Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\) using the rules of this section. $$ y=\frac{1}{2 x}+2 x $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Differentiation Rules Needed

The given function is composed of two terms: \( y = \frac{1}{2x} + 2x \). To differentiate this expression, we will use the quotient rule for the first term \( \frac{1}{2x} \) and the power rule for the second term \( 2x \).

Step 2: Apply the Quotient Rule to the First Term

Recall the quotient rule: if \( u(x) = \frac{v(x)}{w(x)} \), then \( D_x u = \frac{w(x)D_x v - v(x)D_x w}{w(x)^2} \). For \( \frac{1}{2x} \), consider \( v(x) = 1 \) and \( w(x) = 2x \). Then, \( D_x v = 0 \) and \( D_x w = 2 \). Substituting these into the quotient rule gives: \[D_x \left( \frac{1}{2x} \right) = \frac{2 \cdot 0 - 1 \cdot 2}{(2x)^2} = \frac{-2}{4x^2} = \frac{-1}{2x^2}\]

Step 3: Apply the Power Rule to the Second Term

The power rule states that if \( y = ax^n \), then \( D_x y = anx^{n-1} \). For the term \( 2x \), this is \( ax^n \) where \( a = 2 \) and \( n = 1 \). Consequently, \[D_x (2x) = 2 \cdot 1 \cdot x^{0} = 2\]

Step 4: Combine the Results to Find the Derivative

Now, we combine the derivatives of the two terms: The derivative of \( y = \frac{1}{2x} + 2x \) can be expressed as: \[D_x y = D_x \left( \frac{1}{2x} \right) + D_x (2x) = \frac{-1}{2x^2} + 2\]

Step 5: Simplify the Expression

The derivative of the function is already in its simplest form: \[D_x y = \frac{-1}{2x^2} + 2\]

Key Concepts

Quotient Rule

In calculus, the Quotient Rule is a fundamental technique used for differentiating functions that are presented as a ratio of two other functions. It's particularly handy when one function is divided by another.

Consider a situation where a function can be expressed as \( u(x) = \frac{v(x)}{w(x)} \). The Quotient Rule provides a method to find its derivative \( D_xu \) by following this formula:

• \( D_x u = \frac{w(x)D_x v - v(x)D_x w}{(w(x))^2} \)

In the given example, \( u(x) = \frac{1}{2x} \) is split into two components:

• \( v(x) = 1 \), whose derivative \( D_x v = 0 \)
• \( w(x) = 2x \), whose derivative \( D_x w = 2 \)

Using these components within the Quotient Rule, we compute the derivative as:

• \( D_x \left( \frac{1}{2x} \right) = \frac{2 \cdot 0 - 1 \cdot 2}{(2x)^2} = \frac{-2}{4x^2} = \frac{-1}{2x^2} \)

This technique is vital when dealing with more complex functions that involve division.

Power Rule

The Power Rule is another essential tool in calculus for finding derivatives. It's a very direct way of handling polynomials or any term where a variable is raised to a power.

The Power Rule can be stated simply: When you have a function of the form \( y = ax^n \), the derivative \( D_x y \) is calculated as:\( D_x y = anx^{n-1} \).

In the exercise, for the term \( 2x \), we can see it as \( 2x^1 \). We apply the power rule:

• Here, \( a = 2 \) and \( n = 1 \), so the derivative is \( D_x(2x) = 2 \cdot 1 \cdot x^0 = 2 \).

The Power Rule simplifies finding the derivative of terms like \( ax^n \) immensely by reducing complicated operations to basic multiplication and adjustment of exponents.

Calculus

Calculus, at its core, is the mathematical study of change. It's not just about numbers and equations; it's a tool to understand the world around us. Calculus helps describe how quantities change over time or any variable.

One of its foundational components is differentiation, which focuses on finding the rate of change or the slope of a curve at any point. This is where powerful rules like the Quotient Rule and the Power Rule come into play.

By using differentiation in calculus:

• We can find rates of change, much like how velocity reveals how position changes over time.
• We solve problems in fields ranging from physics to economics, dealing with situations where change is a factor.

Through exercises like the one described, students see firsthand how calculus transforms theoretical concepts into practical applications, making it an indispensable part of modern science and engineering.