/*! 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 17 In Exercises \(17-28,\) find \(d... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 17

In Exercises \(17-28,\) find \(d y\) $$ y=x^{3}-3 \sqrt{x} $$

Short Answer

Expert verified
The derivative is \(dy = 3x^2 - \frac{3}{2}x^{-1/2}\).

Step by step solution

01

Differentiate the Power Functions

The first step is to differentiate each term of the function separately. For the term \(x^3\), apply the power rule: \(\frac{d}{dx}(x^n) = nx^{n-1}\). Thus, the derivative of \(x^3\) is \(3x^2\).
02

Differentiate the Square Root Function

Next, differentiate \(-3\sqrt{x}\). Re-write \(\sqrt{x}\) as \(x^{1/2}\) and apply the power rule: \(-3 \cdot \frac{d}{dx}(x^{1/2}) = -3 \cdot \frac{1}{2}x^{-1/2}\). Simplify this to \(-\frac{3}{2}x^{-1/2}\).
03

Combine the Derivatives

Now combine the derivatives from the previous steps. So, the derivative of \(y\) is \(dy = 3x^2 - \frac{3}{2}x^{-1/2}\).

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.

Power Rule
The power rule is a fundamental rule in calculus used for finding the derivative of a function of the form \( x^n \). To apply this rule, you would differentiate by multiplying the power \( n \) by the coefficient of \( x \), and then decrease the exponent by one. This can be expressed as \( \frac{d}{dx}(x^n) = nx^{n-1} \).
This rule is especially handy when dealing with polynomial functions or any equations where functionalities are expressed with exponents.
  • Let's look at an example: \( x^3 \). Applying the power rule, we multiply 3 (the exponent) by 1 (coefficient of \( x \)) to get 3, then reduce the power to get \( 3x^2 \).
  • This results in a simple formula derived using basic arithmetic operations.
By understanding how the power rule works, you can efficiently find derivatives of terms with integer powers, speeding up calculation processes in various calculus problems.
Square Root Function Differentiation
Differentiating square root functions initially might seem tricky, but it becomes easier when thinking in terms of exponents. A square root function like \( \sqrt{x} \) can be rewritten in exponent form as \( x^{1/2} \). This allows us to apply the same principles from the power rule.

When differentiating \( -3\sqrt{x} \), express it as \( -3x^{1/2} \):
  • Applying the power rule, multiply the exponent \( \frac{1}{2} \) by -3, which gives \( -\frac{3}{2} \).
  • Next, reduce the exponent by 1, leading to \( x^{-1/2} \).
This results in the derivative: \( -\frac{3}{2}x^{-1/2} \).
This simplification makes the differentiation process straightforward, turning complex radicals into something more manageable through consistent application of rules.
Derivative Calculation
In derivative calculation, combining individual derivatives into a solution for a compound function is crucial. This involves neatly combining solutions from simpler calculations. For the function \( y = x^3 - 3\sqrt{x} \), differentiate each part separately, then sum them up.
  • From our power rule, the derivative of \( x^3 \) is \( 3x^2 \).
  • For \( -3\sqrt{x} \), we found it to be \( -\frac{3}{2}x^{-1/2} \).

Now, simply combine these results:
\[\frac{dy}{dx} = 3x^2 - \frac{3}{2}x^{-1/2}\]With practice, you'll find that breaking complex problems into smaller parts, and then combining them, simplifies derivative calculation and leads to more accurate results. This approach helps when dealing with multi-part functions in calculus and aids in grasping more abstract mathematical concepts.

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

In Exercises \(57-60,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: $$ \begin{array}{l}{\text { a. Plot the function } f \text { over } I} \\ {\text { b. Find the linearization } L \text { of the function at the point } a \text { . }} \\ {\text { c. Plot } f \text { and } L \text { together on a single graph. }} \\ {\text { d. Plot the absolute error }|f(x)-L(x)| \text { over } I \text { and find its max- }} \\ {\text { imum value. }}\end{array} $$ $$ \begin{array}{l}{\text { e. From your graph in part (d), estimate as large a } \delta>0 \text { as you }} \\ {\text { can, satisfing }}\end{array} $$ $$ \begin{array}{c}{|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon} \\\ {\text { for } \epsilon=0.5,0.1, \text { and } 0.01 . \text { Then check graphically to see if }} \\ {\text { your } \delta \text { -estimate holds true. }}\end{array} $$ $$ f(x)=x^{3}+x^{2}-2 x, \quad[-1,2], \quad a=1 $$

Cardiac output In the late 1860 \(\mathrm{s}\) , Adolf Fick, a professor of physiology in the Faculty of Medicine in Wurzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about 7 \(\mathrm{L} / \mathrm{min.}\) At rest it is likely to be a bit under 6 \(\mathrm{L} / \mathrm{min.}\) If you are a trained marathon runner running a marathon, your cardiac output can be as high as 30 \(\mathrm{L} / \mathrm{min} .\) $$\begin{array}{c}{\text { Your cardiac output can be calculated with the formula }} \\ {y=\frac{Q}{D}}\end{array}$$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{ml} / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{ml} / \mathrm{min}\) and \(D=97-56=41 \mathrm{ml} / \mathrm{L},\) $$y=\frac{233 \mathrm{ml} / \min }{41 \mathrm{ml} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min},$$ fairly close to the 6 \(\mathrm{L} / \mathrm{min}\) that most people have at basal (resting conditions. (Data courtesy of J. Kenneth Herd, M.D. Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?

If the original 24 \(\mathrm{m}\) edge length \(x\) of a cube decreases at the rate of 5 \(\mathrm{m} / \mathrm{min}\) , when \(x=3 \mathrm{m}\) at what rate does the cube's a. surface area change? b. volume change?

Diagonals If \(x, y,\) and \(z\) are lengths of the edges of a rectangular box, the common length of the box's diagonals is \(s=\) \(\sqrt{x^{2}+y^{2}+z^{2}}\) a. Assuming that \(x, y,\) and \(z\) are differentiable functions of \(t\) how is \(d s / d t\) related to \(d x / d t, d y / d t,\) and \(d z / d t\) ? b. How is \(d s / d t\) related to \(d y / d t\) and \(d z / d t\) if \(x\) is constant? c. How are \(d x / d t, d y / d t,\) and \(d z / d t\) related if \(s\) is constant?

Find both \(d y / d x\) (treating \(y\) as a differentiable function of \(x\) ) and \(d x / d y\) (treating \(x\) as a differentiable function of \(y )\) . How do \(d y / d x\) and \(d x / d y\) seem to be related? Explain the relationship geometrically in terms of the graphs. \begin{equation} x^{3}+y^{2}=\sin ^{2} y \end{equation}
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Pure 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.