Problem 32 - Calculus Single Variable

Question

Differentiate the given expression with respect to \(x\). $$ \operatorname{arccsc}(\sin (x)) $$

Step-by-Step Solution

Answer

The derivative of \( \operatorname{arccsc}(\sin(x)) \) is \(-\csc(x) \cdot \cos(x)\).

Step 1: Write Down the Expression

We are given the expression \( \operatorname{arccsc}(\sin(x)) \). We need to differentiate this expression with respect to \( x \).

Step 2: Use the Inverse Trigonometric Function Formula

Recall that the derivative of \( \operatorname{arccsc}(u) \) with respect to \( u \) is \( -\frac{1}{|u|\sqrt{u^2 - 1}} \). In our case, \( u = \sin(x) \).

Step 3: Differentiate the Inner Function

The next step involves differentiating the inner function \( \sin(x) \) with respect to \( x \). The derivative is given by \( \cos(x) \).

Step 4: Apply the Chain Rule

The Chain Rule states that \( \frac{d}{dx}f(g(x)) = f'(g(x))g'(x) \). So the derivative of \( \operatorname{arccsc}(\sin(x)) \) with respect to \( x \) is \( -\frac{1}{|\sin(x)|\sqrt{\sin^2(x) - 1}} \cdot \cos(x) \).

Step 5: Simplify the Derivative

We also know from trigonometric identities that \( \sin^2(x) = 1 - \cos^2(x) \). Thus, \( \sin^2(x) - 1 = -\cos^2(x) \). Therefore, \( \sqrt{\sin^2(x) - 1} = i\cos(x) \).However, since the inverse cosecant only exists for values where \(|u| > 1\), simplifying the imaginary unit is typically not shown except for further theoretical mathematical exploration. We rely on the restriction \( |\sin(x)| \to 1/x \).

Step 6: Final Answer Deduction

Given the nature of \( |u| \) conditions, instead, remember practical deduction: this calculated algebra yields mathematical principles. It simplifies to: -\frac{\cos(x)}{|\sin(x)|\cos(x)} = -csc(x).Essentially, when approached through trigonometric principle evaluations.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are crucial when dealing with trigonometric values in reverse. They allow us to find angles with a given trigonometric value. In this problem, we focus particularly on the inverse cosecant function, denoted as \( \operatorname{arccsc}(x) \).
It is important to remember that the domain for \( \operatorname{arccsc}(x) \) is restricted to \( x \leq -1 \) or \( x \geq 1 \) since the cosecant function itself, \( csc(x) = \frac{1}{\sin(x)} \), does not take values between -1 and 1. Thus, when finding derivatives related to \( \operatorname{arccsc} \), we should be mindful of these restrictions.
The derivative of \( \operatorname{arccsc}(u) \), where \( u \) is a function of \( x \), is given by \( -\frac{1}{|u|\sqrt{u^2 - 1}} \). This result becomes fundamental when applying it to more complex nested functions, as we shall see through the chain rule.

Chain Rule

The chain rule is a powerful tool in calculus that helps differentiate composite functions. Essentially, when you have a function inside another function, the chain rule simplifies the process of finding the derivative.
In mathematical terms, if \( y = f(g(x)) \), then the derivative of \( y \) with respect to \( x \) is \( f'(g(x)) \cdot g'(x) \). This means you differentiate the outer function (\( f \)) with respect to its input (\( g(x) \)), and then multiply by the derivative of the inner function (\( g \)) with respect to \( x \).
For our exercise, by using the chain rule, we recognized that with \( \operatorname{arccsc}(\sin(x)) \), the inner function is \( \sin(x) \), and its derivative is \( \cos(x) \). This recognition allows us to accurately compute the derivative of the composite function efficiently.

Trigonometric Identities

Trigonometric identities are useful mathematical equations that relate the angles and sides of triangles. They help simplify complex trigonometric expressions, especially in calculus problems.
One important identity we use here is \( \sin^2(x) + \cos^2(x) = 1 \). We can rearrange this identity to show that \( \sin^2(x) = 1 - \cos^2(x) \). This fundamental relationship provides the means to simplify the radical expression \( \sqrt{\sin^2(x) - 1} \) into \( i\cos(x) \) under complex number contexts. However, due to the constraints of inverse trigonometric functions, the practical simplifications typically used do not showcase the imaginary component unless dealing in theoretical math realms.

• European trigonometry values and identities allow for clearer manipulation as seen in trigonometric function derivative calculations.

Recognizing the area of application is key, particularly prolonged understanding regarding where and why certain functions do not exist in certain ranges.

Calculus

Calculus serves as the foundation of advanced mathematics, tackling rates of change and the accumulation of quantities, respectively represented by differentiation and integration.
With differentiation, we seek to find the rate of change of a function. In this exercise, we aim to find how \( \operatorname{arccsc}(\sin(x)) \) changes concerning \( x \). By understanding the role of the chain rule, inverse trigonometric functions, and trigonometric identities in this context, calculus becomes a cohesive and comprehensive tool.
Calculus allows us not just to perform these operations but to understand the implications of change within various functions. It's about recognizing patterns, making predictions, and understanding the behavior of equations within psuedo-real number spaces, nurturing a deeper comprehension and appreciation for mathematics.