Problem 21 - Calculus - AP Edition

Question

Find \(d y / d x\). $$y=\ln (\tan x)$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Function

The function we need to differentiate is given as \( y = \ln(\tan x) \). This is a composition of functions where \( y \) is the natural logarithm of \( \tan x \).

Step 2: Apply the Chain Rule

To differentiate \( y = \ln(\tan x) \), we recognize it as a composition of \( u = \tan x \) and \( y = \ln u \). According to the chain rule, \( \frac{dy}{dx} = \frac{du}{dx} \cdot \frac{dy}{du} \).

Step 3: Differentiate the Outer Function

First, differentiate the outer function \( y = \ln u \) with respect to \( u \). The derivative \( \frac{dy}{du} \) is \( \frac{1}{u} \).

Step 4: Differentiate the Inner Function

Next, differentiate the inner function \( u = \tan x \) with respect to \( x \). The derivative \( \frac{du}{dx} \) is \( \sec^2 x \).

Step 5: Combine the Derivatives

By combining the derivatives using the chain rule, we have \( \frac{dy}{dx} = \frac{1}{u} \cdot \sec^2 x = \frac{1}{\tan x} \cdot \sec^2 x \).

Step 6: Simplify the Result

Simplify the expression \( \frac{dy}{dx} = \frac{1}{\tan x} \cdot \sec^2 x \). Noting that \( \sec^2 x = 1 + \tan^2 x \), we have \( \frac{dy}{dx} = \frac{1 + \tan^2 x}{\tan x} = \frac{\sec^2 x}{\tan x} \).

Step 7: Finalize the Derivative Expression

Re-express the final derivative using trigonometric identities. We know \( \sec^2 x = \frac{1}{\cos^2 x} \) and \( \tan x = \frac{\sin x}{\cos x} \). Therefore, the derivative simplifies to \( \frac{dy}{dx} = \frac{1}{\sin x \cdot \cos x} = \frac{1}{\sin(2x)/2} = \frac{2}{\sin(2x)} \).

Key Concepts

Chain Rule

The Chain Rule is a fundamental technique in calculus used to differentiate composite functions. Simply put, it helps make derivatives of functions that are made by the combination of two or more functions. For a composite function like our current exercise, where the function is expressed as \(y = \ln(\tan x)\), you might see that this is a composition of \(u = \tan x\) and \(y = \ln(u)\). Each of these has its own derivative.\\Here's how the Chain Rule works in a few simple steps:

• First, differentiate the outer function. In our case, that function is \(\ln(u)\), which differentiates to \(\frac{1}{u}\).
• Next, take the derivative of the inner function \(u = \tan x\). This gives \(\sec^2 x\).
• Finally, multiply the two derivatives together to find \(\frac{dy}{dx}\).

This multiplication step is what the Chain Rule is all about: it combines the rate of change of the outer layer with the inner layer in a single operation, represented as \(\frac{dy}{dx} = \frac{du}{dx} \cdot \frac{dy}{du}\). This allows us to understand how the entire function changes with respect to \(x\).

Natural Logarithm

Natural logarithms, often denoted as \(\ln(x)\), are a special type of logarithm with base \(e\), where \(e\) is an important constant approximately equal to 2.718. Natural logarithms are a critical part of calculus and many real-world applications, from science to economics.\\When dealing with natural logarithms in differentiation, one of the key rules is that the derivative of \(\ln(u)\) with respect to \(u\) is \(\frac{1}{u}\). That simple rule is at the core of our problem of differentiating \(y = \ln(\tan x)\).\\Here's why natural logarithms are important:

• They transform multiplication into addition, which simplifies complex equations.
• They naturally arise in various models involving growth and decay, such as population dynamics and radioactive decay.
• Their derivative, \(\frac{1}{u}\), provides a straightforward method for finding rates of change in calculus.

Understanding how to differentiate expressions involving \(\ln(x)\) allows you to efficiently handle a wide array of mathematical problems that involve exponential growth patterns and compound interest formulas.

Trigonometric Functions

Trigonometric functions such as \(\sin\), \(\cos\), and \(\tan\) often appear in calculus, physics, and engineering due to their ability to model periodic phenomena like sound and light waves. In differentiation, trigonometric functions let us describe how angles and rotational motion change with time or position.\\In our context, we are looking at the \(\tan\) function. Differentiating \(\tan x\) gives us \(\sec^2 x\). Trigonometric identities can also help simplify complex expressions, as seen in solving the derivative of \(y = \ln(\tan x)\).\\Key points to understand about trigonometric functions in calculus:

• \(\tan x\) is defined as \(\frac{\sin x}{\cos x}\), which is especially useful when converting between different trigonometric expressions.
• The derivative of \(\tan x\), \(\sec^2 x = 1 + \tan^2 x\), is derived from the identity \(\sec x = \frac{1}{\cos x}\).
• These derivatives and identities simplify calculations involving rotations, wave functions, and harmonic motion.

Mastering both differentiation and simplification of trigonometric functions opens up ways to analyze and interpret complex wave patterns and oscillatory motions encountered in physics and engineering disciplines.