Problem 34 - Thomas Calculus

Question

In Exercises \(21-42,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\tan ^{-1}(\ln x) $$

Step-by-Step Solution

Answer

The derivative is \( \frac{1}{x(1+(\ln x)^2)} \).

Step 1: Identify the Function Form

The function given is \( y = \tan^{-1}(\ln x) \). This means that \( y \) is the arctangent of \( \ln x \). Our goal is to find the derivative of \( y \) with respect to \( x \).

Step 2: Use the Chain Rule

To differentiate \( y = \tan^{-1}(\ln x) \), we need to use the chain rule. Recall that if \( y = \tan^{-1}(u) \), then \( \frac{dy}{du} = \frac{1}{1+u^2} \). First, differentiate with respect to the inner function \( u = \ln x \).

Step 3: Differentiate the Outer Function

Differentiate \( y = \tan^{-1}(u) \) with respect to \( u \). We have:\[\frac{dy}{du} = \frac{1}{1+u^2}\]Substitute \( u = \ln x \) into this derivative to get \( \frac{1}{1+ (\ln x)^2} \).

Step 4: Differentiate the Inner Function

Now differentiate \( u = \ln x \) with respect to \( x \). We know \( \frac{du}{dx} = \frac{1}{x} \).

Step 5: Combine Results Using the Chain Rule

Now, apply the chain rule: \[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{1+(\ln x)^2} \cdot \frac{1}{x}\]This simplifies to:\[\frac{dy}{dx} = \frac{1}{x(1+(\ln x)^2)}\]

Step 6: Finalize the Expression

Combine all parts to write the derivative concisely as a single expression. The derivative of \( y \) with respect to \( x \) is:\[\frac{dy}{dx} = \frac{1}{x(1+(\ln x)^2)}\]

Key Concepts

Chain Rule

The chain rule is a vital technique in calculus used to differentiate composite functions. When you have a function within another function, the chain rule helps you find the derivative.
Here's the basic formula: if you have a composite function \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). This means you first differentiate the outer function and then multiply it by the derivative of the inner function.

• First, identify the inner and outer functions.
• Differentiate each function separately.
• Multiply the derivatives as per the chain rule formula.

In this exercise, the outer function is \( y = \tan^{-1}(u) \) and the inner function is \( u = \ln x \).
This structured approach ensures accuracy when working on complex problems.

Inverse Trigonometric Functions

Inverse trigonometric functions allow you to work backward from an angle to find the value of the ratio of sides of a right triangle. The function \( \tan^{-1}(x) \) returns the angle whose tangent is \( x \). These functions are essential in numerous mathematical calculations.
For differentiation, each inverse trigonometric function has a specific formula. For instance, the derivative of \( y = \tan^{-1}(u) \) is \( \frac{1}{1+u^2} \).

• Understand the basic function profile.
• Familiarize with the derivative formulas.

These derivatives are crucial to applying the chain rule in complex functions like \( y = \tan^{-1}(\ln x) \).

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is a logarithm to the base \( e \), where \( e \) is approximately 2.718. It plays a key role in calculus, particularly when dealing with exponential growth and decay.
The derivative of the natural logarithm is relatively simple: \( \frac{d}{dx}(\ln x) = \frac{1}{x} \). Knowing this makes it easier to handle calculus problems involving \( \ln x \).

• Recognize the relationship between exponential functions and natural logs.
• Use the derivative \( \frac{1}{x} \) in chain rule applications.

This understanding is essential when differentiating composite functions like \( \tan^{-1}(\ln x) \).

Differentiation

Differentiation is the process of finding the derivative, or rate of change, of a function with respect to a variable. It helps us understand how a function behaves as its inputs change.
Here are some key points:

• Differentiation finds the slope of a tangent line to the graph of a function.
• It is used to determine maxima, minima, and points of inflection in graphs.
• Effective in solving problems involving motion, growth, and optimization.

In the exercise \( y = \tan^{-1}(\ln x) \), differentiation is performed not just once, but in layers, using the chain rule. This hierarchical approach shows the flexibility and power of differentiation in calculus.

Arctangent

The arctangent, or inverse tangent function, \( \tan^{-1}(x) \), is used to retrieve the angle value from a given tangent ratio. It's useful in scenarios where you need to reverse engineer angles from ratios.
When differentiating, the derivative of \( \tan^{-1}(u) \) is crucial, as seen in the exercise. Knowing that we require \( \frac{1}{1+u^2} \) makes it feasible to apply it in chain rule steps.

• The function generally outputs an angle in radians.
• Vital for trigonometry-related differentiation.

This function's ability to transform a ratio back into an angle, combined with its well-defined derivative, makes it indispensable in calculus, especially within composite functions.