Problem 17 - Essential Calculus Early Transcendentals

Question

Find the derivative of the function. Simplify where possible. \(y=\left(\tan ^{-1} x\right)^{2}\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Function and Apply Chain Rule

The function we need to differentiate is \( y = \left(\tan^{-1} x\right)^2 \). Notice that this is a composite function where the outer function is \( u^2 \) and the inner function is \( u = \tan^{-1} x \). We'll use the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

Step 2: Differentiate the Outer Function

The outer function is \( y = u^2 \). The derivative of \( u^2 \) with respect to \( u \) is \( \frac{dy}{du} = 2u \).

Step 3: Differentiate the Inner Function

The inner function is \( u = \tan^{-1} x \). The derivative of \( \tan^{-1} x \) with respect to \( x \) is \( \frac{du}{dx} = \frac{1}{1+x^2} \).

Step 4: Combine Using the Chain Rule

According to the chain rule: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2u \cdot \frac{1}{1+x^2} = 2\left(\tan^{-1} x \right) \cdot \frac{1}{1+x^2} \].

Step 5: Simplify the Expression

Combining all simplified parts, we can simplify the derivative further if needed to obtain \( \frac{dy}{dx} = \frac{2\tan^{-1} x}{1+x^2} \).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus that helps us find the derivative of composite functions. A composite function is essentially a function nestled inside another function. The trick of the chain rule is to differentiate these layers one at a time. The chain rule formula is:

• Take the derivative of the outer function with respect to the inner function.
• Then, multiply it by the derivative of the inner function with respect to the variable.

In the case of our problem, the outer function is \(u^2\), and the function inside it is \(u = \tan^{-1} x\). This process allows us to tackle complex derivatives with ease, by breaking them down into simpler parts.

Inverse Trigonometric Functions

Inverse trigonometric functions, like \(\tan^{-1} x\), are functions that "undo" their corresponding trigonometric functions. They are essential for solving equations where the functions are inverted to find angles or lengths. When differentiating inverse trigonometric functions, each has its own unique derivative formula. For \(\tan^{-1} x\), the derivative is \(\frac{1}{1 + x^2}\). Notice that this formula is handy because it involves a simple and structured approach to finding derivatives, which can be used in a variety of calculus problems. These functions are particularly important in fields like physics and engineering, where understanding angles and slopes are crucial.

Composite Functions

Composite functions are the combination of two or more functions, where one function is placed inside another. In mathematical terms, if you have two functions, f(x) and g(x), then a composite function is written as \(f(g(x))\). The problem at hand is a perfect example: \(y = (\tan^{-1} x)^2\). Here, \(\tan^{-1} x\) is the inner function, and squaring it is the outer function. When dealing with derivatives of composite functions, each layer must be carefully peeled away using the chain rule. Understanding these layers and using appropriate differentiation techniques for each is key to mastering calculus derivatives.