Problem 42 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\ln \left(x^{2}+4\right)-x \tan ^{-1}\left(\frac{x}{2}\right)$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{2x}{x^2 + 4} - \tan^{-1}\left( \frac{x}{2} \right) - \frac{x}{2 + x^2} \)

Step 1: Identify the Derivative Rules Needed

The given function is a combination of a natural logarithm and an inverse tangent function. To differentiate this, we need the Chain Rule, the derivative of the natural logarithm, and the derivative of the inverse tangent function.

Step 2: Differentiate the Natural Logarithm Component

The natural logarithm component is \( \ln(x^2 + 4) \). The derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = x^2 + 4 \), so \( \frac{du}{dx} = 2x \). Thus, the derivative is \( \frac{1}{x^2 + 4} \cdot 2x = \frac{2x}{x^2 + 4} \).

Step 3: Differentiate the Inverse Tangent Component

The inverse tangent component is \( -x \tan^{-1}\left(\frac{x}{2}\right) \). This requires the Product Rule. Let \( u = x \) and \( v = \tan^{-1}\left(\frac{x}{2}\right) \). The derivative of \( u \) is 1, and the derivative of \( v \) is \( \frac{1}{1+(\frac{x}{2})^2} \cdot \frac{1}{2} = \frac{1}{2+x^2} \). Using the Product Rule, we find: \( -\left[1 \cdot \tan^{-1}\left(\frac{x}{2}\right) + x \cdot \frac{1}{2+x^2}\right] \).

Step 4: Combine the Derivatives

Combine the derivatives from Steps 2 and 3. The derivative \( \frac{dy}{dx} \) is: \[ \frac{dy}{dx} = \frac{2x}{x^2 + 4} - \left[ \tan^{-1}\left(\frac{x}{2}\right) + \frac{x}{2 + x^2} \right] \]

Step 5: Simplify the Expression

Simplify the combined expression from Step 4. The derivative becomes: \[ \frac{dy}{dx} = \frac{2x}{x^2 + 4} - \tan^{-1}\left( \frac{x}{2} \right) - \frac{x}{2 + x^2} \]

Key Concepts

Understanding the Chain Rule

The Chain Rule is a powerful tool in calculus that helps us differentiate composite functions, where one function is nested inside another. It's like peeling layers of an onion: you begin with the outer function and then move inward.

To apply the Chain Rule, follow these steps:

• Identify the outer function and the inner function within your composite function.
• Differentiating the outer function, leave the inside as it is.
• Multiply the derivative of the outer function by the derivative of the inner function.

In our exercise, when we differentiated the natural logarithm component, \( \ln(x^2 + 4) \), we considered \( x^2 + 4 \) to be the inner function \( u \). The Chain Rule helped us find that the derivative is \( \frac{1}{u} \cdot \frac{du}{dx} \). In simpler terms, it's like peeling back a layer and finding what's beneath.

Exploring the Inverse Tangent Derivative

The inverse tangent function, often denoted as \( \tan^{-1}(x) \), behaves uniquely in differentiation. Just like how every unique key has its own lock, the derivative of the inverse tangent has its special formula.

The derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1+x^2} \). This formula shows up every time we differentiate an inverse tangent function since it naturally accounts for the curve's gradient at each point along its path.

In our problem, we needed to differentiate \( \tan^{-1}\left(\frac{x}{2}\right) \). Using the Chain Rule again, we differentiate the outer inverse tangent function and then multiply by the derivative of \( \frac{x}{2} \), the inside component. This gives us \( \frac{1}{2+x^2} \), forming a part of our result.

Grasping the Product Rule

When functions are multiplied together, like \( x \) multiplied by any other function, the Product Rule becomes indispensable. It's a method that helps us find the derivative in such situations.

To apply the Product Rule:

• Identify the two functions that are being multiplied. Let's call them \( u \) and \( v \).
• Differentiate each function separately to get \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).
• The Product Rule formula is \( u'v + uv' \)—basically, derivation in a harmonious duet!

In our exercise, when dealing with \( -x \tan^{-1}\left(\frac{x}{2}\right) \), the \( u \) was \( x \) and the \( v \) was \( \tan^{-1}\left(\frac{x}{2}\right) \). We used the Product Rule to handle their interaction gracefully, forming one part of our derivative expression.