Problem 28 - Calculus Single Variable

Question

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

Step-by-Step Solution

Answer

The derivative of \(\frac{\arctan(x)}{x}\) is \(\frac{x - \arctan(x)(1+x^2)}{x^2(1+x^2)}\).

Step 1: Understand the Problem

We need to differentiate the function \(f(x) = \frac{\arctan(x)}{x}\) with respect to \(x\). This involves using the quotient rule for differentiation, as the function is in the form of a fraction \(u(x)/v(x)\).

Step 2: Identify the Components of the Quotient Rule

The quotient rule states that for a function \(\frac{u(x)}{v(x)}\), the derivative \(f'(x)\) is given by \(\frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}\). Here, \(u(x) = \arctan(x)\) and \(v(x) = x\).

Step 3: Differentiate the Numerator

Find the derivative of \(u(x) = \arctan(x)\). The derivative \(u'(x)\) is \(\frac{d}{dx} [\arctan(x)] = \frac{1}{1+x^2}\).

Step 4: Differentiate the Denominator

Find the derivative of \(v(x) = x\). The derivative \(v'(x)\) is \(\frac{d}{dx} [x] = 1\).

Step 5: Apply the Quotient Rule

Now apply the quotient rule: \[ f'(x) = \frac{x \cdot \frac{1}{1+x^2} - \arctan(x) \cdot 1}{x^2} = \frac{1}{x(1+x^2)} - \frac{\arctan(x)}{x^2}. \]

Step 6: Simplify the Result

Combine the two fractions over a common denominator \(x^2(1+x^2)\): \[ f'(x) = \frac{1 \cdot x - \arctan(x)(1+x^2)}{x^2(1+x^2)} = \frac{x - \arctan(x)(1+x^2)}{x^2(1+x^2)}. \]

Key Concepts

Trigonometric Functions

Trigonometric functions are fundamental in understanding various aspects of mathematics and its applications. These functions include sine, cosine, tangent, and their inverse functions. The inverse trigonometric functions, such as \( \arctan(x)\), are particularly important when you want to retrieve angles from ratio values.
The function \(\arctan(x)\) gives the angle whose tangent is \(x\). When working with trigonometric functions, it's helpful to remember:

• The domain of the tangent function is all real numbers except where \(\tan(x)\) is undefined, such as \(x = \frac{\pi}{2} + n\pi\) for any integer \(n\).
• The range of \(\arctan(x)\) is limited to \((-\frac{\pi}{2}, \frac{\pi}{2})\).
• Inverse trigonometric functions are crucial in calculus, as they often appear in integration and differentiation scenarios.

Understanding these properties allows you to effectively handle various mathematical problems that involve trigonometric inverse functions.

Differentiation Techniques

Differentiation is a powerful tool in mathematics used to find the rate at which a quantity changes. Various techniques can simplify this process, one of the most significant of which is the quotient rule. The quotient rule is essential when differentiating functions presented as one function divided by another, like our example function \(f(x) = \frac{\arctan(x)}{x}\).
The quotient rule formula, \( \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}\), may seem complex at first, but it is straightforward when broken down. Here:

• \(u(x)\) is the numerator function (\(\arctan(x)\))
• \(v(x)\) is the denominator function (\(x\))
• \(u'(x)\) and \(v'(x)\) are the derivatives of those functions

In our problem, the derivative of \(\arctan(x)\) is \(\frac{1}{1+x^2}\), while the derivative of \(x\) is \(1\). These derivatives align with the basic formula components, working together to find the derivative of the entire function.

Derivative of Inverse Functions

Finding the derivative of inverse functions, like \(\arctan(x)\), is a key skill in calculus. The derivative of an inverse function gives insights into the slope of the tangent to the curve at a given point.
To derive \(\arctan(x)\), you apply the formula for the derivative of inverse trigonometric functions. Specifically, \(\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}\). This formula is crucial in calculating the rate of change for the angle \(\theta\) that has a tangent value of \(x\).
While solving derivative problems with inverse functions, keep in mind:

• Inverse functions reverse the operation of a trigonometric function, mapping ratios back to angles.
• You can often leverage properties of trigonometric identities to simplify problems further.
• The derivative signals how quickly the function value changes concerning the change in \(x\), providing direct insight into function behavior.

Understanding these concepts can significantly enhance your ability to solve challenges involving inverse functions in calculus.