/*! 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 6 Find the derivative of \(y=\sqrt... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 6

Find the derivative of \(y=\sqrt{x}\left(x^{2}+4\right)\). \(\frac{d y}{d x}=\) __________

Short Answer

Expert verified
\(\frac{dy}{dx} = \frac{x^2 + 4 + 4x^{3/2}}{2x^{1/2}}\)

Step by step solution

01

Identify the parts of the function

The given function is a product of two functions: \(y = \sqrt{x}(x^2 + 4)\). Consider \(u = \sqrt{x}\) and \(v = x^2 + 4\).
02

Apply the Product Rule

The product rule states that \( \frac{d}{dx} (u \cdot v) = u'v + uv' \). We need to find the derivatives of \(u\) and \(v\).
03

Find the derivatives of \(u\) and \(v\)

First, find \(\frac{du}{dx}\): \(u = \sqrt{x} = x^{1/2}\) \(u' = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}\).Next, find \(\frac{dv}{dx}\): \(v = x^2 + 4\) \(v' = 2x\).
04

Substitute derivatives back into the Product Rule

Using the product rule formula, substitute \(u, u', v,\) and \(v'\): \(\frac{dy}{dx} = u'v + uv'\). This gives\(\frac{dy}{dx} = \frac{1}{2\sqrt{x}}(x^2 + 4) + \sqrt{x}(2x)\).
05

Simplify the expression

Distribute and combine like terms to simplify: \(\frac{dy}{dx} = \frac{x^2 + 4}{2\sqrt{x}} + 2x\sqrt{x}\) Convert the second term for common denominators: \(2x\sqrt{x} = 2x \cdot x^{1/2} = 2x^{3/2}\)Therefore, \(\frac{dy}{dx} = \frac{x^2 + 4}{2x^{1/2}} + \frac{2x^{3/2}}{1} = \frac{x^2 + 4 + 4x^{3/2}}{2x^{1/2}}\).

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.

product rule
The product rule is a fundamental concept in calculus, especially when dealing with derivatives of products of two functions. If we have a function that is the product of two functions, say, and , the product rule helps us to find its derivative. The product rule states: \( \frac{d}{dx} (u \times v) = u'v + uv' \).

Here鈥檚 how it works:
  • First, find the derivative of and call it .
  • Next, find the derivative of and call it .
  • Then, use the product rule formula to combine these results.

Using the product rule makes it easier to find the derivative of complex functions. Remember to carefully identify each part of the product before applying the rule.
derivative
A derivative represents the rate at which a function is changing at any given point. It is a fundamental tool in calculus for analyzing the behavior of functions.

In mathematical notation, the derivative of a function with respect to is written as \( \frac{d y}{d x} \) or \( y' \).

Here鈥檚 a simple breakdown of key points to understand derivatives better:
  • The process of finding a derivative is called differentiation.
  • The power rule, which states that if , then \( y' = nx^{n-1} \)
  • To differentiate more complex functions, we often use rules like the product rule and chain rule.

In our exercise, you saw the need to find derivatives of simpler functions before using the product rule. For example, if , then its derivative is \( \frac{1}{2\sqrt{x}} \). Practice makes perfect, so keep practicing different derivative rules to master them.
simplifying expressions
Simplifying expressions is an essential step in solving calculus problems. It ensures that the final answer is in its simplest, most readable form. Simplification typically involves combining like terms, reducing fractions to their lowest terms, and factoring where possible.

In our exercise, after finding the derivative using the product rule, the next step involved simplifying the result:

  • Firstly, we combined terms such as <\( \frac{x^2 + 4}{2\sqrt{x}} + 2x\sqrt{x}>\).
  • Next, we expressed <2x\sqrt{x}> with a common denominator <2x^{3/2}>.
  • Finally, the expression was written as a single fraction.
    \( \frac{x^2 + 4 + 4x^{3/2}}{2\sqrt{x}} \).

Create practice problems for simplification to get better at this crucial skill. It helps a lot in exams and real-world applications.

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

If a spherical tank of radius 4 feet has \(h\) feet of water present in the tank, then the volume of water in the tank is given by the formula $$ V=\frac{\pi}{3} h^{2}(12-h), $$ a. At what instantaneous rate is the volume of water in the tank changing with respect to the height of the water at the instant \(h=1 ?\) What are the units on this quantity? b. Now suppose that the height of water in the tank is being regulated by an inflow and outflow (e.g., a faucet and a drain) so that the height of the water at time \(t\) is given by the rule \(h(t)=\sin (\pi t)+1,\) where \(t\) is measured in hours \((\) and \(h\) is still measured in feet). At what rate is the height of the water changing with respect to time at the instant \(t=2 ?\) c. Continuing under the assumptions in (b), at what instantaneous rate is the volume of water in the tank changing with respect to time at the instant \(t=2 ?\) d. What are the main differences between the rates found in (a) and (c)? Include a discussion of the relevant units.

Let \(p(z)\) be given by the rule $$ p(z)=\frac{z \tan (z)}{z^{2} \sec (z)+1}+3 e^{z}+1. $$ a. Determine \(p^{\prime}(z)\) b. Find an equation for the tangent line to \(p\) at the point where \(z=0\). c. At \(z=0\), is \(p\) increasing, decreasing, or neither? Why?

Let $$ f(x)=6 \sin ^{-1}\left(x^{4}\right) $$ \(f^{\prime}(x)=\) _______ NOTE: The webwork system will accept \(\arcsin (x)\) or \(\sin ^{-1}(x)\) as the inverse of \(\sin (x)\)

Let \(f\) and \(g\) be differentiable functions for which the following information is known: \(f(2)=5, g(2)=-3, f^{\prime}(2)=-1 / 2, g^{\prime}(2)=2\). a. Let \(h\) be the new function defined by the rule \(h(x)=3 f(x)-4 g(x)\). Determine \(h(2)\) and \(h^{\prime}(2)\) b. Find an equation for the tangent line to \(y=h(x)\) at the point \((2, h(2))\). c. Let \(p\) be the function defined by the rule \(p(x)=-2 f(x)+\frac{1}{2} g(x)\). Is \(p\) increasing, decreasing, or neither at \(a=2\) ? Why? d. Estimate the value of \(p(2.03)\) by using the local linearization of \(p\) at the point \((2, p(2))\).

Let \(f(v)\) be the gas consumption (in liters/km) of a car going at velocity \(v\) (in km/hour). In other words, \(f(v)\) tells you how many liters of gas the car uses to go one kilometer if it is traveling at \(v\) kilometers per hour. In addition, suppose that \(f(80)=0.05\) and \(f^{\prime}(80)=\) \(0.0004 .\) a. Let \(g(v)\) be the distance the same car goes on one liter of gas at velocity \(v\). What is the relationship between \(f(v)\) and \(g(v)\) ? Hence find \(g(80)\) and \(g^{\prime}(80)\). b. Let \(h(v)\) be the gas consumption in liters per hour of a car going at velocity \(v\). In other words, \(h(v)\) tells you how many liters of gas the car uses in one hour if it is going at velocity \(v\). What is the algebraic relationship between \(h(v)\) and \(f(v)\) ? Hence find \(h(80)\) and \(h^{\prime}(80)\). c. How would you explain the practical meaning of these function and derivative values to a driver who knows no calculus? Include units on each of the function and derivative values you discuss in your response.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

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.