Question

Differentiate the following functions. $$ y=\csc x . \quad \frac{d y}{d x}=\csc ^{2} x \cos x $$

Short answer

Question: Find the derivative of the function $$y = \csc x$$ with respect to x. Answer: The derivative of the function $$y = \csc x$$ with respect to x is $$\frac{dy}{dx} = -\csc^2 x \cos x$$.

Step-by-step solution

Rewrite the function in terms of its reciprocal

Since the cosecant function is $$\csc x = \frac{1}{\sin x}$$, we can rewrite the given function as: $$ y = \frac{1}{\sin x} $$

Apply the chain rule

The chain rule states that if a function is a composition of two functions, then the derivative of the function is: $$ \frac{dy}{dx} = \frac{dy}{du} . \frac{du}{dx} $$ Since our function is $$y = \frac{1}{\sin x}$$, we can set $$u = \sin x$$ and $$y = \frac{1}{u}$$. Now, differentiate each part with respect to x and u respectively: $$ \frac{dy}{du} = -\frac{1}{u^2} = -\frac{1}{(\sin x)^2} = -\csc^2 x $$ and $$ \frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x $$ Now, apply the chain rule to find $$\frac{dy}{dx}$$: $$ \frac{dy}{dx} = \frac{dy}{du} . \frac{du}{dx} = -\csc^2 x . \cos x $$

Simplify and write the final answer

The final answer for the derivative of the given function is $$ \frac{dy}{dx} = -\csc^2 x \cos x $$

Key concepts

Chain Rule

Understanding the chain rule is essential in calculus, especially when dealing with complex functions composed of multiple smaller functions. The chain rule allows us to differentiate a compound function where one function is nested inside another. It's like peeling off layers to get to the core, with each layer representing a separate function.

Here's the basic idea:

• Suppose you have a function that can be expressed as \( y = f(g(x)) \), where \( g(x) \) is inside \( f(x) \).
• The chain rule states that the derivative of this function with respect to \( x \) is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \), where \( g(x) \) is the inner function, and \( f(x) \) is the outer function.
• In the context of our exercise, \( y = \frac{1}{\sin x} \) is a composition of functions, where \( u = \sin x \) and \( y = \frac{1}{u} \).
• First, take the derivative of the outer function \( y = \frac{1}{u} \) with respect to \( u \), then multiply it by the derivative of the inner function \( u = \sin x \) with respect to \( x \).

By breaking down the problem using the chain rule, differentiating complex functions becomes more manageable and systematic.

Trigonometric Functions

Trigonometric functions like sine, cosine, and cosecant have unique derivatives that are important to memorize in calculus. These functions frequently appear in calculus problems, especially when we deal with angles and periodic phenomena. Understanding their properties and derivatives is crucial to solving these kinds of problems effectively.

Key points about trigonometric functions:

• The cosecant function, \( \csc x = \frac{1}{\sin x} \), is the reciprocal of the sine function.
• Its derivative can be found using identities and the chain rule, as shown in the exercise.
• The basic trigonometric derivatives that are commonly used include:
  • \( \frac{d}{dx} (\sin x) = \cos x \)
  • \( \frac{d}{dx} (\cos x) = -\sin x \)
  • \( \frac{d}{dx} (\csc x) = -\csc x \cot x \) or, through decomposition, as in our problem.

Understanding these derivatives and how to manipulate trigonometric identities can help simplify calculus problems and lead to accurate solutions.

Calculus Problem Solving

Solving calculus problems often involves strategizing on how to apply various methods and rules effectively. An essential part of solving any differential calculus problem is determining which differentiation rule to use based on the function's structure. There are different types of rules in calculus, such as product rule, quotient rule, and chain rule, each suited for different kinds of function compositions.

Effective calculus problem solving involves:

• Identifying the type of problem or the structure of the function given.
• Choosing the correct rule or method to apply, like the chain rule for nested functions.
• Applying mathematical identities or transformations to simplify the function, such as rewriting \( \csc x \) as \( \frac{1}{\sin x} \) to apply differentiation rules easily.
• Simplifying the derivative to its final form by performing algebraic manipulations.
• Checking if the derivative simplifies to a recognizable form, as shown when simplifying to \( -\csc^2 x \cos x \) in the given solution.

By practicing these strategies, you become more skilled at spotting patterns and choosing the most efficient methods to solve calculus problems. Remember, practice and familiarity with different functions and their derivatives is key to mastering calculus problem solving.