Problem 62 - Calculus for Biology and Medicine

Question

Use the quotient rule to show that $$ \frac{d}{d x} \cot x=-\csc ^{2} x $$

Step-by-Step Solution

Answer

The derivative of \( \cot x \) is \(-\csc^2 x\) using the quotient rule.

Step 1: Understanding the Quotient Rule

The quotient rule is a method used in calculus to find the derivative of a function that is the quotient of two other functions. If you have a function \( y = \frac{u}{v} \), the derivative \( \frac{dy}{dx} \) is given by \( \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).

Step 2: Express \( \cot x \) as a Quotient

The cotangent function can be expressed as a quotient: \( \cot x = \frac{\cos x}{\sin x} \). In this form, \( u = \cos x \) and \( v = \sin x \).

Step 3: Differentiate \( u \) and \( v \)

Differentiate \( u = \cos x \) to get \( \frac{du}{dx} = -\sin x \). Differentiate \( v = \sin x \) to get \( \frac{dv}{dx} = \cos x \).

Step 4: Apply the Quotient Rule

Using the quotient rule, \( \frac{d}{dx} \left( \frac{\cos x}{\sin x} \right) \) is \( \frac{\sin x (-\sin x) - \cos x \cos x}{\sin^2 x} \). Simplify the numerator: \( -\sin^2 x - \cos^2 x \).

Step 5: Simplify Using Trigonometric Identity

Recall the identity \( \sin^2 x + \cos^2 x = 1 \). Thus, the numerator simplifies to \( -1 \). The derivative is \( \frac{-1}{\sin^2 x} \).

Step 6: Express in Terms of Cosecant

The cosecant function \( \csc x = \frac{1}{\sin x} \). Thus, \( \frac{-1}{\sin^2 x} = -\csc^2 x \). This confirms the derivative is \( \frac{d}{dx} \cot x = -\csc^2 x \).

Key Concepts

Derivative of Trigonometric Functions

Trigonometric functions, like sine, cosine, and cotangent, play a crucial role in calculus, especially when dealing with derivatives. The derivative of a function gives us the rate at which the function is changing at any point. Understanding the derivatives of trigonometric functions is key because they establish foundational patterns and rules that you will frequently encounter in calculus problems and applications.

When differentiating trigonometric functions, each function has a unique derivative. For instance, the derivative of \( \sin x \) is \( \cos x \), while the derivative of \( \cos x \) is \(-\sin x \). These derivatives provide a base for solving more complex derivatives involving trigonometric functions.

• \( \frac{d}{dx} \sin x = \cos x \)
• \( \frac{d}{dx} \cos x = -\sin x \)
• \( \frac{d}{dx} \tan x = \sec^2 x \)
• \( \frac{d}{dx} \csc x = -\csc x \cot x \)
• \( \frac{d}{dx} \sec x = \sec x \tan x \)
• \( \frac{d}{dx} \cot x = -\csc^2 x \)

The derivative \( \frac{d}{dx} \cot x = -\csc^2 x \) is particularly interesting because it showcases the interconnectedness of trigonometric functions and their derivatives, a theme common in calculus.

Calculus Differentiation Techniques

Calculus provides several techniques to differentiate functions efficiently, among which the quotient rule is a powerful tool. Differentiation involves finding the rate of change of one quantity concerning another, and several rules help simplify this process for complex functions.

The quotient rule is particularly useful when dealing with the division of two functions. Let's say we have a function \( y = \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions of \( x \). In such cases, the quotient rule states:

• \( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \)

This formula helps to find the derivative of a function that is a quotient of two others. A vital element of applying the quotient rule correctly is to ensure that both the numerator and the denominator functions are differentiated accurately and consistently. This approach maintains the integrity of the differentiation process and exhibits how calculus principles can be applied in various scenarios.

In our example, differentiating \( \cot x = \frac{\cos x}{\sin x} \) involves using the quotient rule efficiently. By identifying \( u = \cos x \) and \( v = \sin x \), and determining their individual derivatives, you're strategically solving the problem and calculating the derivative effectively.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value within their domains. These identities are extremely useful in calculus because they allow you to simplify complex expressions, especially when dealing with derivatives and integrals.

Some fundamental identities include the Pythagorean identity:

• \( \sin^2 x + \cos^2 x = 1 \)

This identity is crucial in simplifying expressions, as we saw when differentiating \( \cot x \). The relationship \( \sin^2 x + \cos^2 x = 1 \) is rearranged to find other useful forms, such as \( \cos^2 x = 1 - \sin^2 x \).

• Cosecant and secant identities: \( \csc x = \frac{1}{\sin x} \) and \( \sec x = \frac{1}{\cos x} \).

In calculus, knowing these identities and how to apply them is pivotal. They help streamline the differentiation of more complex trigonometric expressions and support the process of deriving other identities or results, like showing \( \frac{d}{dx} \cot x = -\csc^2 x \) clearly and succinctly. Recognizing and applying these identities not only simplifies the work but also deepens the understanding of how trigonometric functions interact.