/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 Find the differential and evalua... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the differential and evaluate for the given \(x\) and \(d x\). \(y=x^{3}+2 x+\frac{1}{x}, \quad x=1, \quad d x=0.05\)

Short Answer

Expert verified
The differential is 0.20.

Step by step solution

01

Find the Derivative

The first step is to find the derivative of the function \( y = x^3 + 2x + \frac{1}{x} \) with respect to \( x \). To do this, use the power rule and the derivative of a reciprocal.\[ \frac{dy}{dx} = \frac{d}{dx}(x^3) + \frac{d}{dx}(2x) + \frac{d}{dx}\left(\frac{1}{x}\right) \]The derivative becomes:\[ 3x^2 + 2 - \frac{1}{x^2} \].
02

Express the Differential

The next step is to express the differential \( dy \) in terms of \( dx \). The differential \( dy \) can be expressed as:\[ dy = \frac{dy}{dx} \, dx = (3x^2 + 2 - \frac{1}{x^2}) \, dx \].
03

Substitute the Given Values

Now substitute the given values into the expression for the differential. Given that \( x = 1 \) and \( dx = 0.05 \), substitute these values:\[ dy = (3(1)^2 + 2 - \frac{1}{(1)^2}) \, 0.05 \].
04

Simplify the Expression

Simplify the expression obtained from substitution:\[ dy = (3\times1 + 2 - 1) \, 0.05 \].Calculate this to find:\[ dy = 4 \, \times 0.05 = 0.20 \].

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Key Concepts

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

Derivative
The concept of a derivative is central to differential calculus. It measures how a function changes as its input changes. When we talk about finding the derivative of a function, we're looking at the rate of change or the slope of the tangent line at any point on the function.
The notation for a derivative is usually given as \(\frac{dy}{dx}\), which means the derivative of \(y\) with respect to \(x\). In our exercise, the function is \(y = x^3 + 2x + \frac{1}{x}\), and we found its derivative by determining how each part of the function changes with \(x\).
Understanding how to find derivatives is a powerful tool because it helps us analyze the behavior of functions in diverse fields like physics, economics, and engineering.
  • A derivative is the instantaneous rate of change.
  • It is essential for finding slopes and analyzing curves.
  • Derivatives apply to any function where you can mathematically express rate change.
Power Rule
The power rule is a simple yet powerful tool used to find the derivative of functions with power terms. If you have a function \(f(x) = x^n\), the power rule states that the derivative is \(f'(x) = nx^{n-1}\). This speeds up the process of taking derivatives because you can apply this rule directly to any power of \(x\).
In the given function \(y = x^3 + 2x + \frac{1}{x}\), you can use the power rule for the term \(x^3\), making it easy to find its derivative as \(3x^2\). The power rule also applies to terms like \(2x\) where \(n = 1\).
For learners, mastering the power rule is one of the foundational skills you need in calculus.
  • Quickly determine derivatives for polynomial expressions.
  • Remember: bring the power down, multiply, and reduce the power by one.
  • Useful for simple functions or those including polynomial terms.
Differential
A differential is a concept used to approximate changes in a function. It deals with small increments in the variable of interest. If \(dy\) is the differential of a function \(y\), then it is given by the derivative \(\frac{dy}{dx}\) multiplied by a small change \(dx\) in \(x\).
This is expressed generalely as \(dy = \frac{dy}{dx} \, dx\).
In practice, this helps estimate how much a function's output changes when its input changes slightly. In our exercise, we calculate the differential using known values of \(x\) and \(dx\) to find how much \(y\) changes.
  • Approximates change in function value with small changes in inputs.
  • Uses the derivative to "scale" our small change \(dx\).
  • Essential in practical applications like error estimation.
Reciprocal Function
Reciprocal functions involve the inverse operation of a term, often written as \(\frac{1}{x}\). When differentiating a reciprocal function, we're interested in how this inverse relationship changes with respect to \(x\). The derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\), found using the power rule by rewriting \(\frac{1}{x}\) as \(x^{-1}\).
Understanding the behavior of reciprocal functions and their derivatives is important because they appear frequently in mathematical models, particularly in physics and engineering, where inverses represent rate dependencies or constraints.
  • Key to understanding inverse growth or decay relationships.
  • Involves rewriting terms to apply standard derivative rules.
  • A common component of complex functions in calculus problems.

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