a fundamental concept in calculus. It involves finding the rate of change or the slope of a function at any given point. The derivative of a function gives us insights into its behavior, such as where it is increasing, decreasing, or stationary. In this exercise, we calculated the derivative of the function \( y = \sin x \cos x \). To do this, we used a combination of the product rule and trigonometric identities to simplify our differential expression. Let's break down the calculation steps:

• Recognize the need to use the product rule for the function \( y = \sin x \cos x \) because it is a multiplication of two functions.
• Calculate the derivative of each function: \( u' = \cos x \) and \( v' = -\sin x \).
• Apply the product rule to get: \( (\sin x \cos x)' = \cos^2 x - \sin^2 x \).
• Simplify this using trigonometric identities: \( D_{x}y = \cos 2x \).

In this calculation process, every step builds on knowledge of calculus and algebraic identities, highlighting how interconnected different areas of mathematics can be.