Problem 34 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The derivative is \( \frac{1}{x(1+(\ln x)^2)} \).

Step 1: Identify the Function

The function given is \( y = \tan^{-1}(\ln x) \), which is a composite function. The outer function is \( \tan^{-1}(u) \), where \( u = \ln x \).

Step 2: Use the Chain Rule

To find the derivative of \( y \) with respect to \( x \), we will use the chain rule. The chain rule states that if \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). In this case, \( f(u) = \tan^{-1}(u) \) and \( g(x) = \ln x \).

Step 3: Differentiate the Outer Function

The derivative of \( \tan^{-1}(u) \) with respect to \( u \) is \( \frac{1}{1+u^2} \). Therefore, \( \frac{d}{du}[\tan^{-1}(\ln x)] = \frac{1}{1+(\ln x)^2} \).

Step 4: Differentiate the Inner Function

The derivative of \( u = \ln x \) with respect to \( x \) is \( \frac{d}{dx}[\ln x] = \frac{1}{x} \).

Step 5: Multiply the Derivatives

Using the chain rule, multiply the derivative from Step 3 by the derivative from Step 4. Hence, \( \frac{dy}{dx} = \frac{1}{1+(\ln x)^2} \cdot \frac{1}{x} \).

Step 6: Simplify the Expression

Simplify the expression to find the final derivative: \( \frac{dy}{dx} = \frac{1}{x(1+(\ln x)^2)} \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. When you have a function inside another function, like in the case of \( y = \tan^{-1}(\ln x) \), the chain rule simplifies the process of finding derivatives. It's important to understand how each part of the function contributes to the overall derivative.

To apply the chain rule, consider your function as a composition of two functions: an outer function and an inner function. Here, the outer function is \( f(u) = \tan^{-1}(u) \) and the inner function is \( g(x) = \ln x \). The chain rule formula is given by:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

Simply put, you take the derivative of the outer function while treating the inner function as a variable, then multiply it by the derivative of the inner function.

This method breaks down complex differentiation into manageable steps, making it easier to get to the derivative precisely. In our example, this results in finding: \( \frac{dy}{dx} = \frac{1}{1+(\ln x)^2} \cdot \frac{1}{x} \). This meticulous step-by-step application is what makes the chain rule so effective.

Inverse Trigonometric Functions

Inverse trigonometric functions reverse the trigonometric functions, taking an input and returning an angle. They are denoted using inverse notation, such as \(\tan^{-1}(x)\), which means the angle whose tangent is \(x\).

These functions are often encountered when you need to reverse the trigonometric operations involved in solving equations or taking derivatives. The derivative of an inverse trigonometric function adheres to specific formulas. For \( \tan^{-1}(x) \), the derivative is:

• \( \frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2} \)

This derivative is critical when differentiating composite functions where inverse trigonometric functions appear as the outer layer.

In our exercise, this rule helps us find the derivative of \(\tan^{-1}(\ln x)\) with respect to \(u = \ln x\), resulting in \( \frac{1}{1+(\ln x)^2} \).

Grasping these foundational derivatives allows you to tackle complex calculus problems involving inverse trigonometric functions.

Logarithmic Differentiation

Logarithmic differentiation is a technique used for differentiating functions where the variable appears in either the base or the exponent.Although simpler in straightforward functions with natural logarithms, it serves an essential role when combined with other advanced differentiation techniques like the chain rule.

Taking the natural log of a function simplifies the differentiation process due to properties of logarithms, and that's what appears in our composite function \( \tan^{-1}(\ln x) \).For instance, the derivative of \( u = \ln x \) with respect to \( x \) is:

• \( \frac{d}{dx}[\ln x] = \frac{1}{x} \)

This straightforward derivative is often encountered in calculus and simplifies the inner function differentiation significantly, as seen in the chain rule application of our given problem.

Combining this with the chain rule allows us to tackle more complicated expressions to reach derivatives effectively. Logarithmic differentiation is crucial, especially when paired with inverse trigonometric derivatives as we worked through in the derivative of \( \tan^{-1}(\ln x) \).