Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-20, find \(D_{x} y\). $$ y=\cos ^{3}\left(\frac{x^{2}}{1-x}\right) $$

Step-by-Step Solution

Answer

The derivative is \(-3\cos^2\left(\frac{x^2}{1-x}\right)\sin\left(\frac{x^2}{1-x}\right)\frac{2x-x^2}{(1-x)^2}\)."

Step 1: Recognize the Composition of Functions

The problem involves the function \( y = \cos^3\left(\frac{x^2}{1-x}\right) \). It's important to note that this is a composition of functions. First, find the outer function which is \( u^3 \) where \( u = \cos\left(\frac{x^2}{1-x}\right) \). The derivative of \( u^3 \) with respect to \( u \) will be needed.

Step 2: Differentiate the Outer Function

Find the derivative of the outer function \( u^3 \) with respect to \( u \). The derivative is \( 3u^2 \). This gives us part of the chain rule: \( \frac{d}{du}(u^3) = 3\cos^2\left(\frac{x^2}{1-x}\right) \).

Step 3: Recognize the Intermediate Function

Identify the intermediate function \( u = \cos\left(v\right) \) where \( v = \frac{x^2}{1-x} \). We need to differentiate \( \cos(v) \) with respect to \( v \).

Step 4: Differentiate the Intermediate Function

Find the derivative of \( \cos(v) \) with respect to \( v \). This yields \(-\sin(v) \). Therefore, the derivative is \( -\sin\left(\frac{x^2}{1-x}\right) \).

Step 5: Recognize the Inner Function

Identify the inner function \( v = \frac{x^2}{1-x} \). We must find the derivative of this function with respect to \( x \).

Step 6: Differentiate the Inner Function

Use the quotient rule to differentiate \( \frac{x^2}{1-x} \). The rule states: \( \frac{d}{dx}\left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \). Let \( f(x) = x^2 \) and \( g(x) = 1-x \). The derivatives are \( f'(x) = 2x \) and \( g'(x) = -1 \). This yields: \( \frac{2x(1-x) - x^2(-1)}{(1-x)^2} = \frac{2x - 2x^2 + x^2}{(1-x)^2} = \frac{2x - x^2}{(1-x)^2} \).

Step 7: Apply the Chain Rule

Combining all derivatives using the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dv} \times \frac{dv}{dx} = 3\cos^2\left(\frac{x^2}{1-x}\right) \times -\sin\left(\frac{x^2}{1-x}\right) \times \frac{2x - x^2}{(1-x)^2} \).

Step 8: Simplify the Expression

Simplify the expression: \( \frac{dy}{dx} = -3\cos^2\left(\frac{x^2}{1-x}\right) \sin\left(\frac{x^2}{1-x}\right) \frac{2x - x^2}{(1-x)^2} \).

Key Concepts

Chain Rule

The chain rule is essential in differential calculus when dealing with composite functions. It helps us find the derivative of a complex function by breaking it down into simpler parts.
The general idea is that if you have a function that is composed of two other functions, say \( f(g(x)) \), the derivative can be found using the formula:

• \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)

In the problem given, you're dealing with multiple layers of functions: one inside the other. So, we apply the chain rule repeatedly, distinguishing a function inside another and finding derivatives step by step.
Step by step, it allows us to handle each piece of the function, starting from the outermost layer and moving towards the innermost.

Derivatives

Derivatives are a fundamental concept in calculus representing the rate of change. They are critical for understanding how functions behave and are calculated as the limit of the difference quotient.
For a function \( f(x) \), the derivative \( f'(x) \) gives the slope or steepness of the function at any point \( x \). This is represented by the formula:

• \( f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \)

In this exercise, we've used derivatives in different layers of the composition of functions, from the outermost to the innermost. Always ensure you apply the correct rules, like the power rule, when differentiating specific parts of the function.

Composite Functions

Composite functions are functions within functions, denoted as \( f(g(x)) \). They are prevalent in calculus because many real-world problems require modeling complex interactions.
When dealing with composite functions, note:

• Outer Function: This is the last operation you perform if evaluating the function directly.
• Inner Function: This is done first in the operation order.

In the current problem, \( y = \cos^3\left(\frac{x^2}{1-x}\right) \), we consider \( y \) as a composition of functions, breaking it into individual components such as \( u = \cos(v) \). Understanding composite functions helps us apply rules correctly to find derivatives.

Quotient Rule

The quotient rule is necessary when taking derivatives of functions that are ratios of two differentiable functions. It's expressed as:

• \[ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \]

This formula helps us find the derivative of a ratio by applying derivatives to the numerator and the denominator independently.
For our problem, using the quotient rule helped us differentiate \( \frac{x^2}{1-x} \), finding its contribution to the entire derivative calculation. It ensures precise handling of changes in both the numerator and the denominator, which is crucial for accuracy.