Problem 40 - Calculus Single Variable

Question

A function \(f\) is given. Calculate \(f^{\prime}(x)\). $$ f(x)=1 /\left(1+x^{2}\right) $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{-2x}{(1+x^2)^2} \).

Step 1: Identify the Rule to Use

The function given is a rational function, specifically \( f(x) = \frac{1}{1+x^2} \). To find the derivative, we need to use the quotient rule, which states \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \), where \( u = 1 \) and \( v = 1 + x^2 \).

Step 2: Find the Derivatives of the Numerator and Denominator

The numerator of the function is \( u = 1 \), so its derivative \( u' = 0 \). The denominator is \( v = 1 + x^2 \), and its derivative \( v' = 2x \) since the derivative of \( x^2 \) is \( 2x \).

Step 3: Apply the Quotient Rule

Substitute the values into the quotient rule: \[ f'(x) = \frac{u'v - uv'}{v^2} = \frac{0 \cdot (1+x^2) - 1 \cdot 2x}{(1+x^2)^2} = \frac{-2x}{(1+x^2)^2} \]

Step 4: Simplify the Expression

The expression \( f'(x) = \frac{-2x}{(1+x^2)^2} \) is already simplified, as there are no further common terms to cancel or simplify.

Key Concepts

Quotient Rule

The quotient rule is a fundamental technique in calculus used to find the derivative of a fraction, or quotient, of two functions. When dealing with fractions, you can't simply differentiate the numerator and the denominator separately and expect the result to be the derivative of the whole fraction. This is where the quotient rule comes to the rescue!
For a function written as the quotient of two functions, say \( \frac{u}{v} \), the quotient rule formula is:

• \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)

Here’s what each component represents:

• \( u \) is the numerator function.
• \( v \) is the denominator function.
• \( u' \) is the derivative of \( u \).
• \( v' \) is the derivative of \( v \).

When using this rule, it’s crucial to ensure each step is calculated correctly to avoid errors, especially with the signs. The subtraction \( u'v - uv' \) is key and must be handled carefully, particularly when plugging in these expressions.The quotient rule is an essential tool in calculus, especially for solving problems involving complex fractions. It helps in finding precise rates of change, enabling deeper insights into the behavior of functions.

Derivative of Rational Function

A rational function is a function that can be expressed as the ratio of two polynomials, where the numerator and the denominator are polynomials. In our exercise, the given function is \( f(x) = \frac{1}{1 + x^2} \). To find the derivative of such a rational function, the quotient rule must be employed since it deals with ratios.
In the process of differentiating \( \frac{1}{1 + x^2} \), we identified the numerator and denominator functions:

• The numerator \( u = 1 \) is a constant function, making its derivative \( u' = 0 \).
• The denominator \( v = 1 + x^2 \) is a polynomial function, so its derivative \( v' = 2x \), derived from applying the power rule to \( x^2 \).

With \( u' \), \( v \), and \( v' \) known, you plug them into the quotient rule formula:

• \( f'(x) = \frac{0 \cdot (1+x^2) - 1 \cdot 2x}{(1+x^2)^2} \)
• This simplifies to \( f'(x) = \frac{-2x}{(1+x^2)^2} \)

Understanding how to derive rational functions helps in grasping the dynamic changes in various mathematical models, ensuring the application of calculus principles is sound and precise.

Simplifying Derivatives

Simplifying derivatives involves reducing the expression to its most concise form after applying differentiation rules. In our example, the resulting derivative after applying the quotient rule was \( f'(x) = \frac{-2x}{(1+x^2)^2} \). This fraction cannot be simplified further since there are no common factors between the numerator and the denominator that can be canceled out.
To determine if you can simplify further, you should:

• Check for common factors in the numerator and denominator.
• See if any polynomial identities or factorizations can apply.
• Ensure that everything is correctly differentiated; mistakes in this stage can lead to unnecessary or impossible simplifications.

Simplifying derivatives is crucial for making them easier to work with in further calculus operations, such as integration or solving differential equations. A simplified derivative is not only more aesthetically pleasing, but it also helps in formulating more straightforward conclusions about a function's rate of change and behavior. Understanding the intricacies of simplification ensures clarity and precision, forming a foundational skill in calculus problem-solving.