Problem 29 - Thomas Calculus

Question

In Exercises \(19-30,\) find \(d y\) $$ y=3 \csc (1-2 \sqrt{x}) $$

Step-by-Step Solution

Answer

\( dy = \frac{3 \csc(1 - 2 \sqrt{x}) \cdot \cot(1 - 2 \sqrt{x})}{\sqrt{x}} \cdot dx \)

Step 1: Identify the Outer Function

The given function is \[ y = 3 \csc (1 - 2 \sqrt{x}) \]The outer function here is the cosecant function, \( \csc u \), where \( u = 1 - 2 \sqrt{x} \). We will need to take the derivative of \( \csc u \) with respect to \( u \) using the derivative formula for cosecant.

Step 2: Differentiate the Outer Function

The derivative of the cosecant function \( \csc u \) is \[ \frac{d}{du}(\csc u) = -\csc u \cdot \cot u \]Thus, the derivative of \( y \) with respect to \( u \), considering the outer function, is\[ \frac{dy}{du} = -3 \csc (1 - 2 \sqrt{x}) \cdot \cot (1 - 2 \sqrt{x}) \]

Step 3: Identify the Inner Function

The inner function is \( u = 1 - 2 \sqrt{x} \). We now need to find the derivative of \( u \) with respect to \( x \).

Step 4: Differentiate the Inner Function

To differentiate \( u = 1 - 2 \sqrt{x} \), first note that \( \sqrt{x} = x^{1/2} \). So, the derivative of \( \sqrt{x} \) is \[ \frac{d}{dx}(\sqrt{x}) = \frac{1}{2}x^{-1/2} \]And the derivative of \( 2 \sqrt{x} \) is \[ \frac{d}{dx}(2 \sqrt{x}) = 2 \cdot \frac{1}{2}x^{-1/2} = x^{-1/2} \]So the derivative of \( u \) with respect to \( x \) is\[ \frac{du}{dx} = 0 - x^{-1/2} = -\frac{1}{\sqrt{x}} \]

Step 5: Apply the Chain Rule

Using the chain rule, which states \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \), we substitute the derivatives we found:\[ \frac{dy}{dx} = -3 \csc(1 - 2 \sqrt{x}) \cdot \cot(1 - 2 \sqrt{x}) \cdot \left(-\frac{1}{\sqrt{x}}\right) \]Simplifying gives:\[ \frac{dy}{dx} = \frac{3 \csc(1 - 2 \sqrt{x}) \cdot \cot(1 - 2 \sqrt{x})}{\sqrt{x}} \]

Step 6: Find Differential \(dy\)

Finally, to find \( dy \), we express the result as\[ dy = \frac{3 \csc(1 - 2 \sqrt{x}) \cdot \cot(1 - 2 \sqrt{x})}{\sqrt{x}} \cdot dx \]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in differential calculus that is particularly useful for finding derivatives of composite functions. A composite function is a function that is applied to another function, forming a chain of functions.

If you have a function, say \( y = f(g(x)) \), the chain rule helps by expressing the derivative of \( y \) with respect to \( x \) as a product of derivatives:

• First, take the derivative of \( f \) with respect to \( g \), which gives \( \frac{df}{dg} \).
• Then, take the derivative of \( g \) with respect to \( x \), which gives \( \frac{dg}{dx} \).
• The chain rule formula is \( \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \).

In our exercise, the function \( y = 3 \csc(1 - 2 \sqrt{x}) \) is a composite function consisting of an outer function \( 3 \csc(u) \) and an inner function \( u = 1 - 2 \sqrt{x} \). By applying the chain rule, you differentiate these parts separately and then combine them to find \( \frac{dy}{dx} \).

Trigonometric Functions

Trigonometric functions such as sine, cosine, tangent, and their reciprocals like cosecant, are pivotal in calculus, especially when analyzing curves and angles. They have well-known derivatives that simplify the process of differentiation.

For instance, the derivative of \( \csc u \) is \( -\csc u \cot u \). This comes from their definitions and the fundamental properties of trigonometric identities. When differentiating trigonometric functions:

• Recognize the function and its derivative, such as \( \frac{d}{du}(\csc u) = -\csc u \cot u \) for the cosecant function.
• The derivatives often involve products or powers of trigonometric functions, requiring careful attention to detail.

In our problem, identifying and differentiating the trigonometric part correctly was crucial to finding the derivative of \( y \) with respect to \( x \).

Remember, it's important to understand these derivatives because they frequently appear in physics, engineering, and beyond.

Differential Calculus

Differential calculus is a branch of mathematics concerned with the study of how functions change. It involves understanding and calculating derivatives, which measure how a function's output changes as its input changes.

Derivatives are essential for finding:

• Rates of change, such as velocity and acceleration in physics.
• Slope of the tangent line to a curve at any point.
• Optimizing problems, finding maximums or minimums of functions.

In the exercise, we focused on finding the derivative of a given function with respect to its variable. The goal was to express this change clearly using differential calculus techniques, such as the chain rule and knowledge of trigonometric derivatives.

Mastering differential calculus improves your problem-solving ability across various applications, from technical to everyday situations. It gives you tools to analyze and interpret dynamic changes quantitatively.