Problem 207 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We are given the function to derive: \( rac{d}{d x}( ext{csc } x) \). We need to find its derivative using the quotient rule.

Step 2: Express Cosecant as a Quotient

The cosecant of \( x \) is reciprocally expressed as \( ext{csc } x = \frac{1}{ ext{sin } x} \). Now we can apply the quotient rule for derivatives, which works with terms in the form \( \frac{u}{v} \).

Step 3: Apply the Quotient Rule

The quotient rule states \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \). Here, set \( u = 1 \) and \( v = \sin x \).

Step 4: Derivatives of u and v

Calculate \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):- \( u = 1 \) implies \( \frac{du}{dx} = 0 \).- \( v = \sin x \) implies \( \frac{dv}{dx} = \cos x \).

Step 5: Substitute Values into Quotient Rule

Insert the derivatives and the function values into the quotient rule formula:\[ \frac{d}{dx}\left(\frac{1}{\sin x}\right) = \frac{\sin x \cdot 0 - 1 \cdot \cos x}{(\sin x)^2} \]

Step 6: Simplify the Expression

Simplify the expression:\[ \frac{d}{dx}\left(\text{csc } x\right) = \frac{-\cos x}{(\sin x)^2} \]

Step 7: Express in Terms of Csc and Cot

Rewrite the simplified expression using trigonometric identities:- \( \text{csc } x = \frac{1}{\sin x} \)- \( \text{cot } x = \frac{\cos x}{\sin x} \)Thus,\[ \frac{-\cos x}{(\sin x)^2} = -\text{csc } x \cdot \text{cot } x \]

Step 8: Final Result

The derivative of \( \text{csc } x \) is \( -\text{csc } x \cdot \text{cot } x \). This matches the given expression we needed to derive.

Key Concepts

Trigonometric Derivatives

When it comes to calculus, derivatives of trigonometric functions are essential, especially in analyzing periodic behaviors. Calculus not only helps in understanding the rate of change of such functions but also provides insights into their graph behaviors. In the context of trigonometric derivatives, we navigate through functions such as sine, cosine, tangent, and their reciprocals. Broadly speaking, the derivatives of these functions follow a specific pattern and are vital for applications in physics, engineering, and even in economic models.
To effectively engage with trigonometric derivatives, remember the basic functions:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \( -\sin x \).
• The derivative of \( \tan x \) is \( \sec^2 x \).
• The derivative of \( \csc x \), which we'll explore further.

Understanding these basics, one can derive more complex trigonometric functions, especially those involving reciprocal identities.
Beyond memorizing derivatives, it's essential to understand the underlying rationale, as seen through various calculus rules, which allow us to derive these formulas efficiently. This forms a foundational aspect for students growing in mathematical prowess.

Cosecant Function

The cosecant function, represented as \( \csc x \), is essential in understanding trigonometry's depth. It is the reciprocal of the sine function, defined as \( \csc x = \frac{1}{\sin x} \). This relationship tells us a lot about its properties—the intervals where sine equals zero, \( \csc x \) becomes undefined.
Learning the cosecant function requires recognizing its reciprocal nature. This characteristic plays a crucial role when calculating derivatives, as seen in our original problem. Since it involves trigonometric relationships, the cosecant function surfaces in various practical scenarios, such as wave analysis in physics.
Furthermore, the graphical representation of \( \csc x \) reveals asymptotes where the sine function crosses zero. Understanding these behaviors gives insight into solving differential equations involving trigonometric functions. Hence, the cosecant is indispensable in trigonometry and calculus, adding depth to mathematical analyses.

Derivative Calculation

Finding the derivative of a trigonometric function, such as \( \csc x \), utilizes fundamental calculus tools like the quotient rule. The quotient rule offers a systematic way to differentiate functions expressed as a quotient of two simpler functions. For the cosecant function, recapping the process gives insight into the derivative's behavior:
First, recognize \( \csc x \) as \( \frac{1}{\sin x} \) so it can be differentiated as a quotient. Here, applying the quotient rule is essential. The rule itself is defined as:

• If \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \).

Given \( u = 1 \) and \( v = \sin x \), we find:

• \( \frac{du}{dx} = 0 \) since \( u \) is constant.
• \( \frac{dv}{dx} = \cos x \) since the derivative of \( \sin x \) is \( \cos x \).

Substituting these in the quotient rule yields:

• \( \frac{d}{dx} \left( \frac{1}{\sin x} \right) = \frac{-\cos x}{(\sin x)^2} \).

Simplifying further using trigonometric identities, we reach the derivative \( -\csc x \cdot \cot x \). This precise calculation using calculus principles gives students a robust method to handle derivatives effectively.