Problem 227 - Calculus Volume 1

Question

For each of the following exercises, a. decompose each function in the form \(y=f(u)\) and \(u=g(x),\) and b. find \(\frac{d y}{d x}\) as a function of \(x .\) $$y=-6 \sin ^{-3} x$$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{18 \cos x}{\sin^4 x}\)

Step 1: Identify Composition Functions

To decompose the function in the form \(y = f(u)\) and \(u = g(x)\), we first need to express the given function \(y = -6 \sin^{-3} x\) in terms of a composite function. Here, we can set \(u = \sin x\). Thus, our expression becomes \(y = -6 u^{-3}\) where \(u = g(x) = \sin x\).

Step 2: Differentiate Inner Function

Now differentiate the inner function \(u = g(x) = \sin x\) with respect to \(x\). This yields:\[\frac{du}{dx} = \cos x\].

Step 3: Differentiate Outer Function

Next, differentiate the outer function \(y = f(u) = -6u^{-3}\) with respect to \(u\). This yields:\[\frac{dy}{du} = \frac{d}{du}(-6u^{-3}) = 18u^{-4}\].

Step 4: Apply Chain Rule

To find \(\frac{dy}{dx}\), apply the chain rule:\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\].Substitute the derivatives from the previous steps:\[\frac{dy}{dx} = 18u^{-4} \cdot \cos x\].

Step 5: Substitute Back \(u\) with \(\sin x\)

Return to the original variable \(x\), substituting \(u = \sin x\) back into the derivative:\[\frac{dy}{dx} = 18(\sin x)^{-4} \cdot \cos x = \frac{18 \cos x}{\sin^4 x}\].

Key Concepts

Composite Functions

In calculus, composite functions involve combining two functions where the output of one function becomes the input to another. They can be thought of as functions-within-functions, much like nesting dolls, where one fits inside another.
This concept comes into play when we work with problems like decomposing functions for differentiation.
Consider the function given in the exercise: \( y = -6 \sin^{-3} x \). To break it into composite functions, we identify it as \( y = f(u) \) and \( u = g(x) \), where \( u = \sin x \). The outer function \( f(u) \) is \( -6u^{-3} \), using \( u \) (i.e., \( \sin x \)) as its variable.

• The inner function \( g(x) = \sin x \).
• The outer function \( f(u) = -6u^{-3} \).

Breaking down functions in this manner allows us to tackle complex derivatives using the Chain Rule, one layer at a time.

Differentiation

Differentiation is a fundamental process in calculus that measures how a function changes as its input changes, essentially finding the rate of change or slope at any given point. It is vital for understanding how functions behave and for solving real-world problems involving rates, like speed or growth.
In the context of our exercise, we perform differentiation on both the inner and outer components of the composite function. First, we differentiate the inner function \( u = \sin x \) with respect to \( x \), which gives us the derivative: \( \frac{du}{dx} = \cos x \). This tells us how \( u \) changes as \( x \) changes.
Then, we differentiate the outer function \( y = -6u^{-3} \) with respect to \( u \), providing the result \( \frac{dy}{du} = 18u^{-4} \). This derivative indicates how \( y \) responds to changes in \( u \). These efforts set the stage for applying the Chain Rule to complete the differentiation process.

Trigonometric Functions

Trigonometric functions are crucial in mathematics, often representing periodic phenomena like waves. These functions, such as sine and cosine, appear extensively in calculus, especially when differentiation is involved.
In this exercise, the function \( \sin x \) serves as the inner function in our composite function decomposition. We rely on its derivative for the differentiation process. The derivative of \( \sin x \) is \( \cos x \), a basic yet essential trigonometric identity used to find how rapidly the sine function increases or decreases. This derivative is instrumental in solving our problem, affecting how the entire composite function changes with \( x \).
Using trigonometric functions in calculus not only helps solve academic exercises but also models real-world scenarios, like describing oscillations or wave patterns. Understanding these basic derivatives boosts one's ability to handle more advanced calculus concepts that build upon these foundational ideas.