Problem 33 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The derivative is \( dy = \frac{2x}{1+x^2} \, dx \).

Step 1: Identify the function

The given function is \( y = \ln(1 + x^2) \), where \( y \) is a natural logarithm of the expression \( 1 + x^2 \). Our task is to find the derivative of \( y \), denoted as \( dy \).

Step 2: Apply the derivative of natural logarithm rule

To differentiate \( y \) with respect to \( x \), we use the rule that states \( \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = 1 + x^2 \).

Step 3: Differentiate the inner expression

Now, differentiate the inner function \( u = 1 + x^2 \) with respect to \( x \):\[ \frac{du}{dx} = 2x \]

Step 4: Apply the chain rule

Using the chain rule from Step 2, we differentiate \( y = \ln(1 + x^2) \):\[\frac{dy}{dx} = \frac{1}{1 + x^2} \cdot \frac{du}{dx} = \frac{1}{1 + x^2} \cdot 2x = \frac{2x}{1 + x^2}\]

Step 5: Express the derivative as dy

The derivative \( \frac{dy}{dx} = \frac{2x}{1 + x^2} \) means that the change in \( y \) with respect to \( x \) is given by multiplying both sides by \( dx \) to express it in terms of \( dy \):\[ dy = \frac{2x}{1 + x^2} \, dx \]

Key Concepts

Derivative of Natural Logarithm

The derivative of the natural logarithm function is a key concept in calculus that simplifies many calculations. When working with the natural logarithm, which is denoted as \( \ln(u) \), the derivative with respect to \( x \) is given by the formula \( \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx} \). This rule allows us to derive the logarithm by differentiating the function \( u \) contained within the logarithm.
In our exercise, we want to find the derivative of \( y = \ln(1 + x^2) \). To do so, we recognize that our function inside the logarithm is \( u = 1 + x^2 \). This function behaves as a separate entity with its own derivative, unrelated to the outer logarithmic function. It’s important to notice that we need to consider both the outer \( \ln \, \) function and the inner \( u \) function to successfully differentiate.
By applying the derivative rule for logarithms, we have a solid method for approaching complex expressions involving natural logs.

Chain Rule

The chain rule is a fundamental differentiation technique that allows us to handle compositions of functions, which occur when one function is nested inside another. This rule is essential when you encounter a situation where you need to differentiate a function of a function.
Here’s the formal statement for the chain rule: if you have a composite function \( y = f(g(x)) \), the derivative with respect to \( x \) is given by \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). In simpler words, you differentiate the outer function, keeping the inner function unchanged, then multiply by the derivative of the inner function.
In our problem, we use the chain rule to differentiate \( y = \ln(1 + x^2) \). The outer function is the natural logarithm \( f = \ln(x) \) and the inner function is \( g = 1 + x^2 \).

• First, apply the derivative of the outside \( \ln(u) \), which simplifies to \( \frac{1}{u} \).
• Then, multiply by the derivative of \( u = 1 + x^2 \), which is \( 2x \).

The result is \( \frac{dy}{dx} = \frac{2x}{1 + x^2} \), showing how the chain rule elegantly handles the differentiation of nested functions.

Differentiation Techniques

Differentiation encompasses various methods and rules that simplify the process of finding the derivative of a function. These techniques are crucial in solving many kinds of problems across mathematics and applied sciences.

• **Power Rule:** Often used when differentiating polynomial functions, where \( \frac{d}{dx} (x^n) = nx^{n-1} \).
• **Product Rule:** Necessary when finding the derivative of a product of two functions, expressed as \( (uv)' = u'v + uv' \).
• **Quotient Rule:** Applied when differentiating a division of two functions, given by \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).
• **Chain Rule:** Essential for composite functions, as previously discussed.

Each of these rules supports a systematic way to tackle derivatives, ensuring that even complex functions can be broken down efficiently. Referring to our original problem, the chain rule was combined with the natural logarithm differentiation technique to achieve the solution. Understanding when and how to apply these techniques is pivotal in mastering calculus and proficient problem solving.