Problem 60 - University Calculus: Early Transcendentals

Question

Derive the formula for the derivative with respect to \(x\) of a. \(\sec x\) b. csc \(x\) c. cot \(x\)

Step-by-Step Solution

Answer

a. Derivative of \(\sec x\) is \(\sec x \cdot \tan x\). b. Derivative of \(\csc x\) is \(-\csc x \cdot \cot x\). c. Derivative of \(\cot x\) is \(-\csc^2 x\).

Step 1: Understanding the Derivative of Trigonometric Functions

To find the derivative of trigonometric functions like \(\sec x, \csc x,\) and \(\cot x\), we start by recalling the known derivatives for \(\sin x\) and \(\cos x\). These are the foundation. We have \(\frac{d}{dx} (\sin x) = \cos x\) and \(\frac{d}{dx} (\cos x) = -\sin x\).

Step 2: Derivative of \(\sec x\) Using Quotient Rule

Recall that \(\sec x = \frac{1}{\cos x}\). To differentiate \(\sec x\), use the quotient rule: if \(y = \frac{u}{v}\), then \(\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\). Here, \(u = 1\) and \(v = \cos x\), so \(\frac{du}{dx} = 0\) and \(\frac{dv}{dx} = -\sin x\). Apply the formula: \[\frac{d}{dx}(\sec x) = \frac{\cos x \cdot 0 - 1 \cdot (-\sin x)}{(\cos x)^2} = \frac{\sin x}{\cos^2 x} = \sec x \cdot \tan x.\]

Step 3: Derivative of \(\csc x\) Using Quotient Rule

Recall that \(\csc x = \frac{1}{\sin x}\). Using the quotient rule, with \(u = 1\) and \(v = \sin x\), we find that \(\frac{du}{dx} = 0\) and \(\frac{dv}{dx} = \cos x\). The derivative is: \[\frac{d}{dx}(\csc x) = \frac{\sin x \cdot 0 - 1 \cdot \cos x}{(\sin x)^2} = -\frac{\cos x}{\sin^2 x} = -\csc x \cdot \cot x.\]

Step 4: Derivative of \(\cot x\) Using Quotient Rule

Recall that \(\cot x = \frac{\cos x}{\sin x}\). Applying the quotient rule here, \(u = \cos x\) and \(v = \sin x\), gives \(\frac{du}{dx} = -\sin x\) and \(\frac{dv}{dx} = \cos x\). Calculating the derivative: \[\frac{d}{dx}(\cot x) = \frac{\sin x(-\sin x) - \cos x(\cos x)}{(\sin x)^2} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = -\frac{1}{\sin^2 x} = -\csc^2 x.\]

Step 5: Summary of Results

Based on the calculations, the derivatives of the trigonometric functions are as follows: 1. \(\frac{d}{dx}(\sec x) = \sec x \cdot \tan x\)2. \(\frac{d}{dx}(\csc x) = -\csc x \cdot \cot x\)3. \(\frac{d}{dx}(\cot x) = -\csc^2 x\)

Key Concepts

Trigonometric Identities

Trigonometric identities are the foundational equations that describe various relationships among trigonometric functions. They play a crucial role in simplifying and solving problems involving trigonometric functions, particularly when differentiating. Understanding these identities helps recognize patterns and simplify expressions. Here are some fundamental trigonometric identities that are often used:

• Pythagorean Identity: \( \sin^2 x + \cos^2 x = 1 \)
• Reciprocal Identities:
  • \( \sec x = \frac{1}{\cos x} \)
  • \( \csc x = \frac{1}{\sin x} \)
  • \( \cot x = \frac{1}{\tan x} \)
• Quotient Identities:
  • \( \tan x = \frac{\sin x}{\cos x} \)
  • \( \cot x = \frac{\cos x}{\sin x} \)

These identities are vital whenever calculus work involves trigonometric functions since they make it possible to express functions in alternative forms, which can simplify differentiation.

Quotient Rule

The quotient rule is an essential technique in calculus differentiation when it comes to finding the derivative of a fraction. It's especially useful for functions that are expressed as a ratio, like those involving standard trigonometric functions. The quotient rule formula is:\[\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\]Where \(u\) and \(v\) are functions of \(x\). There are some critical points to remember when applying the quotient rule:

• Always identify which part of the function is \(u\) and which is \(v\).
• Separate the numerator's derivative from the denominator's.
• Pay attention to signs, especially when multiplying through the negative terms.
• Simplify your answer after applying the rule.

The quotient rule provides a framework to systematically handle these operations and derive accurate results.

Calculus Differentiation

Calculus differentiation involves finding the rate at which a function changes at any point. This concept is the backbone of calculus and applies to various fields like physics, engineering, and economics. Differentiation provides insights into the behavior of functions: increasing, decreasing, concavity, and critical points. Trigonometric functions, like any other real-valued function, can be differentiated using standard rules and identities. The process often involves:

• Recognizing the base function type.
• Applying derivative rules, like power, product, chain, and quotient rules.
• Utilizing trigonometric identities to simplify expressions before or after differentiation.

Understanding differentiation principles ensures accurate mathematical modeling and problem-solving capabilities.

Trigonometric Derivatives

Trigonometric derivatives specifically deal with the derivatives of trigonometric functions. These derivatives are crucial since trigonometric functions frequently appear in calculus problems. Here's a quick rundown of some key derivatives:

• The derivative of \(\sin x\) is \(\cos x\).
• The derivative of \(\cos x\) is \(-\sin x\).
• Moving on to other trig functions:
  • \(\sec x = \frac{1}{\cos x}\) leads to its derivative \(\sec x \cdot \tan x\).
  • \(\csc x = \frac{1}{\sin x}\) leads to its derivative \(-\csc x \cdot \cot x\).
  • \(\cot x = \frac{\cos x}{\sin x}\) leads to its derivative \(-\csc^2 x\).

Knowing these derivatives is essential for calculus differentiation problems involving trigonometric functions and aids in solving complex mathematical scenarios at ease.