Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems } 1-44, \text { find } D_{x} y \text { using the rules of this section. } $$ $$ y=\frac{4}{2 x^{3}-3 x} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{-24x^2 + 12}{(2x^3 - 3x)^2} \).

Step 1: Identify the Rule

To find the derivative of the given function \( y = \frac{4}{2x^{3} - 3x} \), we can use the quotient rule, which states that if \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \). Here, set \( u = 4 \) and \( v = 2x^{3} - 3x \).

Step 2: Differentiate the Numerator and Denominator

Differentiate \( u \) and \( v \):- \( \frac{du}{dx} = 0 \) because 4 is constant.- \( \frac{dv}{dx} = \frac{d}{dx}(2x^3 - 3x) = 6x^2 - 3 \).

Step 3: Apply the Quotient Rule

Plug \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule formula:\[ \frac{dy}{dx} = \frac{(2x^3 - 3x)(0) - 4(6x^2 - 3)}{(2x^3 - 3x)^2} \]

Step 4: Simplify the Derivative Expression

Simplify the expression:\[ \frac{dy}{dx} = \frac{-4(6x^2 - 3)}{(2x^3 - 3x)^2} \]\[ \frac{dy}{dx} = \frac{-24x^2 + 12}{(2x^3 - 3x)^2} \]

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that allows us to determine the rate at which a quantity changes. In the context of functions, differentiation gives us the derivative, which is a new function representing the slope of the tangent line to the graph of the original function at any given point.

The process involves applying certain rules to manipulate the function into its derivative form. These rules include the power rule, product rule, and, as seen in our exercise, the quotient rule. Each rule has specific situations for its use:

• The power rule helps differentiate polynomials.
• The product rule applies when two functions are multiplied.
• The quotient rule, used in this exercise, deals with functions where one is divided by another.

Understanding which rule to apply is crucial in differentiation, as is confirming that each part of the function fits the required criteria of the rule.

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It is divided mainly into differential calculus and integral calculus.

Differential calculus involves finding the derivative of a function, which helps calculate rates of change, as seen in our exercise through the derivative of a rational function.

• Derivatives allow us to understand how one quantity changes in response to another, which is vital in fields such as physics, engineering, and even economics.
• Integral calculus, on the other hand, involves finding the area under curves or the accumulated quantity, and it is related to antiderivatives.

Calculus is fundamental for advancing knowledge in mathematics and its applications across sciences, making the understanding of differentiation an essential skill.

Derivative Calculations

Derivative calculations rely on using specified rules, such as the quotient rule, for determining the derivative of functions.

For a function expressed as a quotient, like the one in the problem, the quotient rule is applied as follows: If you have a function \( y = \frac{u}{v} \), the derivative is \( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).

Let's break down the given example step-by-step to clarify this concept:

• First, identify the numerator and denominator: here \( u = 4 \) and \( v = 2x^3 - 3x \).
• Next, differentiate each: since \( u \) is a constant, \( \frac{du}{dx} = 0 \). For \( v \), apply the power rule to get \( \frac{dv}{dx} = 6x^2 - 3 \).
• Substitute these into the quotient rule formula to find the derivative.

Through this systematic procedure, derivative calculations become more straightforward, allowing you to solve a variety of differentiation problems.