Problem 46 - Calculus for Biology and Medicine

Question

In Problems \(40-46\), find \(\frac{d y}{d x}\) by applying the chain rule repeatedly. $$ y=\left(\frac{(2 x+1)^{2}-x}{\left(3 x^{3}+1\right)^{3}-x}\right)^{2} $$

Step-by-Step Solution

Answer

Differentiate using the chain rule and quotient rule to find \( \frac{dy}{dx} = 2u \cdot \frac{v'w - vw'}{w^2} \), where \( u = \frac{(2x+1)^2 - x}{(3x^3+1)^3 - x} \).

Step 1: Identify the outer function

We begin by identifying the outermost function in the given expression. The function given is \( y = \left( (2x+1)^2 - x \big/ (3x^3 + 1)^3 - x \right)^2 \). Here, the outer function is \( u^2 \) where \( u = \frac{(2x+1)^2 - x}{(3x^3+1)^3 - x} \).

Step 2: Differentiate the outer function

Apply the power rule to differentiate the outer function with respect to \( u \). The derivative of \( u^2 \) with respect to \( u \) is \( 2u \).

Step 3: Differentiate the inner function

Differentiate \( u = \frac{(2x+1)^2 - x}{(3x^3+1)^3 - x} \) using the quotient rule. Let \( v = (2x+1)^2 - x \) and \( w = (3x^3+1)^3 - x \). The quotient rule is given by \( \frac{du}{dx} = \frac{v'w - vw'}{w^2} \).

Step 4: Differentiate the numerator of the inner function

Differentiate \( v = (2x+1)^2 - x \). First, differentiate \( (2x+1)^2 \) to get \( 4(2x+1) \). Differentiate \( -x \) to get \( -1 \). Thus, \( v' = 4(2x+1) - 1 = 4x + 3 \).

Step 5: Differentiate the denominator of the inner function

Differentiate \( w = (3x^3+1)^3 - x \). Using the notation \( g = (3x^3 + 1)^3 \), we have \( w = g - x \). Thus, \( w' = 3(3x^3 + 1)^2 \cdot 9x^2 - 1 = 27x^2(3x^3+1)^2 - 1 \).

Step 6: Apply the quotient rule

With \( v' = 4x + 3 \) and \( w' = 27x^2(3x^3+1)^2 - 1 \), apply the quotient rule: \( \frac{v'w - vw'}{w^2} = \frac{(4x + 3)((3x^3+1)^3 - x) - ((2x+1)^2 - x)(27x^2(3x^3+1)^2 - 1)}{((3x^3+1)^3 - x)^2} \).

Step 7: Combine all the derivatives

Now, combine the derivatives into the chain rule. Express \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). We have previously determined \( \frac{dy}{du} = 2u \) and \( \frac{du}{dx} \) from the quotient rule. Substitute \( u \) back in: \( 2 \left( \frac{(2x+1)^2 - x}{(3x^3+1)^3 - x} \right) \cdot \frac{v'w - vw'}{w^2} \).

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental aspect of calculus when dealing with division of two functions. It's especially important when you have a function in the form \( u(x) / w(x) \) and need to find its derivative. The rule is mathematically expressed as:

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

The rule ensures the proper differentiation by accounting for both the top function, \( u(x) \), called the numerator, and the bottom function, \( w(x) \), called the denominator. It's crucial to first differentiate both components separately:

• Find \( u'(x) \), the derivative of the numerator.
• Find \( w'(x) \), the derivative of the denominator.
• Substitute these into the Quotient Rule formula.

In our exercise, we apply this rule to \( u = (2x+1)^2 - x \) and \( w = (3x^3+1)^3 - x \). Differentiating these functions gives us the necessary components to substitute back into the Quotient Rule. One needs to proceed carefully, ensuring calculations for both derivatives are correct and then complete the division within the rule. This results in a smoother computation of the derivative \( \frac{du}{dx} \).

Differentiation Steps

Differentiation is the process of finding a function's derivative, which represents the rate of change. To streamline this process, we break it down into systematic steps.

• **Identify the Functions**: Recognize which segments of the function are nested, or occur one inside the other. As in our example, the outer function is \( u^2 \).
• **Differentiate Outer Function**: Use simple power or exponential rules on the outermost expression. The power rule tells us that \( \frac{d}{du}(u^n) = nu^{n-1} \), which aids in simplifying our derivative calculations.
• **Manage Innermost Functions with Chain & Quotient Rules**: Employ strategies like the Chain Rule for nested functions and the Quotient Rule for division within functions.
• **Combine Derivatives**: Finally, bring together the derivatives of each section, typically by multiplying them, as shown when aligning the derivative of the outer function with that of the inner function.

Following this step-by-step method reduces room for error and makes complex differentiation more manageable. In the given example, the outer differentiation using the power rule yielded \( 2u \), whilst the inner required a careful application of the quotient rule.

Mathematical Derivatives

Mathematical derivatives quantify how a function changes as its input changes. In calculus, they are an essential tool used to understand the dynamic nature of functions, particularly those linking real-world changes such as speed, growth, or decline.

• **The Concept of a Derivative**: At its core, a derivative provides the slope of the tangent line to the function at any point. This slope demonstrates the instantaneous rate of change.
• **Derivative Rules**: Calculus comes equipped with standard rules—Product Rule, Quotient Rule, Chain Rule—which assist in differentiating complex functions. Each serves a purpose depending on the form of the function. For example, the Chain Rule helps with composite functions, and the Quotient Rule is vital for functions that involve division.
• **Importance in Calculus**: Derivatives are powerful in predicting future behavior of functions and solving maximization or minimization problems. They allow us to understand the sensitivity of one variable in relation to another.

Understanding derivatives not only equips individuals to handle mathematical challenges but offers insights into phenomena across sciences and engineering. In our problem, the challenge is to navigate through provided rules effectively to reach a correct derivative reflection of the function.