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Chapter 4: Problem 73

Find the differential and evaluate for the given \(x\) and \(d x\). \(y=\frac{1}{x+1}, \quad x=1, \quad d x=0.25\)

Short Answer

Expert verified
The differential \( dy \) is \(-0.0625\).

Step by step solution

01

Find the Derivative

To find the differential, we first need to find the derivative of the function. The given function is \( y = \frac{1}{x+1} \). Let's find \( \frac{dy}{dx} \). Using the quotient rule, where \( u = 1 \) and \( v = x+1 \), the derivative is:\[ \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} = \frac{(x+1) \cdot 0 - 1 \cdot 1}{(x+1)^2} = -\frac{1}{(x+1)^2} \]
02

Express the Differential

The differential \( dy \) is given by \( dy = \frac{dy}{dx} \cdot dx \). From Step 1, we found \( \frac{dy}{dx} = -\frac{1}{(x+1)^2} \). Therefore, the differential is:\[ dy = -\frac{1}{(x+1)^2} \cdot dx \]
03

Evaluate the Differential at Given Values

Plug in the given values of \( x = 1 \) and \( dx = 0.25 \) into the expression for the differential:\[ dy = -\frac{1}{(1+1)^2} \cdot 0.25 = -\frac{1}{4} \cdot 0.25 \]
04

Simplify to Find the Value of the Differential

Calculate the numerical value:\[ dy = -\frac{1}{4} \cdot 0.25 = -0.0625 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quotient Rule
The quotient rule is a method used in differential calculus to find the derivative of a quotient of two functions.It is especially useful when dealing with rational functions where one function is divided by another.
In general, the quotient rule states that if you have a function \( y = \frac{u(x)}{v(x)} \), the derivative \( \frac{dy}{dx} \) is given by:
  • Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \), the derivatives of the top and bottom functions respectively.
  • Apply the formula \[ \frac{dy}{dx} = \frac{v(x) \cdot \frac{du}{dx} - u(x) \cdot \frac{dv}{dx}}{(v(x))^2} \] to find the derivative of the quotient.
This rule helps calculate the slope of the tangent line to the graph at any given point.By using the quotient rule in our exercise, we were able to differentiate the function \( \frac{1}{x+1} \) without directly multiplying through.
Derivative
A derivative represents the rate of change of a function with respect to a variable.It is a fundamental concept in calculus used to determine the sensitivity of one quantity to a change in another.
When you compute a derivative such as \( \frac{dy}{dx} \), you are essentially finding how much the output \(y\) changes with respect to the input \(x\).
  • For instance, in our example, \( y = \frac{1}{x+1} \), the derivative \[ \frac{dy}{dx} = -\frac{1}{(x+1)^2}\] was calculated using the quotient rule.
  • This expression tells us that as \(x\) changes, \(y\) decreases at a rate of \(-\frac{1}{(x+1)^2}\), which indicates an overall decay in the function value as \(x\) increases.
Understanding derivatives is essential for analyzing how functions behave, such as determining maximum or minimum values and identifying points of inflection.
Differential
A differential is a small change in a function resulting from a small change in the input.It is closely linked to derivatives as it uses them to approximate changes in function values.
In simple terms, if you know the derivative of a function, you can use the differential to predict how the function will change for a tiny alteration in \(x\).
  • For our function \( y = \frac{1}{x+1} \), the differential \( dy = -\frac{1}{(x+1)^2} \cdot dx \) reflects how small changes in \(x\) result in changes in \(y\).
  • Given \( x=1 \) and \( dx = 0.25 \), the differential evaluates to \( dy = -0.0625 \), meaning the function \(y\) decreases by about 0.0625 when \(x\) increases by 0.25.
Differentials are fundamental in approximating solutions and understanding linear approximations, especially when numerical precision is a key requirement.

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Most popular questions from this chapter

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