Problem 61 - Thomas Calculus

Question

In Exercises \(49-70\) , find the derivative of \(y\) with respect to the appropriate variable. $$ y=\ln \left(\tan ^{-1} x\right) $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Composite Functions

The function given is a composition of two functions: the natural logarithm function and the inverse tangent function. The outer function is \(\ln(u)\) and the inner function is \(\tan^{-1}(x)\).

Step 2: Use the Chain Rule

To find the derivative of the composite function \(y = \ln(\tan^{-1}(x))\), apply the chain rule. According to the chain rule, \(\frac{d}{dx}\left[f(g(x))\right] = f'(g(x)) \cdot g'(x)\). Here, \(f(u) = \ln(u)\) and \(g(x) = \tan^{-1}(x)\).

Step 3: Differentiate the Outer Function

The derivative of the outer function \(f(u) = \ln(u)\) with respect to \(u\) is \(f'(u) = \frac{1}{u}\).

Step 4: Differentiate the Inner Function

The derivative of \(g(x) = \tan^{-1}(x)\) with respect to \(x\) is \(g'(x) = \frac{1}{1+x^2}\).

Step 5: Apply the Chain Rule

Using the derivatives obtained, apply the chain rule: \[\frac{dy}{dx} = \frac{1}{\tan^{-1}(x)} \cdot \frac{1}{1+x^2}.\]

Step 6: Simplify the Expression

The expression for the derivative can be written as \[\frac{dy}{dx} = \frac{1}{\tan^{-1}(x)(1+x^2)}.\] This is the simplified form of the derivative of the given function.

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus, essential for finding the derivatives of composite functions. When dealing with a function within another function, like \(y = \ln(\tan^{-1}(x))\), the chain rule helps unravel these layers.
In simple terms, it states:

• If you have a composite function \(f(g(x))\), the derivative \(\frac{d}{dx} f(g(x))\) is obtained by multiplying the derivative of the outer function evaluated at the inner function by the derivative of the inner function.

For our specific example, we identify that:

• The outer function \(f(u)\) is \(\ln(u)\).
• The inner function \(g(x)\) is \(\tan^{-1}(x)\).

Applying the chain rule, we differentiate the outer function first, which is the natural logarithm in this case, and then the inner function, which is the inverse tangent. The rule effectively links these derivations to find the rate of change of the entire composite function.

Inverse Trigonometric Functions

Inverse trigonometric functions, like \(\tan^{-1}(x)\), provide angles whose trigonometric function is the given number. These functions are useful in calculus for describing relationships in various problems, especially those involving right triangles or polar coordinates.
When differentiating an inverse trigonometric function, remember that they have unique formulas. For example:

• The derivative of \(\tan^{-1}(x)\) with respect to \(x\) is \(\frac{1}{1+x^2}\). This stems from the need to consider the range and the restriction of these inverse functions' output angle.

This specific derivative is vital when applying the chain rule to composite functions that include inverse trigonometric components, as it simplifies the process and reduces potential errors.

Logarithmic Functions

Logarithmic functions, such as \(\ln(u)\), are inverses of exponential functions and are essential for many aspects of calculus and real-world applications. They help in scaling multipliers in models and simplifying products into sums, making calculations more manageable.
Differentiating a natural logarithm function is straightforward:

• The derivative of \(\ln(u)\) with respect to \(u\) is \(\frac{1}{u}\). This key property serves as a building block in various derivations and integrations.

In our specific task of differentiating \(y = \ln(\tan^{-1}(x))\), knowing how to handle the logarithmic aspect allows us to compound it with the chain rule. This understanding forms the backbone of combining the results of the inner derivative (using inverse trigonometric rules) with the outer derivative (via the natural logarithm), leading to a full solution.