/*! 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 3 Differentiate (a) \(y=x^{2}-x^... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 3

Differentiate (a) \(y=x^{2}-x^{3}\) (b) \(y=50-\frac{1}{x^{3}}\)

Short Answer

Expert verified
(a) \( y' = 2x - 3x^2 \); (b) \( y' = 3x^{-4} \).

Step by step solution

01

Differentiate Part (a)

Given the function \( y = x^{2} - x^{3} \), use the power rule for differentiation. The power rule states that if \( y = x^n \), then the derivative \( y' = nx^{n-1} \).
02

Apply the Power Rule to Each Term in (a)

Differentiate each term separately: \( \frac{d}{dx}(x^2) - \frac{d}{dx}(x^3) \). Using the power rule, \( \frac{d}{dx}(x^2) = 2x^{1} \) and \( \frac{d}{dx}(x^3) = 3x^{2} \).
03

Simplify the Derivative for (a)

Combine the results to get the final derivative: \( y' = 2x - 3x^2 \).
04

Differentiate Part (b)

Given the function \( y = 50 - \frac{1}{x^3} \), rewrite \( \frac{1}{x^3} \) as \( x^{-3} \). This gives \( y = 50 - x^{-3} \).
05

Apply the Power Rule to Each Term in (b)

Differentiate each term separately: \( \frac{d}{dx}(50) - \frac{d}{dx}(x^{-3}) \). The derivative of the constant 50 is 0, and using the power rule, \( \frac{d}{dx}(x^{-3}) = -3x^{-4} \).
06

Simplify the Derivative for (b)

Combine the remaining term to get the final derivative: \( y' = 0 + 3x^{-4} = 3x^{-4} \).

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 one of the most fundamental tools in differentiation, and it helps us find the derivative of functions with terms like \(x^n\).
If you have a function \(y = x^n\), the power rule states that the derivative of this function is \(y' = nx^{n-1}\).
This rule works for any real number \(n\).

For example, if you need to differentiate the term \(x^2\):
  • According to the power rule, the derivative of \(x^2\) is \(2x^{2-1} = 2x^{1}\) or simply \(2x\).

Similarly, for \(x^3\), using the power rule gives us:
  • The derivative of \(x^3\) as \(3x^{3-1} = 3x^2\).

Understanding the power rule is essential as it simplifies the process of finding derivatives for polynomial functions.
Derivative of Polynomials
When working with polynomials, the process of differentiation involves applying the power rule to each term separately.
For a general polynomial function \(y = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\),
the derivative \(y'\) can be found by taking the derivative of each term:
  • First term: \(y = a_nx^n\), then \(y' = a_n \cdot nx^{n-1}\)
  • Next term: \(a_{n-1}x^{n-1}\), then \(y' = a_{n-1}\cdot (n-1)x^{n-2}\)
  • Constant term: \(a_0\), and its derivative is 0.

So, for example,
with the polynomial function \(y = x^2 - x^3\),
we can apply the power rule to each term:
  • \

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

Plot a graph of each of the following functions over the specified range of values and use these graphs to estimate the coordinates of all of the stationary points. Use calculus to find the exact coordinates of these points. (a) \(y=3 x^{4}-28 x^{3}+84 x^{2}-96 x+30 \quad(0 \leq x \leq 5)\) (b) \(y=x^{4}-8 x^{3}+18 x^{2}-10 \quad(-1 \leq x \leq 4)\) (c) \(y=\frac{x}{x^{2}+1} \quad(-4 \leq x \leq 4)\)

If fixed costs are 15 and the variable costs are \(2 Q\) per unit, write down expressions for \(\mathrm{TC}, \mathrm{AC}\) and \(M C\). Find the value of \(Q\) which minimizes \(A C\) and verify that \(A C=M C\) at this point.

Differentiate $$ y=(5 x+7)^{2} $$ (a) by using the chain rule (b) by first multiplying out the brackets and then differentiating term by term.

Find the output needed to maximize profit given that the total cost and total revenue functions are \(\mathrm{TC}=2 Q \quad\) and \(\quad \mathrm{TR}=100 \ln (Q+1)\) respectively.

The demand function of a good is given by $$ Q=1000 \mathrm{e}^{-0.2 P} $$ If fixed costs are 100 and the variable costs are 2 per unit, show that the profit function is given by \(\pi=1000 P \mathrm{e}^{-0.2 P}-2000 \mathrm{e}^{-0.2 P}-100\) Find the price needed to maximize profit.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

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.