/*! 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 33 Find the first and second deriva... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 33

Find the first and second derivatives of the functions. $$y=\frac{x^{3}+7}{x}$$

Short Answer

Expert verified
First derivative: \( y' = 2x - \frac{7}{x^2} \); Second derivative: \( y'' = 2 + \frac{14}{x^3} \).

Step by step solution

01

Simplify the Function

Rewrite the function in a more convenient format for differentiation. The function is given by \( y = \frac{x^3 + 7}{x} \). Split this into two separate terms: \( y = x^2 + \frac{7}{x} \). This separates the terms so each can be differentiated individually.
02

Differentiate to Find the First Derivative

Find the derivative of each term from the simplified function. For \( x^2 \), the derivative is \( 2x \). For \( \frac{7}{x} \), rewrite it as \( 7x^{-1} \) and use the power rule to find its derivative: the derivative is \( -7x^{-2} \). Therefore, the first derivative \( y' \) is \( 2x - \frac{7}{x^2} \).
03

Differentiate to Find the Second Derivative

Differentiate the first derivative to find the second derivative. The first derivative is \( y' = 2x - 7x^{-2} \). The derivative of \( 2x \) is \( 2 \), and the derivative of \( -7x^{-2} \) is \( 14x^{-3} \) using the power rule. Thus, the second derivative \( y'' \) is \( 2 + \frac{14}{x^3} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

First Derivative
In calculus, the first derivative of a function provides information about the rate of change of the function at any given point. For instance, when you find the first derivative of a function like \( y = \frac{x^3 + 7}{x} \), it helps determine how \( y \) changes as \( x \) changes. The process generally involves differentiating each term of the function separately.
  • Start by rewriting the function in a form that makes differentiation straightforward. In the case of \( y = \frac{x^3 + 7}{x} \), break it down to \( y = x^2 + 7x^{-1} \).
  • Apply the derivative rules to each component. For \( x^2 \), the derivative is simply \( 2x \).
  • The derivative of \( 7x^{-1} \) becomes \( -7x^{-2} \) using the power rule.
Combining these results, the first derivative is given by \( y' = 2x - \frac{7}{x^2} \), indicating how the slope or steepness of the function changes with \( x \).
Second Derivative
The second derivative of a function represents the rate of change of the first derivative. It provides insight into the concavity of the function, essentially showing whether the function is curving upwards or downwards. Let's explore how to get the second derivative from the first derivative.
  • Take the first derivative \( y' = 2x - 7x^{-2} \), which represents the initial slope.
  • Differentiate this expression further. The derivative of \( 2x \) is \( 2 \), as the power rule simplifies it easily.
  • For \( -7x^{-2} \), apply the power rule to find its derivative, resulting in \( 14x^{-3} \).
Thus, the second derivative \( y'' = 2 + \frac{14}{x^3} \) unveils how the steepness is itself changing. A positive second derivative indicates a convex curve, while a negative one implies a concave shape.
Power Rule
The power rule is a fundamental principle in calculus that significantly simplifies the process of differentiation. It's particularly useful when dealing with polynomial terms, where a variable \( x \) is raised to some exponent. Here鈥檚 a brief breakdown of how it works and its application.
  • The power rule states that if you have a function \( f(x) = x^n \), its derivative \( f'(x) \) is \( nx^{n-1} \). This means you bring the exponent down as a coefficient and subtract one from the original exponent.
  • For example, differentiating \( x^3 \) results in \( 3x^2 \), and for \( x^{-1} \), it turns into \( -x^{-2} \).
  • This rule was applied when differentiating terms such as \( 7x^{-1} \), as seen in the solution: \( -7x^{-2} \).
By harnessing the power rule, calculus problems involving differentiation become far more manageable, allowing you to quickly and accurately find derivatives.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=u^{5}+1, \quad u=g(x)=\sqrt{x}, \quad x=1$$

Find \(d y / d t\) $$y=3 t\left(2 t^{2}-5\right)^{4}$

Use a CAS to perform the following steps a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\). c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$x y^{3}+\tan (x+y)=1, \quad P\left(\frac{\pi}{4}, 0\right)$$

\(y=\cos x\) for \(-\pi \leq x \leq 2 \pi .\) On the same screen, graph $$y=\frac{\sin (x+h)-\sin x}{h}$$ for \(h=1,0.5,0.3,\) and \(0.1 .\) Then, in a new window, try \(h=-1,-0.5,\) and \(-0.3 .\) What happens as \(h \rightarrow 0^{+2} \mathrm{As} h \rightarrow 0^{-2}\) What phenomenon is being illustrated here?

Find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=1-\frac{1}{u}, \quad u=g(x)=\frac{1}{1-x}, \quad x=-1$$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.