Problem 73 - Calculus Volume 1

Question

For the following exercises, find the differential and evaluate for the given \(x\) and \(d x .\) $$ y=\frac{1}{x+1}, \quad x=1, \quad d x=0.25 $$

Step-by-Step Solution

Answer

The differential is \(dy = -0.0625\).

Step 1: Understand the Problem

We need to find the differential, denoted as \(dy\), of the function \(y = \frac{1}{x+1}\) and then evaluate it at \(x = 1\) with \(dx = 0.25\). The differential \(dy\) provides the change in \(y\) corresponding to a small change \(dx\) in \(x\).

Step 2: Differentiate the Function

First, we find the derivative of \(y\) with respect to \(x\). Given \(y = \frac{1}{x+1}\), we use the power rule and chain rule. The derivative is:\[\frac{dy}{dx} = -\frac{1}{(x+1)^2}.\]

Step 3: Calculate the Differential

The differential \(dy\) is calculated using the formula \(dy = \frac{dy}{dx} \cdot dx\). Substituting the derivative and \(dx\), we get: \[dy = -\frac{1}{(x+1)^2} \cdot 0.25.\]

Step 4: Substitute Values to Evaluate

Substitute \(x = 1\) into the expression for \(dy\):\[dy = -\frac{1}{(1+1)^2} \cdot 0.25 = -\frac{1}{4} \cdot 0.25.\]

Step 5: Simplify the Expression

Simplify \(-\frac{1}{4} \cdot 0.25\) to find \(dy\):\[dy = -0.0625.\] Thus, the differential evaluated at the given \(x\) and \(dx\) is \(-0.0625\).

Key Concepts

Differentiation Techniques

Differentiation is a key concept in calculus that helps us understand how a function changes. To differentiate the function \(y = \frac{1}{x+1}\), we need to use specific techniques like the power rule and chain rule. Let's explore these.

• Power Rule: If you have a function of the form \(f(x) = x^n\), its derivative is \(f'(x) = nx^{n-1}\). This rule doesn't directly apply to \(y = \frac{1}{x+1}\), but it helps set the foundation for more complex derivatives.
• Chain Rule: This rule is essential when dealing with composite functions. If you have a function within another function, differentiate the outer function first, then multiply it by the derivative of the inner function. For \(y = \frac{1}{x+1}\), think of it as \((x+1)^{-1}\). Apply the power rule to get \(-1(x+1)^{-2}\) and then multiply by the derivative of \(x+1\), which is simply 1.

By applying these rules, we arrive at the derivative \(\frac{dy}{dx} = -\frac{1}{(x+1)^2}\). Differentiation techniques are powerful tools that allow us to understand the rate of change of functions.

Differential Evaluation

Once we've differentiated our function, the next task is to evaluate the differential, denoted as \(dy\). The differential tells us how a small change in \(x\), represented by \(dx\), results in a small change in \(y\).To compute \(dy\), we use the formula:\[dy = \frac{dy}{dx} \cdot dx\]

• First, calculate the derivative \(\frac{dy}{dx}\) which we found in the previous section as \(-\frac{1}{(x+1)^2}\).
• Multiply this by \(dx\), the small change in \(x\), given in the problem as 0.25.

Using the formula, we substitute to get the expression \(dy = -\frac{1}{(x+1)^2} \cdot 0.25\). This step transforms the abstract derivative into something more tangible by evaluating how \(y\) changes for a specific \(dx\). This is what differential evaluation is all about—translating the mathematics into real-world changes.

Derivative Calculation

Calculating a derivative involves the practical application of differentiation techniques. Let's break down the steps we used from the function \(y = \frac{1}{x+1}\) to find the derivative as \(-\frac{1}{(x+1)^2}\).

• Identify the form: Recognize \(y\) as a function in the form of \((x+1)^{-1}\), which suggests the use of the power rule combined with the chain rule.
• Apply the power rule: Differentiate \((x+1)^{-1}\) by bringing the exponent down in front and decreasing the exponent by one, resulting in \(-1(x+1)^{-2}\).
• Use the chain rule: Since \((x+1)\) is the inner function, its derivative is simply 1. Multiply the results to get \(\frac{dy}{dx} = -\frac{1}{(x+1)^2}\).

This derivative calculation is critical because it forms the basis for determining how the function behaves at any point \(x\). Understanding this foundation helps tackle more complex problems in differential calculus.