Problem 30 - Thomas Calculus

Question

In Exercises \(19-30,\) find \(d y\) $$ y=2 \cot \left(\frac{1}{\sqrt{x}}\right) $$

Step-by-Step Solution

Answer

The differential is \( dy = \frac{\csc^2\left(\frac{1}{\sqrt{x}}\right)}{x^{3/2}} dx \).

Step 1: Understand the Function

We need to find the differential \(dy\) of the function \(y = 2 \cot \left( \frac{1}{\sqrt{x}} \right) \). This is done by differentiating \(y\) with respect to \(x\) to find \( \frac{dy}{dx} \) and then multiplying by \(dx\) to find \(dy\).

Step 2: Apply the Chain Rule

To differentiate \(y = 2 \cot \left( \frac{1}{\sqrt{x}} \right)\), apply the chain rule. Let \(u = \frac{1}{\sqrt{x}}\), then \(y = 2 \cot(u)\). The derivative of \(\cot(u)\) with respect to \(u\) is \(- \csc^2(u)\).

Step 3: Differentiate Inner Function

Now find \(\frac{du}{dx}\), where \(u = \frac{1}{\sqrt{x}} = x^{-1/2}\). Differentiate it to get \(\frac{du}{dx} = -\frac{1}{2}x^{-3/2} = -\frac{1}{2x^{3/2}}\).

Step 4: Differentiate Outer Function

Using the chain rule, \( \frac{dy}{dx} = 2 \cdot (-\csc^2(u)) \cdot \frac{du}{dx}\). Substitute \(\csc^2(u) = \csc^2\left(\frac{1}{\sqrt{x}} \right)\) and \(\frac{du}{dx}\) from Step 3.

Step 5: Calculate \(\frac{dy}{dx}\)

Now substitute \( \frac{du}{dx} = -\frac{1}{2x^{3/2}} \) into the derivative of the outer function: \( \frac{dy}{dx} = 2 (- \csc^2\left(\frac{1}{\sqrt{x}}\right)) \left(-\frac{1}{2x^{3/2}}\right) = \frac{\csc^2\left(\frac{1}{\sqrt{x}}\right)}{x^{3/2}}\).

Step 6: Find \( dy \)

Multiply \( \frac{dy}{dx} \) by \(dx\) to find \( dy \). Therefore, \( dy = \frac{\csc^2\left(\frac{1}{\sqrt{x}}\right)}{x^{3/2}} dx \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in differential calculus. It helps us differentiate composite functions, which are functions made up of two or more basic functions. In simple terms, if you have a function that is inside another function, the chain rule teaches you how to differentiate it.
The rule states that to differentiate a composite function, you must:

• Differentiate the outer function based on the inner function.
• Multiply that by the derivative of the inner function with respect to the original variable.

For instance, in the function given, we have an inner function, \( u = \frac{1}{\sqrt{x}} \), and an outer function, \( y = 2 \cot(u) \). The chain rule tells us to first find the derivative of \( \cot(u) \) with respect to \( u \), then multiply that by the derivative of \( u \) with respect to \( x \). This method simplifies the process of differentiation for complex, nested functions.

Differentiation

Differentiation is a process in calculus used to find the rate at which a function changes at any point. Essentially, when you differentiate a function, you're finding its derivative.
The derivative is a valuable tool because it tells you how fast or how slow a function is changing, which is crucial for many applications in science and engineering.
To differentiate a function like \( y = 2 \cot \left(\frac{1}{\sqrt{x}}\right) \), we apply differentiation rules. We use standard derivatives for basic functions, like how the derivative of \( \cot(u) \) with respect to \( u \) is \( -\csc^2(u) \). Every function has a derivative, and mastering these rules can simplify calculus tasks significantly.

Trigonometric Functions

Trigonometric functions are a group of functions that involve angles. They are critical in calculus for solving differential equations and function differentiation. Common trigonometric functions are sine, cosine, and tangent, along with their reciprocals: cosecant (csc), secant (sec), and cotangent (cot).
In the context of the given exercise, the cotangent function, \( \cot(u) \), is used. Its derivative is \( -\csc^2(u) \), and recognizing these derivatives is essential for differentiation.
Understanding how trigonometric functions behave and how to differentiate them forms the foundation for analyzing more complex mathematical models. They often appear in physics, engineering, and computer graphics when modeling waves, circular motion, and periodic behaviors.

Differential Equations

Differential equations involve derivatives and are equations that relate a function with its derivatives. They model many real-world phenomena, such as the growth of populations, heat distribution, and motion.
To solve differential equations, various methods, including separation of variables and integrating factors, can be used depending on the equation's complexity and type.
In simpler problems like the exercise mentioned, which asks for finding the differential \( dy \), understanding the basic process of finding \( \frac{dy}{dx} \) and then obtaining \( dy \) by multiplying it by \( dx \) is crucial.
This is a simplified approach to dealing with situations where the exact behavior of a changing quantity, described by a function, needs to be analyzed. Understanding these fundamentals is key to mastering differential equations in both pure and applied mathematics.