Question

Find \(D_{x} y\). $$ y=\tan ^{-1}\left(\ln x^{2}\right) $$

Short answer

\(D_x y = \frac{2}{x(1+ (\ln x^2)^2)}\)."

Step-by-step solution

Recognize the Function Type

The given function is a composite function involving a natural logarithm and the inverse tangent. We have to find the derivative of \(y = \tan^{-1}(\ln x^2)\).

Differentiate Using Chain Rule

First, note that \(y = \tan^{-1}(u)\) where \(u = \ln x^2\). Use the chain rule to differentiate. Recall that the derivative of \(\tan^{-1}(u)\) with respect to \(u\) is \(\frac{1}{1+u^2}\). Thus, \(\frac{dy}{du} = \frac{1}{1+(\ln x^2)^2}\).

Differentiate \(u = \ln x^2\)

Next, we differentiate \(u = \ln x^2\) with respect to \(x\). Rewrite \(u = \ln (x^2) = 2 \ln x\). The derivative is \(\frac{du}{dx} = \frac{d}{dx}(2 \ln x) = \frac{2}{x}\).

Apply the Chain Rule

Multiply the derivatives from Step 2 and Step 3 according to the chain rule: \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{1+(\ln x^2)^2} \cdot \frac{2}{x}\).

Simplify the Expression

The result is \(D_x y = \frac{2}{x(1+ (\ln x^2)^2)}\). This expression represents the derivative of the given function.

Key concepts

Composite Functions

A composite function is formed by applying one function to the results of another. In the given exercise, we have a function within a function: the inverse tangent, indicated by \( \tan^{-1} \), and the natural logarithm, \( \ln \).
When dealing with composite functions, it is essential to identify each part. This helps when differentiating because each part needs to be considered separately before combining them.
The overall function is \( y = \tan^{-1}(\ln x^2) \), where \( \ln x^2 \) is nested inside \( \tan^{-1} \).

• Inner function: \( u = \ln x^2 \)
• Outer function: \( y = \tan^{-1}(u) \)

Understanding the structure of the composite function sets the stage for using the chain rule effectively.

Chain Rule

The chain rule is a powerful technique for differentiating composite functions. It allows us to take the derivative of a composition of functions by multiplying the derivative of each function.
This is essential when the situation calls for finding the rate of change of one function with respect to another that's embedded within it.
For a composite function like \( y = \tan^{-1}(\ln x^2) \), the chain rule involves two main steps:

• Differentiate the outer function \( \tan^{-1}(u) \) with respect to the inner function \( u \), resulting in \( \frac{dy}{du} = \frac{1}{1+u^2} \).
• Differentiate the inner function \( u = \ln x^2 \) with respect to \( x \), resulting in \( \frac{du}{dx} = \frac{2}{x} \).

Finally, multiply these derivatives together to find the derivative of the composite function with respect to \( x \):\[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = \frac{1}{1+(\ln x^2)^2} \times \frac{2}{x}\].
This method captures how the changes in \( x \) influence the changes in \( y \), taking the indirect route through \( u \).

Inverse Functions

An inverse function "undoes" the effect of the original function. In the context of this exercise, \( \tan^{-1} \) is the inverse function of the tangent function. It means, if \( \tan(\theta) = x \), then \( \tan^{-1}(x) = \theta \).
Inverse functions are crucial because they allow us to express angles or other original values that could otherwise be tough to retrieve from trigonometric ratios.
When differentiating inverse functions like \( \tan^{-1}(u) \), we apply their specific derivative formula. The derivative of \( \tan^{-1}(u) \) with respect to \( u \) is \( \frac{1}{1+u^2} \).
This reflects how a small change in \( u \) translates into a change in \( \tan^{-1}(u) \). Understanding how to handle inverse functions is vital, particularly in solving intricate calculus problems.

Natural Logarithm

The natural logarithm \( \ln \) is a logarithm to the base \( e \) (where \( e \approx 2.718 \)). It arises frequently in various mathematical calculations, especially those involving growth rates and scales.
In this specific exercise, \( \ln x^2 \) is a crucial part of the composite function. It can be rewritten using properties of logarithms as \( 2 \ln x \).
This transformation simplifies differentiation. The derivative of \( \ln x \) is a fundamental formula: \( \frac{d}{dx} \ln x = \frac{1}{x} \). Hence, \( \frac{d}{dx} (2 \ln x) = \frac{2}{x} \) is straightforward.
Understanding the natural logarithm is key to tackling complex differential problems where the logarithmic growth behavior needs modeling.

• The natural logarithm is continuous and differentiable for all \( x > 0 \).
• It naturally integrates and differentiates with ease due to its properties and rules.

Knowledge of these basics is foundational as seen in this exercise to smoothly derive the expression.