Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-20, find \(D_{x} y\). $$ y=\left(2-3 x^{2}\right)^{4}\left(x^{7}+3\right)^{3} $$

Step-by-Step Solution

Answer

Differentiate using product and chain rules: \( D_{x} y = (-24x(2 - 3x^2)^3)(x^7 + 3)^3 + (2 - 3x^2)^4(21x^6)(x^7 + 3)^2 \).

Step 1: Identify the Functions

We have the function \( y \) given as a product of two functions: \( u = (2 - 3x^2)^4 \) and \( v = (x^7 + 3)^3 \). We will apply the product rule for differentiation.

Step 2: Apply the Product Rule

The product rule states that if \( y = uv \), then \( D_{x} y = u'v + uv' \). We need to find the derivatives \( u' \) and \( v' \).

Step 3: Differentiate \( u = (2 - 3x^2)^4 \)

Use the chain rule to differentiate. Let \( g(x) = 2 - 3x^2 \), so \( u = [g(x)]^4 \). The derivative \( u' = 4[g(x)]^3 \cdot g'(x) \). First, find \( g'(x) = -6x \). Thus, \( u' = 4(2 - 3x^2)^3(-6x) = -24x(2 - 3x^2)^3 \).

Step 4: Differentiate \( v = (x^7 + 3)^3 \)

Use the chain rule similarly. Let \( h(x) = x^7 + 3 \), so \( v = [h(x)]^3 \). The derivative \( v' = 3[h(x)]^2 \cdot h'(x) \). First, find \( h'(x) = 7x^6 \). Thus, \( v' = 3(x^7 + 3)^2(7x^6) = 21x^6(x^7 + 3)^2 \).

Step 5: Combine Using the Product Rule

Substitute \( u' \) and \( v' \) back into the product rule: \[D_{x} y = ((-24x)(2 - 3x^2)^3)(x^7 + 3)^3 + (2 - 3x^2)^4(21x^6)(x^7 + 3)^2.\] Simplify the expression to finalize.

Key Concepts

Product Rule

The product rule is a fundamental differentiation technique used when dealing with the derivative of a product of two functions. If you have two functions, say \( u(x) \) and \( v(x) \), and you want to differentiate their product \( y = u \times v \), you can’t simply differentiate each one separately and then multiply. Instead, the product rule states:

• If \( y = uv \), then the derivative \( D_{x} y = u'v + uv' \).

This rule helps incorporate the changes in both functions as they contribute to \( y \). When applying this rule,- First, develop the derivative of each function separately: \( u' \) and \( v' \).- Next, substitute into the product rule formula to obtain \( D_{x} y \).
By using this, we capture the interaction between both functions' rates of change.

Chain Rule

The Chain Rule is pivotal for differentiating composite functions, where one function is inside another. Suppose you have a function \( y = f(g(x)) \), the chain rule allows you to differentiate it by multiplying the derivative of \( f \) with respect to \( g \) by the derivative of \( g \) with respect to \( x \). Symbolically, it's expressed as:\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]For example, when finding the derivative of \( u = (2 - 3x^2)^4 \):- Let \( g(x) = 2 - 3x^2 \), so \( u = [g(x)]^4 \).- Then find \( u' = 4[g(x)]^3 \cdot g'(x) \), where \( g'(x) = -6x \).
This allows us to smoothly separate complex layers to differentiate effectively.

Polynomial Differentiation

Polynomial differentiation involves rules to find derivatives of polynomial expressions. A polynomial is generally expressed as \( ax^n \) where \( a \) is a constant and \( n \) is a non-negative integer. The basic rule is straightforward:

• The derivative of \( ax^n \) is \( nax^{n-1} \).

This rule derives from the sum rule in calculus, which states that the derivative of a sum of functions is the sum of their derivatives. Therefore, to differentiate a polynomial:- Take each term individually, and apply the power rule.- For \( ax^n \), multiply \( a \) by \( n \) and reduce the power by one.
By applying this rule to each term, you find the derivative of the entire polynomial. It simplifies calculations, especially within larger composite functions.

Functions and Derivatives

Functions and their derivatives are central concepts in calculus, representing the relationship between a function and its rate of change. Every function has a derivative, which provides insight into how the function behaves as its input changes.- A function \( f(x) \) depicts how outputs relate to inputs.- The derivative, \( f'(x) \), measures how \( f(x) \) changes as \( x \) varies.In other words, derivatives tell us the slope of the function at any given point. This is especially crucial when optimizing functions or understanding motion in physics. In the problem above, both the product and chain rule work harmoniously to form complex derivatives, revealing the dynamic changes of the original function's behavior as \( x \) shifts. Through practice, navigating between functions and derivatives becomes consistent, unveiling underlying patterns and trends.