Problem 39 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\tan ^{-1} \sqrt{x^{2}-1}+\csc ^{-1} x, \quad x>1$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Functions

We need to differentiate two inverse trigonometric functions, which are \( y = \tan^{-1} \sqrt{x^2 - 1} \) and \( y = \csc^{-1} x \). Use their differentiation formulas in the next steps.

Step 2: Differentiate \(\tan^{-1}\sqrt{x^2 - 1}\)

For \( y_1 = \tan^{-1} \sqrt{x^2 - 1} \), use the chain rule. Let \( u = \sqrt{x^2 - 1} \). Then \( y_1 = \tan^{-1} u \). The derivative is \( \frac{d}{dx}(y_1) = \frac{1}{1+u^2} \cdot \frac{du}{dx} \). Calculating \( \frac{du}{dx} \), we have \( \frac{1}{2\sqrt{x^2 - 1}} \cdot 2x = \frac{x}{\sqrt{x^2 - 1}} \). So, \( \frac{dy_1}{dx} = \frac{1}{1+x^2-1} \cdot \frac{x}{\sqrt{x^2-1}} = \frac{x}{x^2 \sqrt{x^2 - 1}} \).

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

For \( y_2 = \csc^{-1} x \), use the formula \( \frac{d}{dx}(\csc^{-1} x) = -\frac{1}{x \sqrt{x^2-1}} \).

Step 4: Combine the Derivatives

Combine the derivatives from Step 2 and Step 3: \( \frac{dy}{dx} = \frac{x}{x^2 \sqrt{x^2 - 1}} - \frac{1}{x \sqrt{x^2-1}} \). Factor the common denominator: \( \frac{dy}{dx} = \frac{x - 1}{x^2 \sqrt{x^2 - 1}} \).

Step 5: Simplify the Expression

The final expression can be simplified if needed, but here it is already in a relatively simple form. The derivative of \( y \) with respect to \( x \) is \( \frac{x - 1}{x^2 \sqrt{x^2 - 1}} \).

Key Concepts

Chain Rule

When differentiating complex functions like the inverse trigonometric functions, one crucial tool is the Chain Rule. The Chain Rule is a formula for calculating the derivative of a composite function. A composite function is a function made up of other functions, like when you substitute one function into another.

In the exercise, we saw the function \(y = \tan^{-1} \sqrt{x^2 - 1}\). This is a composite function because it contains the square root function inside the arctangent function. To find its derivative, we employ the Chain Rule.

Let's break it down:

• Identify the inner function \(u = \sqrt{x^2 - 1}\).
• The outer function is \(y = \tan^{-1} u\).

To find the derivative, follow these steps:

• First, find the derivative of the outer function with respect to \(u\): \( \frac{d}{du}(\tan^{-1} u) = \frac{1}{1 + u^2} \).
• Next, find the derivative of the inner function \(u\) with respect to \(x\): \( \frac{du}{dx} = \frac{x}{\sqrt{x^2 - 1}} \).

Finally, apply the Chain Rule by multiplying these derivatives together to get the final result for the derivative of the composite function. This process is fundamental when solving problems involving inverse trigonometric differentiation.

Inverse Trigonometric Differentiation

Inverse trigonometric functions, such as \( \tan^{-1} x \) and \( \csc^{-1} x \), have specific formulas for their derivatives, which are essential to solve problems like the one in the exercise. These derivatives are critical tools for calculus problems involving angles and trigonometric identities.

The derivative of an inverse tangent function, \( \tan^{-1} x \), is \( \frac{1}{1 + x^2} \). But when you have a composition, such as \( \tan^{-1} \sqrt{x^2 - 1} \), always remember to apply the Chain Rule thereafter.

Similarly, for \( \csc^{-1} x \), the derivative formula is \( -\frac{1}{x\sqrt{x^2-1}} \). This formula is directly used because it is already in the form that considers the particular inverse function properties.

When approaching differentiation with inverse functions:

• Identify the specific inverse trigonometric function and apply its corresponding differentiation formula directly.
• If the function is composite, remember to further utilize the Chain Rule.

Knowing these rules and how to apply them effectively allows you to handle more advanced topics in calculus, especially those involving trigonometric identities and transformations.

Simplification of Derivatives

Simplifying derivatives is a valuable skill in calculus because it often leads to cleaner and more usable results, especially when dealing with complex expressions. In this exercise, the initial derivative expressions are combined and then simplified.

Starting with the expression from the differentiation process: \( \frac{dy}{dx} = \frac{x}{x^2 \sqrt{x^2 - 1}} - \frac{1}{x \sqrt{x^2-1}} \). It's crucial to combine these fractions effectively to simplify the derivative.

This usually involves:

• Finding a common denominator: Factor common terms to help reduce fractions to a single expression.
• Combining terms carefully: Ensure each term is combined correctly to maintain equality.
• Looking for any additional simplifications: Sometimes, factors can cancel each other out, making the expression even simpler.

After following these steps, the exercise solution yields \( \frac{dy}{dx} = \frac{x - 1}{x^2 \sqrt{x^2 - 1}} \). This final step of simplification results in a neat and more manageable expression, which is easier to interpret and use in further calculations or solutions.