Problem 7 - Calculus - AP Edition

Question

Find \(d y / d x\). $$y=\ln \left(\frac{x}{1+x^{2}}\right)$$

Step-by-Step Solution

Answer

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

Step 1: Apply the Logarithm Properties

The function given is a natural logarithm: \( y = \ln \left( \frac{x}{1+x^2} \right) \). We can use the properties of logarithms to simplify this expression. Recall that \( \ln \left( \frac{a}{b} \right) = \ln(a) - \ln(b) \). Apply this property:\[y = \ln(x) - \ln(1+x^2)\]

Step 2: Differentiate the Natural Logarithm Function

Now, we differentiate each term with respect to \( x \). Using the derivative formula for natural logarithms, \( \frac{d}{dx}[\ln(u)] = \frac{1}{u} \frac{du}{dx} \), differentiate each component:1. Differentiate \( \ln(x) \) with respect to \( x \):\[\frac{d}{dx}[\ln(x)] = \frac{1}{x}\]2. Differentiate \( \ln(1+x^2) \) with respect to \( x \):\[\frac{d}{dx}[\ln(1+x^2)] = \frac{1}{1+x^2} \cdot \frac{d}{dx}[1+x^2] = \frac{1}{1+x^2} \cdot 2x = \frac{2x}{1+x^2}\]

Step 3: Combine the Derivatives

Subtract the derivative of the second term from the derivative of the first term to find \( \frac{dy}{dx} \):\[\frac{dy}{dx} = \frac{1}{x} - \frac{2x}{1+x^2}\]

Step 4: Simplify the Expression (Optional)

You can express the derivative in a single rational function if needed. Find a common denominator:The common denominator of \( x \) and \( 1+x^2 \) is \( x(1+x^2) \). Rewrite the derivative:\[\frac{dy}{dx} = \frac{1(1+x^2) - 2x^2}{x(1+x^2)} = \frac{1 + x^2 - 2x^2}{x(1+x^2)} = \frac{1 - x^2}{x(1+x^2)}\]The expression \( \frac{1 - x^2}{x(1+x^2)} \) is a simplified version of the derivative.

Key Concepts

Understanding Natural Logarithm Differentiation

Differentiating natural logarithms involves a special formula that makes the process straightforward. To differentiate an expression like \( \ln(u) \), where \( u \) is a function of \( x \), we use the rule:

• \( \frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx} \)

This formula tells us that you take the reciprocal of the natural log argument \( u \) and multiply it by the derivative of \( u \).
For example, when differentiating \( \ln(x) \) with respect to \( x \), it's quite simple: the derivative is just \( \frac{1}{x} \).
In the exercise, we also differentiate \( \ln(1+x^2) \). Here, our \( u \) is \( 1+x^2 \). The derivative of this function is \( 2x \), so applying the formula gives us:

• \( \frac{1}{1+x^2} \cdot 2x = \frac{2x}{1+x^2} \)

Mastering differentiation of natural logarithms helps in solving various calculus problems efficiently.

Logarithmic Differentiation Explained

Logarithmic differentiation is a powerful tool for differentiating products or quotients. It simplifies the differentiation process by using the properties of logarithms. In the exercise, we start with the function \( y = \ln \left( \frac{x}{1+x^2} \right) \).
Using the logarithm property that \( \ln \left( \frac{a}{b} \right) = \ln(a) - \ln(b) \), the function can be rewritten to make differentiation easier. So \( y = \ln(x) - \ln(1+x^2) \). Now, you can differentiate this expression term by term.
This method is especially useful in dealing with complex functions where direct differentiation might be challenging or cumbersome. It breaks down the problem into more manageable parts by converting the multiplication or division in the function into addition and subtraction, making calculus operations simpler.

Insight into Rational Functions in Calculus

Rational functions are ratios of two polynomials. In calculus, understanding and simplifying them is crucial since they often appear in various problems.
The derivative we calculated in this exercise, \( \frac{dy}{dx} = \frac{1 - x^2}{x(1+x^2)} \), is an example of a rational function. Such functions can usually be simplified by factoring out common terms where possible.
To work with these functions, it's important to find the common denominator and simplify the expression, as done here:

• First, find a common denominator which is usually the product of the denominators.
• Combine the fractions to create a single expression.
• Simplify the expression by combining like terms.

Becoming comfortable with these techniques is essential for tackling both derivative and integral calculus problems, making rational functions a key topic to master in your calculus toolkit.