Problem 60 - Thomas Calculus in SI Units

Question

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=(\tan \theta) \sqrt{2 \theta+1}$$

Step-by-Step Solution

Answer

Derivative is \((\tan \theta) \sqrt{2 \theta+1} \left( \frac{\sec^2 \theta}{\tan \theta} + \frac{1}{2\theta+1} \right)\).

Step 1: Apply Logarithm to Both Sides

Take the natural logarithm of both sides of the equation. This makes it easier to differentiate since we're dealing with a product:\[ \ln y = \ln ((\tan \theta) \sqrt{2\theta + 1}) \] Utilizing properties of logarithms, this can be expanded to:\[ \ln y = \ln (\tan \theta) + \ln (\sqrt{2\theta + 1}) \]

Step 2: Simplify the Expression

Simplify the second term using the logarithmic property for square roots:\[ \ln y = \ln (\tan \theta) + \frac{1}{2} \ln (2\theta + 1) \]

Step 3: Differentiate Both Sides

Differentiate both sides with respect to \(\theta\). Use implicit differentiation for the left side:\[ \frac{1}{y} \frac{dy}{d\theta} = \frac{d}{d\theta} \ln (\tan \theta) + \frac{1}{2} \frac{d}{d\theta} \ln (2\theta + 1) \] The derivative of \(\ln (\tan \theta)\) is:\[ \frac{1}{\tan \theta} \cdot \sec^2 \theta = \frac{\sec^2 \theta}{\tan \theta} \]The derivative of \(\frac{1}{2} \ln (2\theta + 1)\) is:\[ \frac{1}{2} \cdot \frac{1}{2\theta + 1} \cdot 2 = \frac{1}{2\theta + 1} \]

Step 4: Solve for \(\frac{dy}{d\theta}\)

Combine the derivatives from step 3:\[ \frac{1}{y} \frac{dy}{d\theta} = \frac{\sec^2 \theta}{\tan \theta} + \frac{1}{2\theta + 1} \]Multiply the entire expression by \(y\) to solve for \(\frac{dy}{d\theta}\):\[ \frac{dy}{d\theta} = y \left( \frac{\sec^2 \theta}{\tan \theta} + \frac{1}{2\theta + 1} \right) \]Since \(y = (\tan \theta) \sqrt{2 \theta + 1}\), substitute back:\[ \frac{dy}{d\theta} = (\tan \theta) \sqrt{2 \theta + 1} \left( \frac{\sec^2 \theta}{\tan \theta} + \frac{1}{2\theta + 1} \right) \]

Key Concepts

Derivative Calculation

Calculating derivatives involves finding how a function changes as its input changes. It's a fundamental concept in calculus. In this exercise, we're using logarithmic differentiation for the calculation, which simplifies the process when dealing with complex forms. Logarithmic differentiation is handy when you have a function that's a product or a quotient, particularly involving powers or roots. Here's how it helps:

• Take the natural logarithm of the function you're differentiating. This turns products into sums, making them easier to work with.
• Differentiate both sides of the logarithmic equation. This involves implicit differentiation, where you treat the variable as a function of another, and the rules of logarithms simplify the expression.
• Simplify and solve for the derivative. This often involves substituting back the original function you're finding the derivative of.

Understanding each step in this process helps you deal with complex differentiation problems more readily.

Implicit Differentiation

Implicit differentiation is a technique where you differentiate an equation with respect to a variable that isn’t isolated on one side of the equation. This is very useful in cases where separating variables isn't straightforward. In the context of logarithmic differentiation, it allows us to differentiate the logarithm of a function.The process involves:

• Differentiating both sides of the equation with respect to the variable of interest.
• Remembering to apply the chain rule where necessary, since in implicit differentiation, derivatives of functions with respect to themselves might appear.
• Solving a differential expression where derivatives of both the dependent and independent variables appear.

In our example, you see this when we differentiate \( \ln y \) with respect to \( \theta \), resulting in \( \frac{1}{y}\frac{dy}{d\theta} \). This step is crucial for handling complex functions that aren't naturally split into isolated variables on either side.

Trigonometric Functions

Trigonometric functions like \( \tan \theta \) and \( \sec \theta \) are crucial in calculus, especially in differential equations. In our example, these functions are included in the expression we're differentiating. Some key points about their differentiation include:

• The derivative of \( \tan \theta \) is \( \sec^2 \theta \). Recognizing this helps handle expressions where \( \tan \theta \) is part of a function needing differentiation.
• Properties of trigonometric functions and their derivatives help simplify otherwise complex expressions. For example, knowing that \( \frac{1}{\tan \theta} \cdot \sec^2 \theta = \frac{\sec^2 \theta}{\tan \theta} \), allows you to manage expressions succinctly.
• Using these derivatives within logarithmic differentiation blends the rules of trigonometry and logarithms to solve more challenging calculus problems.

Practicing these differentiations can enhance your proficiency in dealing with various calculus scenarios.