Problem 70 - Thomas Calculus

Question

In Exercises \(49-70\) , 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

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

Step 1: Understand the Function

The function provided is \( y = \ln(x^2 + 4) - x \tan^{-1}\left(\frac{x}{2}\right) \). We need to differentiate this function with respect to \( x \). The function is composed of two main terms: the natural logarithm term and the product of \( x \) and the inverse tangent.

Step 2: Differentiate the Logarithmic Term

For the first term \( \ln(x^2 + 4) \), use the chain rule. The derivative of \( \ln(u) \) with respect to \( x \) is \( \frac{1}{u} \frac{du}{dx} \). Here, \( u = x^2 + 4 \) so \( \frac{du}{dx} = 2x \). The derivative of the first term is \( \frac{1}{x^2 + 4} \cdot 2x = \frac{2x}{x^2 + 4} \).

Step 3: Differentiate the Product Term using Product Rule

The second term is \( -x \tan^{-1}\left(\frac{x}{2}\right) \). Use the product rule \( u\cdot v \to u'v + uv' \). Let \( u = x \) and \( v = \tan^{-1}\left(\frac{x}{2}\right) \). The derivative \( u' = 1 \). For \( v' \), use the chain rule: \( v = \tan^{-1}(u) \) gives \( \frac{1}{1+u^2} \frac{du}{dx} \), where \( u = \frac{x}{2} \) and \( \frac{du}{dx} = \frac{1}{2} \). Thus, \( v' = \frac{1}{1+(\frac{x}{2})^2} \cdot \frac{1}{2} = \frac{1}{2} \cdot \frac{1}{1+(\frac{x^2}{4})} = \frac{1}{2 + x^2} \).

Step 4: Calculate Derivative of the Product

Now apply the product rule: \( -[1 \cdot \tan^{-1}\left(\frac{x}{2}\right) + x \cdot \frac{1}{2 + x^2}] = -\tan^{-1}\left(\frac{x}{2}\right) - \frac{x}{2 + x^2} \).

Step 5: Combine the Derivatives

Combine the derivatives of the two terms: \( \frac{dy}{dx} = \frac{2x}{x^2 + 4} - \left( \tan^{-1}\left(\frac{x}{2}\right) + \frac{x}{2 + x^2} \right) \).

Step 6: Simplify the Derivative

Upon combining and simplifying, the derivative is: \( \frac{dy}{dx} = \frac{2x}{x^2 + 4} - \tan^{-1}\left(\frac{x}{2}\right) - \frac{x}{2 + x^2} \).

Key Concepts

Natural Logarithm Differentiation

Natural logarithms play a crucial role in calculus, especially when it comes to differentiation. The natural logarithm function, denoted by \(\ln(x)\), differentiates according to a very specific rule. If you have a composition where \( y = \ln(u) \), then the derivative \( \frac{dy}{dx} \) is calculated using:

• The derivative of the natural logarithm function: \( \frac{1}{u} \)
• The chain rule to determine \( \frac{du}{dx} \), the derivative of the inside function \( u \) with respect to \( x \)

This results in the expression: \[ \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \]Applying this to our exercise, where \( y = \ln(x^2 + 4) \), we first identify \( u = x^2 + 4 \). Differentiating \( u \) with respect to \( x \) gives \( \frac{du}{dx} = 2x \). Therefore, the derivative of the natural logarithm part is:\[ \frac{dy}{dx} = \frac{2x}{x^2 + 4} \]Using this straightforward method of differentiation allows us to handle logarithmic expressions with ease.

Inverse Trigonometric Functions

Inverse trigonometric functions, like the inverse tangent \( \tan^{-1}(x) \), also known as arctan, can seem daunting. However, differentiating these functions becomes manageable once you know the rules. The derivative of the inverse tangent function is: \[ \frac{d}{dx} \tan^{-1}(x) = \frac{1}{1+x^2} \] In our exercise, the term involves \( \tan^{-1}(\frac{x}{2}) \). To differentiate this, we again need to employ the chain rule:

• First, recognize \( u = \frac{x}{2} \)
• Then, \( \frac{du}{dx} = \frac{1}{2} \)

This means our derivative becomes: \[ \frac{1}{1+(\frac{x}{2})^2} \cdot \frac{1}{2} = \frac{1}{2 + x^2} \] By understanding these relationships, handling inverse trigonometric functions in differentiation processes becomes simpler.

Product Rule

The product rule is unavoidable when differentiating functions that are products of two simpler functions. It states that if you have a product \( y = u \cdot v \), then the derivative is given by:\[ \frac{dy}{dx} = u'v + uv' \] This rule allows us to break down the differentiation of complex products into manageable pieces. In the exercise, we encounter the product involving \( x \) and \( \tan^{-1}(\frac{x}{2}) \):

• Here, \( u = x \) and \( v = \tan^{-1}(\frac{x}{2}) \).
• Calculate \( u' = 1 \).
• Substitute the previously found \( v' = \frac{1}{2 + x^2} \).

Applying the product rule thus gives us:\[ 1 \cdot \tan^{-1}(\frac{x}{2}) + x \cdot \frac{1}{2 + x^2} = \tan^{-1}(\frac{x}{2}) + \frac{x}{2 + x^2} \]This structured method simplifies the differentiation of terms that appear complicated initially.

Chain Rule

The chain rule is a powerful tool in calculus, used to differentiate compositions of functions. It enables us to tackle more complex expressions by breaking them down into simpler differentiated parts. Consider a function composed as \( y = f(g(x)) \). Here, the chain rule states that the derivative \( \frac{dy}{dx} \) is: \[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]In our differentiated expression, we used the chain rule to differentiate both the natural logarithm \( \ln(x^2 + 4) \) and the inverse tangent \( \tan^{-1}(\frac{x}{2}) \).

• For the natural logarithm, \( g(x) = x^2 + 4 \), so \( g'(x) = 2x \).
• For the inverse tangent, \( g(x) = \frac{x}{2} \), leading to \( g'(x) = \frac{1}{2} \).

The application of the chain rule allowed us to unravel the derivatives of nested functions seamlessly, demonstrating its utility and necessity in differentiation.