Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=(3 x-2)^{2}\left(3-x^{2}\right)^{2} $$

Step-by-Step Solution

Answer

Use the product rule and simplify: \( D_x y = 6(3x-2)(3-x^2)^2 - 4x(3x-2)^2(3-x^2) \).

Step 1: Identify the Function Structure

The function is a product of two parts: \((3x-2)^2\) and \((3-x^2)^2\). We will use the product rule to differentiate it.

Step 2: Apply the Product Rule

According to the product rule, if \(y = u imes v\), then \(D_x y = u'v + uv'\). Here, \(u = (3x-2)^2\) and \(v = (3-x^2)^2\).

Step 3: Differentiate the First Part (u)

Set \(u = (3x-2)^2\). Differentiate \(u\) with respect to \(x\) using the chain rule: \(u' = 2(3x-2) \cdot 3 = 6(3x-2)\).

Step 4: Differentiate the Second Part (v)

Set \(v = (3-x^2)^2\). Differentiate \(v\) with respect to \(x\) using the chain rule: \(v' = 2(3-x^2) \cdot (-2x) = -4x(3-x^2)\).

Step 5: Substitute into the Product Rule Formula

Substitute \(u\), \(v\), \(u'\), and \(v'\) into the product rule formula: \(D_x y = [6(3x-2)](3-x^2)^2 + (3x-2)^2[-4x(3-x^2)]\).

Step 6: Simplify the Expression

Distribute and combine like terms in the expression obtained from the product rule to simplify: First Term: \(6(3x-2)(3-x^2)^2\) Second Term: \(-4x(3x-2)^2(3-x^2)\). Expand and simplify both terms, then combine them carefully to get the final expression for \(D_x y\).

Step 7: Final Answer

After simplification, the derivative \(D_x y\) is: \[ D_x y = 6(3x-2)(3-x^2)^2 - 4x(3x-2)^2(3-x^2) \]. Further factor if possible.

Key Concepts

Chain Rule

In calculus, the chain rule is a fundamental technique for differentiating composite functions. It helps you find the derivative of a function that is nested within another function. Consider a function of the form \( h(x) = f(g(x)) \). Here, the function \( g(x) \) is inside the function \( f \), making it a composite function.
The chain rule states that you differentiate \( h(x) \) with respect to \( x \) by multiplying the derivative of the outer function \( f \) evaluated at \( g(x) \) by the derivative of the inner function \( g(x) \). The formula is:

• \( h'(x) = f'(g(x)) \cdot g'(x) \)

This rule is quite useful, especially when dealing with powers or other operations between functions, helping us express derivatives when functions are intertwined.

Differentiation

Differentiation is the process used in calculus to compute the derivative of a function. It measures how a function changes as its input changes, giving us the function's rate of change or slope at any point. In practical terms, the derivative tells us how steep a curve is at any specific point along the graph of a function.
When applying differentiation, we often utilize various rules:

• Constant Rule: The derivative of a constant is zero.
• Power Rule: For \( x^n \), the derivative is \( nx^{n-1} \).
• Sum Rule: The derivative of a sum of functions is the sum of their derivatives.
• Product Rule: Used when dealing with the product of two functions, described in detail below.

Differentiation is essential for finding tangents, optimizing functions, and solving complex problems across various fields.

Derivatives

Derivatives represent a cornerstone concept in calculus, symbolizing the instantaneous rate of change of a function with respect to one of its variables. Derivatives are not just limited to simple expressions but apply across polynomials, trigonometric, exponential functions, and beyond.
The notation \( f'(x) \) or \( \frac{df}{dx} \) is often used to denote the derivative of a function \( f \). The idea of taking a derivative is derived from the concept of a limit, which allows us to see how a function behaves as it approaches a certain point.
Understanding derivatives is critical for:

• Determining the slope at a given point on a graph.
• Identifying local extrema (i.e., maximums and minimums).
• Analysing real-world scenarios, like velocity in physics or marginal cost in economics.

Function Structure

The structure of a function fundamentally affects the approach used to differentiate it. When encountering a function that is the product of two or more expressions, examining its structure helps us apply the correct differentiation rules.
In multicomponent functions like those presented in the exercise, identifying the function's constituents is essential:

• Identify each part of the function, for example, \( u \) and \( v \).
• Utilize appropriate differentiation rules, e.g., product rule or chain rule, as required.
• Re-check the structure, ensuring correct application of the relevant derivative rules and simplification method.

Understanding a function's composition impacts each step of the differentiation process, allowing us to apply various techniques properly and ultimately solve complex functions efficiently.