Problem 45 - Calculus - AP Edition

Question

Find \(d y / d x\). $$y=\sin ^{-1}(1 / x)$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -\frac{1}{x \sqrt{x^2-1}} \).

Step 1: Identify the Function to Differentiate

The given function is \( y = \sin^{-1}(1/x) \). We need to find the derivative of this function with respect to \( x \).

Step 2: Apply the Chain Rule

To differentiate \( y = \sin^{-1}(1/x) \), we will use the chain rule. Let \( u = 1/x \). Then \( y = \sin^{-1}(u) \).

Step 3: Differentiate the Outer Function

The derivative of \( y = \sin^{-1}(u) \) with respect to \( u \) is \( \frac{dy}{du} = \frac{1}{\sqrt{1-u^2}} \).

Step 4: Differentiate the Inner Function

Differentiate \( u = 1/x \) with respect to \( x \). We have \( \frac{du}{dx} = -1/x^2 \).

Step 5: Apply the Chain Rule

Combine the derivatives using the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\sqrt{1-(1/x)^2}} \cdot \left(-\frac{1}{x^2}\right) \).

Step 6: Simplify the Expression

Simplify the expression for \( \frac{dy}{dx} \). First, simplify the term inside the square root: \( 1-(1/x)^2 = 1-1/x^2 = \frac{x^2-1}{x^2} \). Thus, the expression becomes \( \frac{dy}{dx} = -\frac{1}{x^2 \sqrt{\frac{x^2-1}{x^2}}} = -\frac{1}{x \sqrt{x^2-1}} \).

Key Concepts

Understanding the Chain Rule

The chain rule is an essential differentiation technique used when dealing with composite functions. It enables us to differentiate functions that are nested within one another. A composite function can be thought of as a function within another function. For example, in the expression \( y = \sin^{-1}(1/x) \), there is an outer function (the inverse sine function) and an inner function (\( 1/x \)).

To apply the chain rule, it's vital to differentiate each function individually and then multiply their derivatives together. You start by identifying these two separate functions:

• The outer function: \( y = \sin^{-1}(u) \)
• The inner function: \( u = 1/x \)

The rule states that if \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) is found as \( \frac{dy}{du} \cdot \frac{du}{dx} \).

This method breaks down complex functions into simpler parts, making them easier to differentiate systematically. Thus, the chain rule is a fundamental stepping-stone in calculus, particularly useful for addressing functions with layers of expressions.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \( \sin^{-1}, \cos^{-1}, \) and \( \tan^{-1} \), are used to find angles given trigonometric ratios. In calculus, it's crucial to understand their properties to differentiate them correctly.

For \( y = \sin^{-1}(u) \), the derivative is expressed as \( \frac{dy}{du} = \frac{1}{\sqrt{1 - u^2}} \). This formula helps us compute the rate of change of the angle as \( u \) changes, and it derives from implicitly differentiating the equation \( u = \sin(y) \).

These derivatives are crucial when dealing with inverse trigonometric functions in calculus because they allow us to handle scenarios where we need to reverse standard trigonometric processes.

• \( \sin^{-1} \) gives an angle whose sine is a specific value.
• The range of \( \sin^{-1} \) is restricted to \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

Understanding these fundamentals is necessary to carry out differentiation tasks accurately and ensure the correct application of rules and formulas.

Differentiation Techniques as a Whole

Differentiation techniques are a set of strategies and rules that allow us to find derivatives of functions efficiently. Mastering these techniques is fundamental in calculus to solve more complicated problems. Common methods include:

• The power rule: Used for differentiating terms like \( x^n \).
• The product rule: Applied to products of functions.
• The quotient rule: For differentiating quotients of functions.

The chain rule, which we applied earlier, is another vital technique, especially for functions composed of other functions.

Each differentiation technique addresses specific function types. For example, the power rule quickly helps find derivatives of polynomial functions, while the product and quotient rules are indispensable for more complex algebraic expressions.

Having a strong grasp of these techniques allows you to approach calculus problems methodically and effectively, utilizing the best strategy to simplify and solve derivatives. With practice and understanding, you'll be able to tackle a wide array of differentiations, from basic to advanced.