Problem 39 - Calculus Early Transcendentals

Question

Find \(d y / d x\) $$ y=\frac{(2 x+3)^{3}}{\left(4 x^{2}-1\right)^{8}} $$

Step-by-Step Solution

Answer

Apply the quotient and chain rules to differentiate: \(d y/d x = \frac{(4x^2-1)^8 (6(2x+3)^2) - (2x+3)^3 (64x(4x^2-1)^7)}{((4x^2-1)^8)^2}\)."

Step 1: Identify the Differentiation Rule

The function given is a quotient of two functions, so we need to use the quotient rule for differentiation, which states that if you have a function given by \(\frac{u}{v}\), then its derivative is \(\frac{v \cdot u' - u \cdot v'}{v^2}\). Here \(u=(2x+3)^3\) and \(v=(4x^2-1)^8\).

Step 2: Differentiate the Numerator Using Chain Rule

To find \(u'\), we use the chain rule on \(u=(2x+3)^3\). Let \(g(x) = 2x+3\), then \(u = g(x)^3\). The derivative using chain rule is \(u' = 3g(x)^2 \cdot g'(x) = 3(2x+3)^2 \cdot 2 = 6(2x+3)^2\).

Step 3: Differentiate the Denominator Using Chain Rule

To find \(v'\), apply the chain rule to \(v=(4x^2-1)^8\). Let \(f(x) = 4x^2-1\), then \(v = f(x)^8\). Using the chain rule, \(v' = 8f(x)^7 \cdot f'(x) = 8(4x^2-1)^7 \cdot 8x = 64x(4x^2-1)^7\).

Step 4: Apply the Quotient Rule

Substitute \(u\), \(u'\), \(v\), and \(v'\) into the quotient rule formula: \[\frac{d y}{d x} = \frac{(4x^2-1)^8 \cdot 6(2x+3)^2 - (2x+3)^3 \cdot 64x(4x^2-1)^7}{((4x^2-1)^8)^2}\] Simplify the expression wherever possible, which will give the derivative of the original function.

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental technique in calculus used for finding the derivative of a quotient of two functions. When you have a function that is divided by another, such as \( \frac{u}{v} \), the quotient rule is your go-to method. It is essential because it helps to efficiently compute derivatives in situations where one function is nested within another.
Here's how it works:

• Consider \( u = (2x+3)^3 \) and \( v = (4x^2-1)^8 \).
• The quotient rule states that, \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} \).
• This means we need to find the derivatives of both \( u \) and \( v \), represented by \( u' \) and \( v' \) respectively.

Breaking down the derivation according to this rule allows us to find the derivative step by step while ensuring accuracy.

Chain Rule

The Chain Rule is another key tool in the differentiator's toolkit, particularly useful when dealing with composite functions—functions within other functions. Imagine peeling an onion layer by layer; the chain rule applies this concept to function differentiation.
For our problem:

• We need to apply the chain rule to find \( u' \) and \( v' \).
• For \( u = (2x+3)^3 \), we let \( g(x) = 2x+3 \), thus \( u = (g(x))^3 \).
• So \( u' = 3(g(x))^2 \cdot g'(x) = 6(2x+3)^2 \).
• For \( v = (4x^2-1)^8 \), insights are drawn by letting \( f(x) = 4x^2-1 \), hence \( v = (f(x))^8 \).
• Thus, \( v' = 8(f(x))^7 \cdot f'(x) = 64x(4x^2-1)^7 \).

This step-wise application of the chain rule simplifies the differentiation process, cutting through complex compositions with ease.

Derivative Simplification

After applying derivative techniques like the quotient and chain rules, it's crucial to simplify the result. Simplifying derivative expressions not only makes the solution clearer, but can also reduce computational workload in subsequent calculations.
Following the differentiation of both the numerator and the denominator, their expressions are then incorporated into the quotient rule formula:

• Substitute \(u = (2x+3)^3\), \(u' = 6(2x+3)^2\), \(v = (4x^2-1)^8\), and \( v' = 64x(4x^2-1)^7\).
• The formula becomes \[ \frac{(4x^2-1)^8 \cdot 6(2x+3)^2 - (2x+3)^3 \cdot 64x(4x^2-1)^7}{((4x^2-1)^8)^2} \]
• Now, simplify where possible. This might involve expanding terms or factoring to make the solution more understandable and manageable.

Each simplification step makes the final derivative expression more elegant and easier to interpret.