Problem 40 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

The derivative is 0.

Step 1: Understand the problem

The given function is a composition of the inverse cotangent and tangent functions: \( y = \cot^{-1} \left( \frac{1}{x} \right) - \tan^{-1}(x) \). Our goal is to find \( \frac{dy}{dx} \).

Step 2: Differentiate \( \cot^{-1} \left( \frac{1}{x} \right) \)

Use the chain rule to differentiate \( \cot^{-1} \left( \frac{1}{x} \right) \). Let \( u = \frac{1}{x} \), then \( y_1 = \cot^{-1}(u) \). Using \( \frac{du}{dx} = -\frac{1}{x^2} \) and \( \frac{dy_1}{du} = -\frac{1}{1+u^2} \), we combine to get \( \frac{dy_1}{dx} = \frac{1}{x^2 + 1} \).

Step 3: Differentiate \(-\tan^{-1} x\)

Use the derivative formula for the inverse tangent: \( \frac{d}{dx} [\tan^{-1} x] = \frac{1}{1+x^2} \). Thus, \( \frac{dy_2}{dx} = -\frac{1}{1+x^2} \).

Step 4: Combine the derivatives

Add the derivatives found to get the derivative of the whole function: \( \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} = \frac{1}{x^2 + 1} - \frac{1}{1+x^2} = 0 \). Thus, the derivative simplifies to zero.

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate compositions of functions. In the context of differentiating functions like \( \cot^{-1} \left( \frac{1}{x} \right) \), the chain rule becomes essential. We apply the chain rule when we have a function inside another function. In our case, \( \cot^{-1}(u) \) is one function, and \( u=\frac{1}{x} \) is another.

Here's how the chain rule works in practice:

• First, differentiate the outer function, \( \cot^{-1}(u) \), with respect to \( u \).
• Second, differentiate the inner function \( u \) with respect to \( x \), which yields \( \frac{du}{dx} = -\frac{1}{x^2} \).
• Finally, multiply these derivatives: \( \frac{dy}{dx} = \frac{dy_1}{du} \times \frac{du}{dx} \).

The chain rule allows us to systematically break down complex derivatives into easier-to-handle pieces, providing clear steps to solve the problem.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \( \cot^{-1} \) and \( \tan^{-1} \), are important in calculus because they allow us to reverse the process of the usual trigonometric functions. When dealing with inverse functions, care must be taken in differentiation due to their unique formulas.

For inverse cotangent, \( \cot^{-1}(u) \), the derivative is:\[\frac{d}{du} [\cot^{-1}(u)] = -\frac{1}{1+u^2}\]Similarly, for inverse tangent, \( \tan^{-1}(x) \), the derivative is:\[\frac{d}{dx} [\tan^{-1}(x)] = \frac{1}{1+x^2}\]These formulas are derived based on the principal values of angles and are key to solving the derivative accurately. Understanding these derivatives is crucial for handling problems involving inverse trigonometric functions.

Differentiation Techniques

Differentiation techniques refer to various strategies used to find the derivative of functions. In this exercise, we utilized several techniques to find \( \frac{dy}{dx} \).

Here's a brief overview of the techniques applied:

• We used the chain rule to handle the function \( \cot^{-1} \left( \frac{1}{x} \right) \). This required composing derivatives of the inner and outer functions.
• The subtraction of \( \tan^{-1}(x) \) prompted the use of the standard derivative rule for inverse trigonometric functions \( \frac{d}{dx} [\tan^{-1} x] = \frac{1}{1+x^2} \).
• Combine the two derivatives obtained through addition and subtraction techniques to find the overall derivative of \( y \).

Each technique serves a specific purpose, leading us to accurately find that the derivative simplifies to zero. Mastery of these methods is essential for tackling a wide variety of differentiation problems.