Problem 288 - Calculus Volume 1

Question

Find \(\frac{d y}{d x}\) for the given function. $$ y=x \cdot \csc ^{-1} x $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \csc^{-1} x - \frac{1}{\sqrt{x^2 - 1}} \)

Step 1: Identify the Function Composition

We notice that the function provided is a product of two functions: 1. The linear function \( f(x) = x \)2. The inverse cosecant function \( g(x) = \csc^{-1} x \).The relationship can be expressed as \( y = x imes \csc^{-1} x \), which is suitable for differentiation using the product rule.

Step 2: Apply the Product Rule

To differentiate the function \( y = x \csc^{-1} x \), apply the product rule:\[ \frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x) \]Here, let \( u(x) = x \) and \( v(x) = \csc^{-1} x \).

Step 3: Differentiate Each Component

Differentiate \( u(x) = x \)- The derivative of \( u(x) = x \) is \( u'(x) = 1 \).Differentiate \( v(x) = \csc^{-1} x \)- The derivative of the inverse cosecant function \( \csc^{-1} x \) is \( v'(x) = -\frac{1}{x \sqrt{x^2 - 1}} \).

Step 4: Substitute into Product Rule Formula

Substitute the derivatives into the product rule:\[ \frac{dy}{dx} = u'(x) v(x) + u(x) v'(x) = (1)(\csc^{-1} x) + x \left(-\frac{1}{x \sqrt{x^2 - 1}}\right) \]

Step 5: Simplify the Expression

Simplify the expression:\[ \frac{dy}{dx} = \csc^{-1} x - \frac{x}{x \sqrt{x^2 - 1}} \]The \( x \) in the numerator and denominator cancels out, so the expression becomes:\[ \frac{dy}{dx} = \csc^{-1} x - \frac{1}{\sqrt{x^2 - 1}} \]

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential in calculus, particularly when dealing with angles and triangles. These functions are the inverses of the standard trigonometric functions like sine, cosine, and tangent.

For example, the inverse cosecant function, denoted as \( \csc^{-1}(x) \), is the inverse of the cosecant function.
These functions help us find angles when we already know the value of a trigonometric ratio.

• Inverse trigonometric functions return angles as outputs.
• They are very useful in integration and differentiation, especially in solving real-world problems like signal processing.

When differentiating inverse trigonometric functions, it's important to remember their specific derivative formulas. These formulas may initially seem challenging, but with practice, they can become intuitive. For inverse cosecant, the derivative is \( -\frac{1}{x \sqrt{x^2 - 1}} \), highlighting that inverse functions often involve square roots and the original variable \( x \).

Product Rule

The product rule is a fundamental theorem in calculus used to differentiate products of two functions. Given two functions \( u(x) \) and \( v(x) \), the product rule states:\[\frac{d}{dx} [u(x) \cdot v(x)] = u'(x) v(x) + u(x) v'(x)\]This formula is vital when a function is composed of multiplying two simpler functions. It assures us that the derivative can be found by taking the derivative of the first function, multiplying it by the second, and adding it to the multiplication of the first function with the derivative of the second.

• Essential for functions with products, ensuring none of the accompanying functions are overlooked.
• Ensures accuracy in differentiating, especially in instances where functions interrelate closely.

In our exercise, the product rule allows us to differentiate \( y = x \cdot \csc^{-1}(x) \) by breaking it into the components of \( x \) and \( \csc^{-1}(x) \).

Derivative

The concept of a derivative is at the heart of calculus, representing the rate of change of a function with respect to a variable. It essentially measures how a function’s output value changes concerning changes in the input value. The derivative of a function \( f(x) \), denoted \( f'(x) \) or \( \frac{df}{dx} \), is fundamental in understanding and predicting variable trends.

• Makes finding tangent slopes at any point possible.
• Helps in understanding the behavior of physical and economic systems.

Calculating derivatives involves using various rules including the product rule, chain rule, and quotient rule, depending on the function's composition. In calculus, having a clear understanding of how to calculate derivatives allows us to solve complex problems involving instantaneous rates of change, which are essential in physics, engineering, and other applied sciences.

Calculus Problem Solving

Problem-solving in calculus involves applying different concepts and rules to find solutions to mathematical problems. This skill is developed through understanding, practice, and applying different calculus techniques tailored to specific problems.

In calculus, problems can involve limits, derivatives, integrals, and differential equations. Each problem requires a distinct set of tools and an understanding of which rules or theorems apply.

• Identify each component of the given problem.
• Decide on the appropriate rules or strategies based on component analysis.
• Use step-by-step methods to apply the rules efficiently.

In the original exercise, identifying how to use the product rule correctly was crucial. Each step built on the previous understanding, leading to the simplification of the derivative. Understanding problem-solving steps and practicing them continuously improves accuracy and efficiency in tackling calculus problems.