Problem 6 - Calculus

Question

Find \(D_{x} y .\) \(y=\csc x=1 / \sin x\)

Step-by-Step Solution

Answer

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

Step 1: Understanding the Function

The function given is the cosecant, which is the reciprocal of the sine function, defined as \( y = \csc x = \frac{1}{\sin x} \). Our task is to find the derivative of this function with respect to \(x\).

Step 2: Using the Derivative Rule for Reciprocals

We can start by finding the derivative of a reciprocal function \( y = \frac{1}{u(x)} \). The derivative is \(-\frac{u'(x)}{(u(x))^2} \), where \( u(x) = \sin x \) in this problem.

Step 3: Finding the Derivative of Sine

The derivative of \( \sin x \) is \( \cos x \). So, \( u'(x) = \cos x \).

Step 4: Substituting into the Derivative Formula

By substituting \( u(x) = \sin x \) and \( u'(x) = \cos x \) into the reciprocal derivative formula \(-\frac{u'(x)}{(u(x))^2} \), we get \(-\frac{\cos x}{(\sin x)^2} \).

Step 5: Simplifying the Expression

The expression \(-\frac{\cos x}{(\sin x)^2} \) can be simplified by recognizing that \( \frac{1}{(\sin x)^2} = \csc^2 x \). Therefore, the simplified form of the derivative is \( -\cos x \cdot \csc^2 x \).

Step 6: Conclusion

By using the rules of derivatives combined with trigonometric identities, we've found that the derivative of \(y = \csc x\) is \(- \cos x \cdot \csc^2 x \), which is equivalent to \(- \csc x \cdot \cot x \).

Key Concepts

Cosecant Function

The cosecant function, abbreviated as \( \csc x \), is one of the six primary trigonometric functions. It is defined as the reciprocal of the sine function. In mathematical terms, this is expressed as \( \csc x = \frac{1}{\sin x} \).

Understanding the cosecant function is important, particularly when dealing with its derivative. It bears a unique relationship with other trigonometric functions which can be quite useful. While \( \sin x \) is a fundamental trigonometric function representing the y-coordinate of a point on the unit circle, \( \csc x \) is used when dealing with right triangles in trigonometry.

One important thing to note is that the domain of the cosecant function excludes values where \( \sin x = 0 \), because division by zero is undefined. These values occur at integer multiples of \( \pi \), where the sine is zero.

Reciprocal Functions

Reciprocal functions are a significant concept in calculus and trigonometry. They involve taking one function and creating a new function by taking the reciprocal, which is \( \frac{1}{\underline{\phantom{xx}}} \) of the original function.

In the case of trigonometric functions, taking reciprocals can provide tools for simplification and solving problems. For instance:

• \( \csc x \) is the reciprocal of \( \sin x \)
• \( \sec x \) is the reciprocal of \( \cos x \)
• \( \cot x \) is the reciprocal of \( \tan x \)

When differentiating reciprocal functions, a specific rule is used. The rule for the derivative of a reciprocal function \( y = \frac{1}{u(x)} \) is \( -\frac{u'(x)}{(u(x))^2} \), where \( u(x) \) is a function of \( x \). Applying this rule allows us to find the derivative of \( \csc x \) by replacing \( u(x) \) with \( \sin x \).

This highlights a broader application in calculus, where finding the derivative of reciprocal functions is a useful skill.

Trigonometric Identities

Trigonometric identities play a critical role when simplifying expressions involving trigonometric functions, and they are essential when dealing with derivatives, like that of the cosecant function.

These identities help in rewriting and simplifying trigonometric expressions, making calculations more manageable. Some fundamental trigonometric identities include:

• \( \sin^2 x + \cos^2 x = 1 \)
• \( 1 + \tan^2 x = \sec^2 x \)
• \( 1 + \cot^2 x = \csc^2 x \)

In the original problem, understanding that \( \frac{1}{(\sin x)^2} \) simplifies to \( \csc^2 x \) through these identities allows us to rewrite the derivative of \( \csc x \) in terms of familiar trigonometric functions.

Applying these identities is crucial for simplification, and makes connections between different trigonometric functions clear and applicable in various problems in calculus and beyond.