Problem 26 - Calculus Single Variable

Question

Differentiate the given expression with respect to \(x\). $$ \arctan (2 / x) $$

Step-by-Step Solution

Answer

The derivative is \(-\frac{2}{x^2+4}\).

Step 1: Identify the Function and Variable

The function to differentiate is \( \arctan(2/x) \). The external function is the arctangent, \( \arctan(u) \), where \( u = \frac{2}{x} \). Our task is to find the derivative \( \frac{d}{dx}(\arctan(2/x)) \).

Step 2: Use the Derivative of Arctangent

The derivative of \( \arctan(u) \) with respect to \( u \) is \( \frac{1}{1+u^2} \). So for \( \arctan(\frac{2}{x}) \), the derivative with respect to \( u \) is \( \frac{1}{1+(\frac{2}{x})^2} \).

Step 3: Simplify the Derivative

Simplify the expression \( \frac{1}{1+(\frac{2}{x})^2} \) resulting in \( \frac{1}{1+\frac{4}{x^2}} \). Further simplification gives \( \frac{x^2}{x^2+4} \).

Step 4: Chain Rule Application

Using the chain rule, differentiate \( u = \frac{2}{x} \) with respect to \( x \). The derivative of \( u \) is \( \frac{d}{dx}\left(\frac{2}{x}\right) = -\frac{2}{x^2} \).

Step 5: Combine the Results

Finally, combine the derivatives using the chain rule: \[ \frac{d}{dx}(\arctan(2/x)) = \frac{1}{1+(2/x)^2} \cdot \frac{d}{dx}(2/x) = \frac{x^2}{x^2+4} \cdot \left(-\frac{2}{x^2}\right) = -\frac{2}{x^2+4}.\]

Key Concepts

Chain Rule

The chain rule is a critical concept in calculus for finding the derivative of compositions of functions. When you have a function that is composed of two or more other functions, the chain rule allows you to differentiate it efficiently.

In our exercise, we are dealing with the function \( \arctan\left(\frac{2}{x}\right) \). To differentiate this function, we need to recognize it as a composition of the arctangent function and the expression \( \frac{2}{x} \). Here's a step-by-step view of applying the chain rule:

• First, identify the inner function, which is \( u = \frac{2}{x} \).
• The outer function, or the function which is acting on \( u \), is \( \arctan(u) \).

The chain rule states that the derivative of a composition of functions \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \). This means that we first find the derivative of the outer function \( \arctan(u) \) with respect to its input, \( u \), and then multiply it by the derivative of the inner function, \( \frac{2}{x} \), with respect to \( x \).

Applying the chain rule correctly requires a clear understanding of the functions involved, which is why breaking the expression into its components is such an important first step.

Arctangent Function

The arctangent function is one of the most commonly used inverse trigonometric functions. It is the inverse of the tangent function and is denoted as \( \arctan(x) \) or sometimes \( \tan^{-1}(x) \).

This function is defined for all real numbers and produces angle values in the range \(-\frac{\pi}{2} < y < \frac{\pi}{2}\). In the context of differentiation, the arctangent function shows up often when dealing with angle calculations or problems involving right triangles.

When differentiating a function involving \( \arctan(u) \), we use its well-known derivative formula:

• The derivative of \( \arctan(u) \) with respect to \( u \) is \( \frac{1}{1+u^2} \).

This formula is particularly important in solving calculus problems that involve inverse trigonometric functions, allowing us to break down and differentiate complex expressions like the one in our exercise.

Derivative of Inverse Trigonometric Functions

Inverse trigonometric functions, including arctangent, arcsine, and arccosine, have specific derivative rules that simplify their differentiation.

These derivatives are fundamental tools in calculus, particularly because they appear in various applications across scientific fields.

Let's look at their derivative formulas more closely:

• The derivative of \( \arcsin(u) \) with respect to \( u \) is \( \frac{1}{\sqrt{1-u^2}} \).
• The derivative of \( \arccos(u) \) with respect to \( u \) is \( -\frac{1}{\sqrt{1-u^2}} \).
• The derivative of \( \arctan(u) \) is \( \frac{1}{1+u^2} \), which is used in our exercise.

For our problem, applying the arctangent derivative formula was a key step to simplifying the differentiation process. These derivative formulas enable us to tackle complex differentiations with ease, providing a foundation for solving more advanced calculus problems efficiently.