Problem 20 - Essential Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the outer function and inner function

The given function is \( y = \tan^{-1}(u) \) where \( u = x - \sqrt{1 + x^2} \). We need to apply the chain rule here.

Step 2: Find the derivative of the outer function

The derivative of \( \tan^{-1}(u) \) with respect to \( u \) is \( \frac{1}{1 + u^2} \). This will be used in the chain rule.

Step 3: Find the derivative of the inner function

The inner function is \( u = x - \sqrt{1+x^2} \). To differentiate it, we use the derivative of \( x \), which is 1, and for \( -\sqrt{1+x^2} \), we use the chain rule to obtain \(-\frac{x}{\sqrt{1+x^2}} \). Thus the derivative of \( u \) is \( 1 - \frac{x}{\sqrt{1+x^2}} \).

Step 4: Apply the chain rule

Using the chain rule, the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = \frac{1}{1 + (x - \sqrt{1+x^2})^2} \cdot \left(1 - \frac{x}{\sqrt{1+x^2}} \right) \).

Step 5: Simplify the expression

Simplify the expression inside the square in the denominator: \((x - \sqrt{1+x^2})^2 = x^2 - 2x\sqrt{1+x^2} + (1+x^2)\). This simplifies to \( 2x^2 + 1 - 2x\sqrt{1+x^2} \). Simplifying further does not lead to an evident simple form.

Key Concepts

Trigonometric Functions

Trigonometric functions are essential in calculus and mathematics, especially when dealing with periodic functions. These functions relate angles of triangles to the ratios of their sides.
Examples include:

• Sine (\( \sin \theta \)
• Cosine (\( \cos \theta \)
• Tangent (\( \tan \theta \)

Each trigonometric function has its respective inverse that helps to find the angle when you know the ratio. In this exercise, we are particularly interested in the inverse tangent function, \( \tan^{-1}(x) \), which provides the angle whose tangent is \( x \).
This function has a range of \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). Using trigonometric functions in calculus often involves differentiating or finding the slope of these functions.

Chain Rule

The chain rule is a fundamental concept in calculus used for differentiating composite functions. A composite function is formed when one function is applied to the result of another function. In this exercise, our function is a composition of the inverse tangent and a square root function.
Mathematically, if you have a function \( y = f(g(x)) \), then the derivative is found using:

• Find the derivative of the outer function \( f(x) \) with respect to \( g(x) \): \( f'(g(x)) \)
• Find the derivative of the inner function \( g(x) \), which is \( g'(x) \)

Then, apply the chain rule: \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
In our case, the outer function is \( \tan^{-1}(u) \) and the inner function is \( u = x - \sqrt{1 + x^2} \). The chain rule helps in effectively calculating the derivative of these composite functions.

Inverse Trigonometric Functions

Inverse trigonometric functions are used to determine the angle when given a trigonometric ratio. These functions are crucial when solving trigonometric equations and finding derivatives in calculus. In our function, \( \tan^{-1}(x) \) is used to denote the angle whose tangent equals \( x \).
These inverse functions have specific derivatives, which are necessary for applying the chain rule. For \( \tan^{-1}(x) \), the derivative with respect to \( x \) is:

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

Recognizing this derivative helps streamline calculations when differentiating composite functions that include inverse trigonometric expressions.

Function Differentiation

Function differentiation is the process of finding how a function changes at any given point. It's fundamental in calculating rates of change and understanding the behavior of functions. Differentiation results in a derivative, which gives the slope or rate of change.
Steps in differentiation include:

• Identifying parts of the function requiring different rules: like power chain or product rules.
• Applying rules methodically, such as the chain rule for composite functions.
• Simplifying results to make them more interpretable or applicable in calculations.

In this exercise, an important part was simplifying the result of differentiating \( y = \tan^{-1}(x - \sqrt{1+x^2}) \). Derivatives tell us how functions like this change with \( x \), providing insights into function graphs and behaviors.