Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \begin{array}{l} \text { . Use the definition of the derivative to show that }\\\ D_{x}(\sin 5 x)=5 \cos 5 x \end{array} $$

Step-by-Step Solution

Answer

The derivative of \( \sin(5x) \) is \( 5 \cos(5x) \) using the definition of a derivative.

Step 1: Recall the definition of a derivative

The derivative of a function \( f(x) \) at a point \( x \) is defined as \[ f'(x) = \lim_{{h \to 0}} \frac{{f(x + h) - f(x)}}{h}. \]

Step 2: Identify the function

In this case, the function \( f(x) = \sin(5x) \). We want to find \( D_{x}(\sin(5x)) \).

Step 3: Apply the derivative definition

Substitute \( f(x) = \sin(5x) \) into the derivative formula: \[ D_{x}(\sin(5x)) = \lim_{{h \to 0}} \frac{{\sin(5(x+h)) - \sin(5x)}}{h}. \]

Step 4: Use the trigonometric identity

Apply the trigonometric identity for the difference of sines: \[ \sin(A) - \sin(B) = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right). \]For \( A = 5(x+h) \) and \( B = 5x \), the expression becomes: \[ \sin(5(x+h)) - \sin(5x) = 2 \cos(5x + \frac{5h}{2}) \sin(\frac{5h}{2}). \]

Step 5: Simplify the derivative expression

Substitute the identity into the limit: \[ \lim_{{h \to 0}} \frac{2 \cos(5x + \frac{5h}{2}) \sin(\frac{5h}{2})}{h}. \]Rewriting gives: \[ \lim_{{h \to 0}} 2 \cos(5x + \frac{5h}{2}) \cdot \frac{\sin(\frac{5h}{2})}{h}. \]

Step 6: Simplify further using substitution

Let \( u = \frac{5h}{2} \); then \( h = \frac{2u}{5} \), and as \( h \to 0, u \to 0 \). The expression becomes: \[ \lim_{{u \to 0}} 2 \cos(5x + u) \cdot \frac{\sin(u)}{\frac{2u}{5}} \left(\frac{5}{2}\right). \]This simplifies to: \[ 5 \lim_{{u \to 0}} \cos(5x + u) \cdot \frac{\sin(u)}{u}. \]

Step 7: Evaluate the limit

The essential limit \( \lim_{{u \to 0}} \frac{\sin(u)}{u} = 1 \). Thus, \[ 5 \cdot 1 \cdot \cos(5x) = 5 \cos(5x). \]

Step 8: Conclusion

Thus, the derivative of \( \sin(5x) \) with respect to \( x \) using the definition of the derivative is indeed \( 5 \cos(5x) \).

Key Concepts

Definition of Derivative

The derivative is a fundamental concept in calculus, representing the instantaneous rate of change of a function with respect to one of its variables. For a function \( f(x) \), its derivative at a particular point \( x \) is defined mathematically as:\[f'(x) = \lim_{{h \to 0}} \frac{{f(x + h) - f(x)}}{h}.\]This definition essentially measures how much the function \( f(x) \) changes as \( x \) changes by a tiny amount, \( h \). By letting \( h \) approach zero, we find the exact slope of the tangent line to the function at the point in question. This is crucial for understanding how functions behave at very local levels, and it establishes the basic foundation for differential calculus.

Trigonometric Identities

Trigonometric identities are equations that relate trigonometric functions to one another. These identities are essential tools in calculus, particularly when dealing with derivatives involving trigonometric functions. One such identity is the sine difference formula:

• \( \sin(A) - \sin(B) = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \)

This identity helps simplify expressions when differentiating functions like \( \sin(5x) \). By converting the difference of sines into a product involving cosine and sine, we can more easily apply limits and calculate derivatives. Understanding and using these identities accurately allows us to manipulate and solve complex problems in trigonometry and calculus.

Limit of a Function

The concept of a limit is central to calculus and mathematical analysis. A limit describes the behavior of a function as its input approaches a certain value. Formally, the limit of a function \( f(x) \) as \( x \) approaches a point \( c \) is written as:\[\lim_{{x \to c}} f(x) = L,\]where \( L \) is the value that \( f(x) \) approaches. In the context of derivatives, limits are used to define and compute the derivative, ensuring that expressions are evaluated at points of infinitesimal change. For example, the evaluation:

• \( \lim_{{u \to 0}} \frac{\sin(u)}{u} = 1 \)

is a key limit often used when differentiating trigonometric functions. Recognizing and applying limits accurately are crucial skills for solving problems in calculus.

Evaluating Derivatives

Evaluating derivatives involves determining the rate of change or the slope of the tangent line of a function at any given point. To compute the derivative of trigonometric functions, one often uses a combination of the definition of the derivative, trigonometric identities, and limit properties.Consider the function \( \sin(5x) \) and evaluating its derivative using:

• The definition of the derivative \( f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \)
• Trigonometric identity for simplifying \( \sin(5(x+h)) - \sin(5x) \)
• Using the limit \( \lim_{{u \to 0}} \frac{\sin(u)}{u} = 1 \)

These tools allow us to methodically approach complex derivative problems. For \( \sin(5x) \), the process involves applying the difference formula, simplifying, and using limits effectively to arrive at the derivative \( 5 \cos(5x) \). These steps illustrate the seamless integration of mathematical concepts to solve calculus problems efficiently.