Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \underline{\phantom{xxx}} \text { find } D_{x} y . $$ $$ y=\tan x=\frac{\sin x}{\cos x} $$

Step-by-Step Solution

Answer

The derivative of \( y = \tan x \) is \( D_{x} y = \sec^2 x \).

Step 1: Identify the Function

The given function is expressed as a tangent function: \[ y = \tan x = \frac{\sin x}{\cos x} \].

Step 2: Apply the Quotient Rule

To find the derivative of \( y = \frac{\sin x}{\cos x} \), we will use the quotient rule: \[ D_{x} \left( \frac{u}{v} \right) = \frac{v \cdot D_{x}u - u \cdot D_{x}v}{v^2} \].

Step 3: Define Components of the Quotient Rule

Identify \( u = \sin x \) and \( v = \cos x \). Then we calculate the derivatives: \( D_{x}u = \cos x \) and \( D_{x}v = -\sin x \).

Step 4: Substitute into the Quotient Rule

Substitute the derivatives into the quotient rule formula:\[ D_{x}y = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{(\cos x)^2} \].

Step 5: Simplify the Expression

Simplify the expression from Step 4: \[ D_{x}y = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \].

Step 6: Use Trigonometric Identity

Recall the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \). Hence the expression becomes: \[ D_{x}y = \frac{1}{\cos^2 x} \].

Step 7: Rewrite the Final Derivative

Recognize that \( \frac{1}{\cos^2 x} \) is the same as \( \sec^2 x \), hence,\[ D_{x} y = \sec^2 x \].

Key Concepts

Quotient Rule

The quotient rule is a fundamental tool in calculus differentiation for finding the derivative of a quotient of two functions. When you have a function written as a division of two other functions, like \( y = \frac{u}{v} \), the quotient rule helps us find the derivative efficiently.

Imagine \( u \) as the numerator and \( v \) as the denominator. For this rule, it's crucial to understand the derivative of each part:

• \( D_{x}u \) is the derivative of the numerator.
• \( D_{x}v \) is the derivative of the denominator.

The quotient rule formula is:\[ D_{x} \left( \frac{u}{v} \right) = \frac{v \cdot D_{x}u - u \cdot D_{x}v}{v^2} \]This formula may seem complex at first, but it breaks down the problem into manageable parts by focusing on the derivatives of the top and bottom of the fraction separately.

To illustrate this, consider the example given: \( y = \frac{\sin x}{\cos x} \). By applying the quotient rule, we derive the function using the formula, ensuring each derivative is correctly calculated.

Trigonometric Identities

Trigonometric identities are essential in simplifying expressions involving trigonometric functions. One of the most crucial identities is the Pythagorean identity, which states that for any angle \( x \):

• \( \sin^2 x + \cos^2 x = 1 \)

This identity comes in handy when differentiating functions like the tangent. By recognizing this identity in the quotient rule application, you can simplify the derivative significantly.

In our example, after applying the quotient rule, we end up with an expression \( \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \). By employing the identity \( \sin^2 x + \cos^2 x = 1 \), we simplify this to \( \frac{1}{\cos^2 x} \).

Such identities not only make calculations simpler but also enhance understanding by showing the interrelations between different trigonometric functions. Recognizing these early can help minimize errors and improve problem-solving efficiency.

Calculus Differentiation

Calculus differentiation is the process of finding the rate at which a function is changing at any point. It is a critical concept in calculus, providing insights into the behavior of functions.

Differentiation involves several rules, and knowing when to apply each is essential. The derivative of the tangent function \( \tan x = \frac{\sin x}{\cos x} \) demonstrates how the quotient rule and trigonometric identities play pivotal roles.

To differentiate \( \tan x \), you use the quotient rule because it’s expressed as a division of two functions. After finding the derivative, the use of trigonometric identities simplifies it into \( \sec^2 x \), showing how functions transform through differentiation.

The result, \( D_{x}y = \sec^2 x \), illustrates not only the power of differentiation but also highlights its application in understanding various mathematical concepts and real-life phenomena.