Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The derivative is \(D_{x} y = \frac{x \sin x - 1 + \cos x}{x^2}.\)

Step 1: Recognize the Type of Differentiation Needed

The function given is a quotient, so we'll apply the quotient rule of differentiation which states that for a function \(y = \frac{u}{v}\), the derivative \(D_{x} y\) is \( \frac{u'v - uv'}{v^2} \). Here, \(u = 1 - \cos x\) and \(v = x\).

Step 2: Find the Derivatives of the Numerator and Denominator

Calculate \(u'\) and \(v'\):- \(u = 1 - \cos x\), so \(u' = 0 + \sin x = \sin x\).- \(v = x\), so \(v' = 1\).

Step 3: Apply the Quotient Rule

Substitute \(u, u', v\), and \(v'\) into the quotient rule formula.\[D_{x} y = \frac{(\sin x) \cdot x - (1 - \cos x) \cdot 1}{x^2} = \frac{x \sin x - (1 - \cos x)}{x^2}\]

Step 4: Simplify the Expression

Rearrange and simplify the terms in the numerator:\[D_{x} y = \frac{x \sin x - 1 + \cos x}{x^2}.\]

Key Concepts

Quotient Rule

The process of differentiation involves finding how a function changes as its input changes. When dealing with a fraction-like function, we use the Quotient Rule. It's a handy tool in calculus that allows us to differentiate expressions where one function is divided by another. This is crucial for functions like the one given:

• when the function is expressed as a quotient, such as \( y = \frac{1 - \cos x}{x} \).

The general formula for the quotient rule is given by:\[D_x \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}\]

• \(u\) represents the numerator.
• \(v\) represents the denominator.
• \(u'\) and \(v'\) are the derivatives of \(u\) and \(v\) respectively.

This rule helps dissect the derivative into more manageable parts for ease of calculation.

Numerator and Denominator

In any fraction, the top part is called the numerator, and the bottom part is the denominator. These components are vital in applying the quotient rule. For our function:

• Numerator \( u = 1 - \cos x \)
• Denominator \( v = x \)

Each part, the numerator and the denominator, will have its derivative calculated separately. Understanding their roles in the context of differentiation guides us to apply the rule correctly. Calculating these derivatives involves:

• For \( u = 1 - \cos x \), the derivative involves trigonometric formulas.
• For \( v = x \), the derivative is straightforward (since the derivative of \( x \) with respect to \( x \) is 1).

This breakdown makes it simpler when plugging values into the quotient rule formula.

Derivative Calculation

Calculating derivatives is at the heart of differentiation. For the quotient rule to work effectively, we need the derivatives of both the numerator and the denominator. Remember:

• \( u' = \sin x \)
• \( v' = 1 \)

These derivatives are calculated by applying basic derivative rules. The function \( u = 1 - \cos x \) turns into \( \sin x \) because the derivative of \( -\cos x \) is \( \sin x \). The derivative shows the rate at which \( u \) and \( v \) change, which is crucial for applying the quotient rule.With these derivatives, plug them into the quotient formula:\[D_x y = \frac{u'v - uv'}{v^2} = \frac{(\sin x) \cdot x - (1 - \cos x) \cdot 1}{x^2}\]

Simplification

Once the quotient rule is applied, simplification helps make the expression more usable and easier to understand. The initial output from the quotient rule is:\[D_x y = \frac{x \sin x - (1 - \cos x)}{x^2}\]We aim to simplify the numerator, the top part of the fraction, to make our derivative neat and tidy. This involves algebraic manipulation such as:

• Distributing and combining like terms.
• Simplifying expressions like \(-1 + \cos x\).
• Keeping an eye for common factors.

The goal is to end up with a simpler, equivalent form:\[D_x y = \frac{x \sin x - 1 + \cos x}{x^2}\]Simplifying makes it easier to interpret the derivative and potentially further utilize in solving other problems or equations.