Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The derivative is found using the product and chain rules, then simplified.

Step 1: Identify the Derivative Rule

The expression \( y = (3x - 2)^2(3 - x^2)^2 \) is a product of two functions. Thus, the product rule should be used, which states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then the derivative \( D_x (uv) = u'v + uv' \).

Step 2: Differentiate the First Function

Let \( u(x) = (3x - 2)^2 \). We will use the chain rule to find \( u'(x) \). The chain rule for derivatives states \( \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x) \).Set \( f(t) = t^2 \) and \( g(x) = 3x - 2 \). First, compute \( f'(t) = 2t \) and \( g'(x) = 3 \). So,\[u'(x) = 2(3x - 2) \cdot 3 = 6(3x - 2)\].

Step 3: Differentiate the Second Function

Let \( v(x) = (3 - x^2)^2 \). Again, we will use the chain rule to find \( v'(x) \). Same as before, set \( f(t) = t^2 \), and \( g(x) = 3 - x^2 \). Compute \( f'(t) = 2t \) and \( g'(x) = -2x \). So,\[v'(x) = 2(3 - x^2)(-2x) = -4x(3 - x^2)\].

Step 4: Apply the Product Rule

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

Step 5: Simplify the Expression

Expand and simplify the derivative expression:1. Compute \(6(3x - 2)(3 - x^2)^2\).2. Compute \((3x - 2)^2(-4x)(3 - x^2)\).3. Combine the expressions by performing algebraic simplification to achieve the simplest form of the derivative. This may require factorization or direct expansion based on further expression details.

Key Concepts

Product Rule in Calculus Differentiation

To tackle problems involving the derivative of a product of two functions, we utilize the product rule. The product rule is an essential calculus concept when differentiating expressions where two functions are multiplied together. The rule states:

• If you have two functions, say \( u(x) \) and \( v(x) \), their derivative is \( D_x (uv) = u'v + uv' \).

In simple words, to find the derivative of a product, you differentiate the first function while keeping the second function unchanged. Then, keep the first function unchanged and differentiate the second one. Finally, you add these two results together.

For our exercise, we need to find the derivative of \( y = (3x - 2)^2(3 - x^2)^2 \) using the product rule.
This means we break it into two parts: let \( u(x) = (3x - 2)^2 \) and \( v(x) = (3 - x^2)^2 \). By differentiating each of these separately and following the product rule, we'll find the derivative of the entire expression.

Understanding the Chain Rule

The chain rule is another fundamental concept in calculus differentiation, pivotal for working with composite functions. A composite function is when you have one function nested inside another, like \( f(g(x)) \). The chain rule provides a way to differentiate such functions. It states:

• For a composite function \( f(g(x)) \), the derivative is \( \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x) \).

To understand this better, let's consider our function \( u(x) = (3x - 2)^2 \). We can set \( f(t) = t^2 \) (where \( t = (3x - 2) \)) and apply the chain rule:

• First find \( f'(t) = 2t \).
• Then find \( g'(x) = \frac{d}{dx}(3x - 2) = 3 \).

The derivative \( u'(x) = 2(3x - 2) \cdot 3 = 6(3x - 2) \) is thus obtained.

Similarly, you would apply the chain rule to \( v(x) = (3 - x^2)^2 \) to find \( v'(x) \). This process involves differentiating the inner function and the outer function, then combining them. By understanding the chain rule, you gain tools to deal with more complex differentiation scenarios efficiently.

Derivative Simplification Techniques

Once you have applied the product rule and the chain rule to find derivatives, simplifying these complex expressions is the next step. This is often needed for presenting your answer cleanly and clearly. Simplification involves combining like terms, expanding products, factoring where possible, and sometimes using algebra to rearrange terms more intuitively.

In our step-by-step solution, we've reached the derivative expression:

• \( D_x y = 6(3x - 2)(3 - x^2)^2 + (3x - 2)^2(-4x(3 - x^2)) \)

Here, simplification would involve unpacking these terms so you can clearly see the structure of the expression. Begin by computing each of the terms separately. After expanding and combining like terms, you derive a simplified version of the derivative, which is easier to interpret and use in further calculations.

Simplification is not just about making an expression shorter. It also aids in understanding the dynamics of how functions change. It's a step yielding not only aesthetic completeness but also revealing insights into the behavior of the mathematical model you are dealing with.