Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(y=f(x)=x^{3}\). Find the value of \(d y\) in each case. (a) \(x=0.5, d x=1\) (b) \(x=-1, d x=0.75\)

Step-by-Step Solution

Answer

(a) \(dy = 0.75\); (b) \(dy = 2.25\).

Step 1: Understanding the Problem

The exercise requires us to find the differential \(dy\) of the function \(y = f(x) = x^3\) for given values of \(x\) and \(dx\), meaning small changes in \(x\).

Step 2: Find the Derivative of the Function

To find \(dy\), we need the derivative \(f'(x)\) of the function \(y = x^3\). We use the power rule: \(f'(x) = 3x^2\). This derivative tells us how \(y\) changes with respect to \(x\).

Step 3: Calculate \(dy\) for Part (a)

For \(x = 0.5\) and \(dx = 1\), first calculate the derivative at \(x = 0.5\): \(f'(0.5) = 3(0.5)^2 = 3 \/ \times \/ 0.25 = 0.75\). Then use the formula \(dy = f'(x) \, dx\) to find \(dy = 0.75 \, \times \, 1 = 0.75\).

Step 4: Calculate \(dy\) for Part (b)

For \(x = -1\) and \(dx = 0.75\), calculate the derivative at \(x = -1\): \(f'(-1) = 3(-1)^2 = 3 \/ \times \/ 1 = 3\). Then, \(dy = f'(x) \, dx\), so \(dy = 3 \, \times \, 0.75 = 2.25\).

Key Concepts

Power Rule

In differential calculus, the power rule is an essential tool for finding derivatives of functions expressed as powers of a variable. To apply the power rule, if we have a function of the form \( y = x^n \), where \( n \) is any real number, the derivative \( f'(x) \) is found by:

• Multiplying the variable \( x \) raised to the power \( n-1 \) by \( n \).

Therefore, the power rule can be expressed simply as \( f'(x) = nx^{n-1} \).
This rule makes it straightforward to find the rate at which the function's value changes. For example, as shown in the original exercise, when \( y = x^3 \), we apply the power rule to obtain the derivative \( f'(x) = 3x^2 \).
This derivative will help us understand how tiny changes in \( x \) affect changes in \( y \). This power rule significantly simplifies the process, especially for polynomial functions.

Derivatives

The concept of a derivative is crucial in understanding changes in functions with respect to their variables. Simply put, the derivative provides the slope of the tangent line to the curve of a function at any point. It helps in determining how the function's output value \( y \) alters as its input \( x \) changes.
In the context of the exercise, the derivative \( f'(x) \) of \( y = x^3 \) is \( 3x^2 \). This function tells us the rate at which \( y \) is changing with respect to \( x \) for any specific value of \( x \). For instance:

• At \( x = 0.5 \), the derivative is \( 0.75 \).
• At \( x = -1 \), it becomes \( 3 \).

These calculations allow us to predict the changes in \( y \) whenever there is a small change in \( x \), giving insight into the function's behavior.
Understanding derivatives is crucial for solving real-world problems involving motion, growth, and other changable scenarios.

Differential of a Function

The differential of a function represents a small change in the function output, \( dy \), resultant from a small change in the input, \( dx \). It effectively captures how a function responds to tiny variations in its input.
The formula for the differential is \( dy = f'(x) \, dx \), where \( f'(x) \) is the derivative already calculated. This relationship links changes in \( x \) directly to changes in \( y \), emphasizing their interconnectedness.

• For example, at \( x = 0.5 \) and \( dx = 1 \), the differential \( dy \) becomes \( 0.75 \times 1 = 0.75 \).
• Similarly, at \( x = -1 \) and \( dx = 0.75 \), \( dy \) is \( 3 \times 0.75 = 2.25 \).

Understanding differentials is especially useful in approximations and solving many applied mathematics problems where exact solutions aren’t feasible or necessary. It offers a way to better understand the sensitivity of a system to changes in its parameters.