Problem 205 - Calculus Volume 1

Question

For the following exercises, use the quotient rule to derive the given equations. $$\frac{d}{d x}(\cot x)=-\csc ^{2} x$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Function to Derive

The function given is \(\cot x\), which is also written as \(\frac{\cos x}{\sin x}\). Therefore, we will use the quotient rule to find its derivative.

Step 2: Recall the Quotient Rule Formula

The quotient rule states that for functions \(u(x)\) and \(v(x)\), the derivative of \(\frac{u}{v}\) is given by:\[\frac{d}{d x}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\] where \(u(x) = \cos x\) and \(v(x) = \sin x\).

Step 3: Find the Derivatives of the Numerator and Denominator

Calculate \(\frac{du}{dx}\) and \(\frac{dv}{dx}\):- \(\frac{du}{dx} = \frac{d}{dx}(\cos x) = -\sin x\)- \(\frac{dv}{dx} = \frac{d}{dx}(\sin x) = \cos x\)

Step 4: Apply the Quotient Rule

Substitute \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\) into the quotient rule:\[\frac{d}{dx}(\cot x) = \frac{\sin x \cdot (-\sin x) - \cos x \cdot \cos x}{\sin^2 x}\] This simplifies to:\[\frac{-\sin^2 x - \cos^2 x}{\sin^2 x}\]

Step 5: Simplify Using the Pythagorean Identity

Recall the Pythagorean identity: \(\sin^2 x + \cos^2 x = 1\). Rewrite the expression as:\[\frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = \frac{-1}{\sin^2 x}\] This gives:\(-\csc^2 x\), because \(\csc x = \frac{1}{\sin x}\).

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus. They measure how a function changes as its input changes. In simple terms, a derivative tells us the slope of the tangent line to the graph of a function at any given point. This concept is essential because it allows us to analyze and predict the behavior of functions in a precise way.

• The derivative of a function at a given point informs how the function's value will increase or decrease near that point.
• The process of finding a derivative is called differentiation.
• Using standard rules for differentiation, we can find derivatives for many types of functions, including polynomial, exponential, logarithmic, and trigonometric functions.

Understanding derivatives helps solve problems in various fields like physics, economics, engineering, and beyond. For example, in physics, derivatives can represent velocity and acceleration when dealing with motion.

Trigonometric Functions and Their Properties

Trigonometric functions are mathematical functions related to the angles and ratios found in triangles. The most common trigonometric functions include sine (\( sin x \)), cosine (\( cos x \)), and tangent (\( tan x \)). Cotangent (\( cot x \)) is another important trigonometric function, defined as the reciprocal of the tangent.

• Key identities involving trigonometric functions aid in simplifying complex expressions. A well-known identity is the Pythagorean identity: \( sin^2 x + cos^2 x = 1 \).
• Cotangent can be expressed in terms of sine and cosine: \( cot x = \frac{cos x}{sin x} \).
• Cosecant (\( csc x \)) is the reciprocal of sine: \( csc x = \frac{1}{sin x} \).

Trigonometric functions are not just theoretical; they are widely used in modeling periodic phenomena such as sound waves, light waves, and seasonal temperatures. Mastery of these functions and their properties is crucial for advanced studies in mathematics and physics.

Exploring Differentiation Techniques: The Quotient Rule

Differentiation techniques allow us to find the derivatives of more complex functions, especially when simple rules like the power rule or product rule are not applicable. The quotient rule is particularly useful when differentiating functions that are divided by each other, like in this exercise.

• The quotient rule states:\( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).
• This rule is applied when you have a function that is the quotient of two differentiable functions, \( u(x) \) and \( v(x) \).
• In practice, apply the rule by differentiating the numerator and the denominator separately, then substituting these derivatives back into the formula.
• Simplifying the resulting expression often involves identities or algebraic manipulation to express the derivative in a standard form.

The power of the quotient rule lies in its ability to handle divisions in complex functions without having to simplify them first. Understanding how and when to use this rule efficiently is crucial for solving challenging calculus problems.