Problem 46 - Calculus - AP Edition

Question

Find the differential \(d y\). (a) \(y=\frac{1}{x^{3}-1}\) (b) \(y=\frac{1-x^{3}}{2-x}\)

Step-by-Step Solution

Answer

(a) \( dy = \frac{-3x^2}{(x^3 - 1)^2} \, dx \); (b) \( dy = \frac{2x^3 - 6x^2 + 1}{(2-x)^2} \, dx \).

Step 1: Differentiate the given function for (a)

Given the function for part (a), \( y = \frac{1}{x^3 - 1} \). We need to find \( \frac{dy}{dx} \) by differentiating it. Use the quotient rule, where \( u = 1 \) and \( v = x^3 - 1 \). The quotient rule states: \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]. Here, \( \frac{du}{dx} = 0 \) and \( \frac{dv}{dx} = 3x^2 \). Substitute and simplify accordingly to find \( \frac{dy}{dx} \).

Step 2: Apply the quotient rule for (a)

Substitute the values into the quotient rule: \[ \frac{dy}{dx} = \frac{(x^3 - 1) \times 0 - 1 \times 3x^2}{(x^3 - 1)^2} = \frac{-3x^2}{(x^3 - 1)^2} \]. This is the derivative of \( y \).

Step 3: Compute the differential for (a)

The differential \( dy \) is given by \( dy = \frac{dy}{dx} \cdot dx \). So, from step 2, \[ dy = \frac{-3x^2}{(x^3 - 1)^2} \, dx \].

Step 4: Differentiate the given function for (b)

For part (b), given \( y = \frac{1-x^3}{2-x} \), differentiate using the quotient rule again. Here, \( u = 1-x^3 \) and \( v = 2-x \). Compute \( \frac{du}{dx} = -3x^2 \) and \( \frac{dv}{dx} = -1 \). Then, use \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \].

Step 5: Apply the quotient rule for (b)

Substitute the derivative values into the quotient rule: \[ \frac{dy}{dx} = \frac{(2-x)(-3x^2) - (1-x^3)(-1)}{(2-x)^2} \]. Simplify to find \( \frac{dy}{dx} \).

Step 6: Simplify the expression for (b)

Expanding the terms, we have: \[ \frac{dy}{dx} = \frac{-6x^2 + 3x^3 + 1 - x^3}{(2-x)^2} = \frac{2x^3 - 6x^2 + 1}{(2-x)^2} \].

Step 7: Compute the differential for (b)

The differential \( dy \) for this part is \( dy = \frac{dy}{dx} \cdot dx \). Thus, \[ dy = \frac{2x^3 - 6x^2 + 1}{(2-x)^2} \, dx \].

Key Concepts

Quotient Rule

In differential calculus, the quotient rule is a formula for finding the derivative of a function that is the ratio of two differentiable functions. If a function is given in the form \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the quotient rule helps you find \( \frac{dy}{dx} \), the derivative of \( y \) with respect to \( x \). The formula is:

• \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \)

This means you take the derivative of the numerator function \( u \) and multiply it by the original denominator \( v \), subtract the product of the original numerator \( u \) and the derivative of the denominator \( v \), and then divide everything by the square of the original denominator \( v^2 \).
It is crucial to ensure that \( v eq 0 \) when applying the quotient rule because division by zero is undefined. This technique is particularly useful when simplifying complex functions before differentiating them directly, making it easier to manage calculations.

Derivative of a Function

The derivative of a function is a fundamental concept in calculus that measures how a function's value changes as its input changes. In practical terms, it tells us the rate at which \( y \) changes with respect to \( x \). It provides crucial insights into the behavior of functions, such as determining slopes of tangent lines, finding local maxima and minima, or describing motion.
The notation for the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} \) or \( f'(x) \). When finding the derivative of a function, you often use rules like the power rule, product rule, or the quotient rule, depending on the function's form. Finding derivatives accurately requires understanding these rules and the ability to manipulate algebraic expressions effectively.
Understanding derivatives further lets us construct the tangent line to a curve at any given point. The slope of this tangent line is exactly the value of the derivative at that point, providing a linear approximation of the curve around the point. This notion makes derivatives not only a tool for calculation but also a glimpse into the behavior and trend of functions.

Differential

A differential represents a small change in a function's output relative to a small change in its input. It's denoted often by \( dy \), where \( dy = \frac{dy}{dx} \cdot dx \). Here, \( \frac{dy}{dx} \) is the derivative of the function, indicating the rate of change, while \( dx \) symbolizes a small change in \( x \).
In the context of an exercise, such as finding the differential of a given function, it involves first determining the derivative \( \frac{dy}{dx} \) using methods like the quotient rule. Then, simply multiply this derivative by \( dx \) to obtain the differential \( dy \).

• For example, if \( y = \frac{1}{x^3-1} \), the process outputs \( dy = \frac{-3x^2}{(x^3 - 1)^2} \, dx \).

This simple multiplication represents a crucial transition from finding instantaneous rates of change (derivatives) to approximating small increments in function values (differentials). Differentials are essential in applications involving approximation and error estimation, lending power and versatility to calculus in both theoretical and applied contexts.