Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\sin ^{-1}\left(2 x^{2}\right) $$

Step-by-Step Solution

Answer

The derivative is \( \frac{4x}{\sqrt{1-4x^4}} \).

Step 1: Understand the Objective

We are tasked with finding the derivative of the function \( y = \sin^{-1}(2x^2) \) with respect to \( x \). This involves using the chain rule and the derivative of the inverse sine function.

Step 2: Identify the Derivative of Inverse Sine

Recall that the derivative of \( \sin^{-1}(u) \) with respect to \( u \) is \( \frac{1}{\sqrt{1-u^2}} \). Here, our function \( u \) is \( 2x^2 \).

Step 3: Apply the Chain Rule

The chain rule states that \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). First, find \( \frac{dy}{du} \), which is \( \frac{1}{\sqrt{1-(2x^2)^2}} \).

Step 4: Differentiate the Inner Function

Now differentiate \( u = 2x^2 \) with respect to \( x \). The derivative \( \frac{du}{dx} = 4x \).

Step 5: Combine the Results

Using the chain rule, substitute \( \frac{dy}{du} \) and \( \frac{du}{dx} \) into \( \frac{dy}{dx} = \frac{1}{\sqrt{1-(2x^2)^2}} \cdot 4x \).

Step 6: Simplify the Expression

Simplify \( \sqrt{1-(2x^2)^2} \) to \( \sqrt{1-4x^4} \), thus the derivative is \( \frac{4x}{\sqrt{1-4x^4}} \).

Key Concepts

Chain Rule in Calculus

The chain rule is a fundamental technique in calculus used to find the derivative of a composite function. In simpler terms, a composite function is one where two or more functions are combined, like in our original problem, where you have the function inside as \( u = 2x^2 \) and the outer as \( y = \sin^{-1}(u) \).

Here’s how the chain rule works step-by-step:

• First, find the derivative of the outer function with respect to the inner function. In this instance, the outer function is \( \sin^{-1}(u) \).
• Second, find the derivative of the inner function with respect to \( x \). Here, that inner function is \( u = 2x^2 \).
• Finally, multiply these derivatives together to get the overall derivative \( \frac{dy}{dx} \).

Using the chain rule allows you to efficiently differentiate complex expressions by breaking them down into simpler components. It becomes an essential tool in solving calculus problems!

Inverse Sine Function

The inverse sine function, denoted as \( \sin^{-1}(x) \) or \( \arcsin(x) \), is used to find an angle whose sine is a given number. This function is important because it allows us to work backwards from the sine of an angle to determine the angle itself.

Key properties of the inverse sine function include:

• Its range is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), representing the angles for which it will return results.
• It is defined for \( x \) values ranging from \(-1\) to \( 1 \).

In differentiation, knowing the derivative of \( \sin^{-1}(u) \) is crucial. This derivative is \( \frac{1}{\sqrt{1-u^2}} \), and it forms part of our solution strategy when working with inverse trigonometric functions.

Recognizing how these derivatives function is instrumental in calculating the derivatives of more elaborate functions like \( \sin^{-1}(2x^2) \). It provides a direct path to handle any inverse sine functions you might encounter.

Differentiation Techniques

Differentiation is all about finding the rate at which one quantity changes with respect to another. In calculus, several techniques can be employed to differentiate functions effectively.

Here are some crucial techniques and tools often used:

• Basic Derivatives: Knowing the derivatives of fundamental functions such as polynomials, sine, cosine, etc., serves as the first step in finding more complex derivatives.
• Chain Rule: As discussed earlier, it’s extremely useful for differentiating composite functions, allowing for the breakdown and simplification of problems.
• Product Rule/Quotient Rule: Essential for handling functions where there’s multiplication or division of two functions.
• Inverse Trigonometric Functions: These derivatives expand your toolkit especially when dealing with trigonometric expressions in reverse form.

Choosing the right differentiation technique makes solving calculus problems both efficient and manageable. Understanding when and how to apply such techniques, like with our example of finding the derivative of \( \sin^{-1}(2x^2) \), can greatly simplify your work.