Problem 55 - Thomas Calculus

Question

In Exercises \(55-68,\) use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$ y=\sqrt{x(x+1)} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{(2x+1)}{2\sqrt{x(x+1)}} \).

Step 1: Express the Function Equation

The given function is \( y = \sqrt{x(x+1)} \). To simplify differentiation, rewrite it in exponential form: \( y = [x(x+1)]^{1/2} \). This will allow us to use logarithmic differentiation more easily.

Step 2: Apply Logarithms to Both Sides

Take the natural logarithm of both sides of the equation: \( \ln y = \ln([x(x+1)]^{1/2}) \). Using the property of logarithms, you can rewrite this as \( \ln y = \frac{1}{2} \ln(x(x+1)) \).

Step 3: Simplify Using Logarithm Properties

Simplify the right side using the logarithmic identity \( \ln(ab) = \ln a + \ln b \): \( \ln y = \frac{1}{2}(\ln x + \ln(x+1)) \).

Step 4: Differentiate Both Sides

Differentiate both sides with respect to \( x \). The derivative of the left side is \( \frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx} \). The derivative of the right side is \( \frac{1}{2} \left(\frac{1}{x} + \frac{1}{x+1}\right) \).

Step 5: Solve for dy/dx

After differentiation, you have: \( \frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{2} \left(\frac{1}{x} + \frac{1}{x+1}\right) \). Multiply both sides by \( y = [x(x+1)]^{1/2} \) to solve for \( \frac{dy}{dx} \). Thus, \( \frac{dy}{dx} = \frac{1}{2}[x(x+1)]^{1/2} \left( \frac{1}{x} + \frac{1}{x+1} \right) \).

Step 6: Simplify the Derivative

Simplify the expression for \( \frac{dy}{dx} \). Combine the fractions inside the parentheses: \( \frac{dy}{dx} = \frac{1}{2}[x(x+1)]^{1/2} \cdot \frac{(x+1) + x}{x(x+1)} \). Simplify further to get \( \frac{dy}{dx} = \frac{1}{2}[x(x+1)]^{1/2} \cdot \frac{2x+1}{x(x+1)} \).

Key Concepts

Understanding Derivatives

In calculus, the derivative is a measure of how a function changes as its input changes. It is a fundamental tool used to understand how quantities change in relation to one another. The derivative of a function at a point gives the rate of change of the function at that point.

Derivatives can be thought of as the 'slope' of a function at any given point. This is analogous to finding the slope of a tangent line to the curve of a function. If you graphically represent a function, the derivative at a point is the slope of the tangent to the curve at that point.

• Notation: Common notations for a derivative include \( f'(x) \, \, \text{and} \, \frac{dy}{dx} \, \text{for functions expressed as} \, y = f(x) \).
• Physical Interpretation: In physics, derivatives are often used to determine velocity. For example, if a function represents the position of an object over time, its derivative with respect to time represents the object's velocity.

When dealing with complex functions, advanced techniques like logarithmic differentiation can be applied, which simplifies the process and is particularly useful when the function is a product or quotient of other functions.

The Natural Logarithm

The natural logarithm, often denoted as \( \ln(x) \, \text{or} \, \log_{e}(x) \), is a logarithm to the base \( e \, \text{where} \, e \approx 2.718 \), an irrational constant. It is a fundamental mathematical concept that frequently appears in calculus and other mathematical fields.

There are several important properties of the natural logarithm you should be familiar with:

• Logarithm of a Product: \( \ln(ab) = \ln a + \ln b \, \text{This property is heavily used in logarithmic differentiation to simplify expressions.} \)
• Logarithm of a Quotient: \( \ln(\frac{a}{b}) = \ln a - \ln b \, \text{This helps in breaking down complex rational functions.} \)
• Logarithm of a Power: \( \ln(a^b) = b \ln a \, \text{Useful for bringing down exponents in differentiation problems.} \)

Natural logarithms are often introduced in calculus to simplify the differentiation of complicated functions, making challenging derivatives more manageable.

Role of Calculus in Problem Solving

Calculus is the branch of mathematics that helps us analyze change and motion, providing us tools such as derivatives and integrals to solve a myriad of problems.

One of the main uses of calculus is in finding the rate at which one quantity changes relative to another. This is particularly useful in fields such as physics, engineering, and economics. By offering a way to model and predict changes, calculus effectively transforms complicated real-world problems into simpler mathematical forms.

• Rate of Change: Derivatives help determine the rate of change of a function. This is applied in scenarios like determining the speed of a car or the marginal cost in economics.
• Optimization Problems: Calculus assists in finding maximum and minimum values of functions, which is useful in many real-life contexts such as minimizing cost or maximizing profit.
• Understanding Accumulations: Integrals, another key concept in calculus, are used to find the total amount collected or accumulated over time, like determining the area under a curve.

Through techniques such as logarithmic differentiation, calculus enables us to work with complex functions that would otherwise be very challenging to differentiate.