Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\sec ^{-1}\left(x^{3}\right) $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{3}{x \sqrt{x^6 - 1}} \) for \( x > 0 \).

Step 1: Understand the Problem

You are asked to find the derivative of the function \( y = \sec^{-1}(x^3) \) with respect to \( x \). To do this, we'll use the derivative formula for the inverse secant function \( \sec^{-1}(u) \) with respect to \( u \).

Step 2: Recall the Derivative Formula for Inverse Secant

The derivative of \( \sec^{-1}(u) \) with respect to \( u \) is \( \frac{1}{|u|\sqrt{u^2 - 1}} \). We need to apply this formula to find \( \frac{dy}{du} \) where \( u = x^3 \).

Step 3: Differentiate Using Chain Rule

Use the chain rule to differentiate \( y = \sec^{-1}(x^3) \). We have:\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]where \( \frac{du}{dx} = 3x^2 \) since \( u = x^3 \).

Step 4: Apply the Inverse Secant Derivative Formula

Substitute our expression for \( u \) in the inverse secant derivative formula:\[ \frac{dy}{du} = \frac{1}{|x^3| \sqrt{(x^3)^2 - 1}} = \frac{1}{|x^3| \sqrt{x^6 - 1}} \]

Step 5: Combine Results Using Chain Rule

Combine the derivative results to find:\[ \frac{dy}{dx} = \frac{1}{|x^3| \sqrt{x^6 - 1}} \cdot 3x^2 \]This simplifies to:\[ \frac{dy}{dx} = \frac{3x^2}{|x^3| \sqrt{x^6 - 1}} \]

Step 6: Simplify the Expression

Notice that \( |x^3| = x^3 \) when \( x > 0 \), and \( |x^3| = -x^3 \) when \( x < 0 \). Incorporate the absolute value into the expression if needed. Thus:\[ \frac{dy}{dx} = \frac{3}{x \sqrt{x^6 - 1}} \] for \( x > 0 \).For general \( x \), consider the absolute value in context depending on the problem domain, which may result in additional conditions.

Key Concepts

Chain Rule

When tackling differentiation problems, understanding the chain rule is crucial. It acts like a bridge between functions, allowing us to differentiate composite functions seamlessly. The basic idea behind the chain rule is to break down complex functions into simpler parts. For a function given as a composition of two functions, say \( y = f(g(x)) \), the derivative can be determined using

• \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)

Here, \( u = g(x) \), making it easier to substitute and find the derivatives step-by-step.
In our example, \( y = \sec^{-1}(x^3) \), we first find \( \frac{dy}{du} \), where \( u = x^3 \), then multiply it by \( \frac{du}{dx} = 3x^2 \). The chain rule essentially helps navigate through layers of functions to get the result neatly.

Inverse Secant Function

The inverse secant function is a bit of an enigma in calculus. It is not as commonly encountered as other trigonometric functions, but knowing how to deal with it is important. The inverse secant, denoted as \( \sec^{-1}(x) \), is the inverse function of the secant function within its restricted domain. Since it is an inverse function, finding its derivative uses a unique formula.

• The derivative for \( \sec^{-1}(u) \) is \( \frac{1}{|u| \sqrt{u^2 - 1}} \)

This formula captures the essence of the function's behaviour in tangent to its curve. In the problem, applying this formula allows us to capture how \( y = \sec^{-1}(x^3) \) changes with respect to \( x \). Adjustments to account for additional variable transformations, like \( u = x^3 \), are often required.

Differentiation Techniques

Differentiation is the crux of calculus problem-solving. It provides tools to determine how functions change, which is vital in understanding dynamic systems in maths and real life. Several techniques exist, and knowing when to use each can simplify complex problems.

• Basic Rules: Power rule, product rule, and quotient rule.
• Advanced Techniques: Chain rule, implicit differentiation, and derivatives of inverse functions.

In the given exercise, the combination of differentiating an inverse function and a polynomial necessitates a strategic approach:
1. Recognize the function as a composite (one inside another).
2. Use the chain rule to connect different parts of the function's composition.
3. Implement the specific derivative rules for inverse trigonometric functions.

Calculus Problem-Solving

Solving calculus problems effectively involves integrating multiple concepts at once. Each problem is like a puzzle, where identifying the right approach is key. For students, familiarity with various calculus tools greatly enhances problem-solving skills.

• Step-by-step analysis often helps decode what initially seems complex.
• Understanding each component of a function sheds light on its derivative.

For \( y = \sec^{-1}(x^3) \), understanding the inverse secant's properties and applying the chain rule correctly are pivotal steps.
Many issues students face stem from overlooking the domain restrictions or misapplying rules without considering how they interact. By practicing similar exercises, students bolster their calculus toolkit, becoming adept at navigating seemingly difficult problems.