Problem 10 - Calculus Single Variable

Question

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

Step-by-Step Solution

Answer

The derivative is \( \cot(x) - x \csc^2(x) \).

Step 1: Rewrite the Expression

First, rewrite the given expression \( \frac{x}{\tan(x)} \) as a product. Since \( y = \frac{x}{\tan(x)} \), it can be rewritten as \( y = x \cdot \cot(x) \), where \( \cot(x) = \frac{1}{\tan(x)} \).

Step 2: Product Rule Introduction

The expression is a product of two functions, \( x \) and \( \cot(x) \). To differentiate it, we need to use the product rule which states if \( y = u \cdot v \), then \( \frac{dy}{dx} = u'v + uv' \). Identify \( u = x \) and \( v = \cot(x) \).

Step 3: Differentiate Each Function

Differentiate \( u \) and \( v \) with respect to \( x \):- \( u = x \) implies \( u' = 1 \).- \( v = \cot(x) \) implies \( v' = -\csc^2(x) \) using the derivative of cotangent.

Step 4: Apply the Product Rule

Now apply the product rule: \[ \frac{dy}{dx} = u'v + uv' = (1)(\cot(x)) + (x)(-\csc^2(x)) \] This simplifies to:\[ \frac{dy}{dx} = \cot(x) - x \csc^2(x) \]

Step 5: Simplify the Result

The simplified derivative of the expression \( \frac{x}{\tan(x)} \) is \( \cot(x) - x \csc^2(x) \). Ensure that this expression is simplified and all trigonometric identities are correctly applied.

Key Concepts

Understanding the Product Rule

The product rule is a fundamental tool in differentiation, especially useful when handling expressions where two functions are multiplied together. In essence, it allows us to find the derivative of a product of two simpler functions. The product rule states that if you have a function \( y = u \cdot v \), where \( u \) and \( v \) are both functions of \( x \), then its derivative is given by

• \( \frac{dy}{dx} = u'v + uv' \)

Here, \( u' \) is the derivative of \( u \) with respect to \( x \), and \( v' \) is the derivative of \( v \) with respect to \( x \). For example, if we consider the problem where \( y = x \cdot \cot(x) \), we identify \( u = x \) and \( v = \cot(x) \). Differentiating \( u \) gives \( u' = 1 \). Differentiating \( v \) yields \( v' = -\csc^2(x) \). Applying the product rule helps us combine these results to find the total derivative. This rule is especially helpful when expressions are not straightforward products of constants and variables.

Exploring Trigonometric Functions

Trigonometric functions are a vital part of calculus, particularly when dealing with periodic functions and their derivatives. Common trigonometric functions include sine, cosine, and tangent, along with their reciprocals: cosecant (\( \csc(x) \)), secant (\( \sec(x) \)), and cotangent (\( \cot(x) \)), which are frequently seen in calculus problems. Understanding their derivatives is crucial:

• The derivative of \( \tan(x) \) is \( \sec^2(x) \).
• The derivative of \( \cot(x) \) is \( -\csc^2(x) \).

These identities are important for simplifying expressions and finding derivatives in problems like differentiating \( \frac{x}{\tan(x)} \). Recognizing and using these derivatives makes it easier to break down and solve complex expressions, particularly when combined with other calculus rules, like the product or chain rule.

The Chain Rule in Differentiation

The chain rule is an essential technique in calculus, used for differentiating compositions of functions, i.e., functions within functions. It comes into play when differentiating expressions where a function is nested inside another. The chain rule is expressed as follows:

• If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

This means we first differentiate the outer function, leaving the inner function alone, then multiply by the derivative of the inner function. In contexts involving trigonometric functions, the chain rule is employed when these functions are part of a larger composite function. Though the original exercise involving \( \frac{x}{\tan(x)} \) primarily utilized the product rule, the chain rule pairs effectively in more complex scenarios where multiple layers of nested functions exist.
Understanding both the product and chain rules, alongside the derivatives of trigonometric functions, equips students with a robust toolkit for tackling a wide range of differentiation problems effectively.