Problem 20 - Calculus of a Single Variable

Question

Finding a Differential In Exercises \(11-20\) , find the differential \(d y\) of the given function. $$ y=\frac{\sec ^{2} x}{x^{2}+1} $$

Step-by-Step Solution

Answer

The differential \(dy\) of the function \(y=\frac{\sec ^{2} x}{x^{2}+1}\) is given by \(dy= 2\sec^2(x) [ tan(x)(x+1) - x] dx / (x^{2}+1)^2\)

Step 1: Understand the problem and necessary rules

Given the function \(y=\frac{\sec ^{2} x}{x^{2}+1}\), students need to find the differential \(dy\) or in other words, the derivative of the function with respect to \(x\). The rules that will be used here include the quotient rule \( (f/g)' = f'g - fg'/g^2 \) for differentiation and the derivative of \( sec^{2}(x) \), which is \( 2sec(x)sec(x)tan(x) \).

Step 2: Apply the quotient rule

Using the quotient rule for differentiation, let \(f = \sec ^{2} x\) and \(g = x^{2}+1\). Therefore, \(f' = 2\sec(x)sec(x)tan(x)\) and \(g' = 2x\). The derivative of y with respect to x, \(dy/dx\) is given by: \[dy/dx=(f'g-fg')/g^2=(2\sec(x)sec(x)tan(x)(x^{2}+1) - \sec^2(x)*2x) / (x^{2}+1)^2\]

Step 3: Further simplify the expression

Simplify the expression obtained in step 2 to get the derivative in its simplest form. Proceed as follows:\[dy/dx=(2x\sec^2(x) tan(x) + 2\sec^2(x) tan(x) - 2x\sec^2(x)) / (x^{2}+1)^2\]\[dy/dx=(2\sec^2(x) tan(x) (x+1) -2x\sec^2(x)) / (x^{2}+1)^2\]\[dy/dx=2\sec^2(x) [ tan(x)(x+1) - x] / (x^{2}+1)^2\]

Step 4: Find the differential \(dy\)

To find the differential \(dy\), express it in terms of \(dx\). This is achieved by multiplying both sides by \(dx\). Hence, the differential \(dy\) is\[dy= 2\sec^2(x) [ tan(x)(x+1) - x] dx / (x^{2}+1)^2\]

Key Concepts

Quotient Rule

The quotient rule is a fundamental differentiation tool used in calculus when dealing with functions presented as the division of two separate functions.
Simply put, if you have a function in the form \( \frac{f(x)}{g(x)} \), the derivative is not as straightforward as differentiating each part separately and dividing.
The quotient rule provides a systematic approach: it states \( (\frac{f}{g})' = \frac{f'g - fg'}{g^2} \).

• \(f'\) is the derivative of the numerator \(f\).
• \(g'\) is the derivative of the denominator \(g\).

When applying this rule, you first find the derivatives of both the numerator and the denominator, and then systematically apply them into the formula. This keeps your work organized and accurate, especially when dealing with complex functions like the one in the original exercise.
The ability to effectively use the quotient rule is essential, as it frequently appears when working with rational functions in calculus.

Derivative of Secant

Understanding the derivative of the secant function is central when dealing with problems involving trigonometric functions.
In this context, the secant function is \( \sec(x) \), and its derivative is directly connected to other trigonometric functions.

• The basic derivative of \( \sec(x) \) is \( \sec(x)\tan(x) \).
• For \( \sec^2(x) \), the derivative becomes \( 2\sec(x)\sec(x)\tan(x) \), which simplifies to \( 2\sec^2(x)\tan(x) \).

The understanding of these derivatives is essential as they simplify ongoing calculations where secant functions are involved.
When combined with the quotient rule, as in the given exercise, it helps in systematically finding the derivative of more complex functions.

Differentiation Rules

Differentiation rules are the set of fundamental principles used to determine the derivatives of functions. These rules simplify the process of finding derivatives by providing straightforward methods for applying calculus principles.

• The basic differentiation rules include the power rule, product rule, quotient rule, and chain rule.
• Each rule addresses specific function types, allowing for systematic and logical application of differentiation techniques.

For instance, in the given exercise, the quotient rule is specifically utilized because we are dealing with a division of functions.
Differentiation rules like these are essential for accurately and efficiently finding derivatives, and mastering them leads to better problem-solving capabilities in calculus-related tasks.

Finding Differentials

Finding differentials is a practical aspect of calculus that extends beyond simply determining derivatives.
A differential, often denoted as \(dy\) or \(dx\), represents an infinitesimally small change in a function's value in relation to a change in its variable.
To find a differential like \(dy\), you typically take the derivative \( \frac{dy}{dx} \) and multiply it with \(dx\).

• This is valuable in estimating changes in function values over small intervals.
• It provides a framework for approximating real-world changes and is useful in applications such as physics and engineering.

In the original exercise, after determining the derivative, expressing it as a differential by multiplying by \(dx\) illustrates how these concepts are more than academic—they are tools for solving real problems.
This approach transitions from abstract differentiation into practical applications, enhancing understanding and application of calculus in diverse situations.