Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{3(1 + \sin^{-1} x)^2}{\sqrt{1-x^2}} \).

Step 1: Understand the Problem

We have the function \( y = (1 + \sin^{-1} x)^3 \) and want to find the derivative of \( y \) with respect to \( x \), which is denoted as \( D_x y \) or \( \frac{dy}{dx} \).

Step 2: Identify the Differentiation Rules Needed

We will use the chain rule which states that \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \). Additionally, we will need the derivative of \( \sin^{-1} x \), which is \( \frac{1}{\sqrt{1-x^2}} \).

Step 3: Apply the Chain Rule

First, let \( u = 1 + \sin^{-1} x \). Then, \( y = u^3 \). Use the chain rule: \( \frac{dy}{dx} = \frac{d}{du}(u^3) \cdot \frac{du}{dx} \).

Step 4: Differentiate \( u^3 \) with Respect to \( u \)

The derivative of \( u^3 \) with respect to \( u \) is \( 3u^2 \).

Step 5: Differentiate \( u = 1 + \sin^{-1} x \) with Respect to \( x \)

The derivative of \( u \) with respect to \( x \) is \( 0 + \frac{1}{\sqrt{1-x^2}} = \frac{1}{\sqrt{1-x^2}} \).

Step 6: Combine the Results

According to the chain rule, we have \( \frac{dy}{dx} = 3u^2 \cdot \frac{1}{\sqrt{1-x^2}} \). Substitute \( u = 1 + \sin^{-1} x \) back into the expression to get \( \frac{dy}{dx} = 3(1 + \sin^{-1} x)^2 \cdot \frac{1}{\sqrt{1-x^2}} \).

Key Concepts

chain rule

The chain rule is a fundamental tool in calculus differentiation that simplifies finding derivatives of composite functions. In simpler terms, a composite function is where one function is nested inside another. To break it down, imagine a function like \( y = f(g(x)) \). Here, \( g(x) \) is inside \( f \), so it's a composite function.
The chain rule helps us differentiate these kinds of functions swiftly. Here's a pro-tip: think of it as peeling off layers, one at a time, like an onion!

• The rule states: \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \).
• First, differentiate the outside function \( f \) concerning \( g(x) \), leaving the inside function unchanged.
• Next, multiply by the derivative of the inside function \( g \) with respect to \( x \).

This method is invaluable when dealing with chains of functions, paving the way to finding elusive derivatives easily.

inverse trigonometric functions

Inverse trigonometric functions are the backward versions of the familiar sine, cosine, and tangent functions. When working with these, you can find an angle when you know the trigonometric ratio. In calculus, understanding these functions opens up a world of solving interesting problems.
For instance, \( \sin^{-1}(x) \) is an example of an inverse trigonometric function. If \( \sin(\theta) = x \), then \( \theta = \sin^{-1}(x) \). It's like asking, "What angle has this sine value?"
Using these functions:

• You'll often encounter restrictions on their ranges to ensure they each remain a true function. For example, the principal range of \( \sin^{-1} \) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
• These are handy in scenarios where solutions involve angles, arcs, or oscillatory patterns.

In calculus, solving derivatives involving these means we get to apply the chain rule, making math a bit more fascinating!

derivative of inverse sine

The derivative of the inverse sine function is an essential element in calculus that often pops up in problems involving shapes, oscillations, and waves. Let’s unwrap this concept with our calculator-like curiosity!
The function \( \sin^{-1}(x) \) tells us what angle \( \theta \) has its sine equal to \( x \). The derivative of this function is crucial when you're tasked with finding rates of change.
For \( \sin^{-1}(x) \):

• The derivative is \( \frac{1}{\sqrt{1-x^2}} \).
• This stems from the Pythagorean identity and enables us to capture the rate of change of angles with precision.

Understanding this derivative allows us to effectively apply it in more complex scenarios, such as in composite functions, where it often finds its way inside a larger expression.
By leveraging this knowledge, differentiating functions containing inverse sine becomes a gratifying puzzle in the broader game of calculus!