Problem 16 Find $f^{\prime}(x).$ $$f(x)... [FREE SOLUTION]

Chapter 3: Problem 16

Find $f^{\prime}(x).$ $$f(x)=\left(2-x-3 x^{3}\right)\left(7+x^{5}\right)$$

Short Answer

Expert verified

The derivative $f'(x)$ is $-7 - 63x^2 + 10x^4 - 6x^5 - 24x^7$."

Step by step solution

Identify the Product Rule

The function is given as a product of two functions: $ u(x) = 2 - x - 3x^3 $ and $ v(x) = 7 + x^5 $. We will use the product rule for differentiation: $(uv)' = u'v + uv'$.

Differentiate $u(x)$

Differentiate $u(x) = 2 - x - 3x^3$ term by term. The derivative is $u'(x) = 0 - 1 - 9x^2 = -1 - 9x^2$.

Differentiate $v(x)$

Differentiate $v(x) = 7 + x^5$. The derivative is $v'(x) = 0 + 5x^4 = 5x^4$.

Apply the Product Rule

Using the product rule, substitute $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into $(uv)' = u'v + uv'$. We get: - $u'v = (-1 - 9x^2)(7 + x^5)$- $uv' = (2 - x - 3x^3)(5x^4)$

Expand $u'v$

Compute $(-1 - 9x^2)(7 + x^5)$:- $-1 \times 7 = -7$- $-1 \times x^5 = -x^5$- $-9x^2 \times 7 = -63x^2$- $-9x^2 \times x^5 = -9x^7$So, $u'v = -7 - x^5 - 63x^2 - 9x^7$.

Expand $uv'$

Compute $(2 - x - 3x^3)(5x^4)$:- $2 \times 5x^4 = 10x^4$- $-x \times 5x^4 = -5x^5$- $-3x^3 \times 5x^4 = -15x^7$So, $uv' = 10x^4 - 5x^5 - 15x^7$.

Combine Terms

Add $u'v$ and $uv'$:- $( -7 - x^5 - 63x^2 - 9x^7) + (10x^4 - 5x^5 - 15x^7)$Combine like terms:- Constant term: $-7$- $x^2$ term: $-63x^2$- $x^4$ term: $10x^4$- $x^5$ term: $-x^5 - 5x^5 = -6x^5$- $x^7$ term: $-9x^7 - 15x^7 = -24x^7$Thus, $f'(x) = -7 - 63x^2 + 10x^4 - 6x^5 - 24x^7$.

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 vital technique in calculus differentiation used when you need to find the derivative of a product involving two functions. Suppose you have two differentiable functions, $u(x)$ and $v(x)$. The product rule states that the derivative of their product $f(x) = u(x) v(x)$ is given by $f'(x) = u'(x)v(x) + u(x)v'(x)$.

Function $u(x)$ is differentiated to give $u'(x)$.
Function $v(x)$ is differentiated to give $v'(x)$.
Apply the formula: the derivative of $f(x)$ is the sum of $u'(x)$ times $v(x)$ and $u(x)$ times $v'(x)$.

The understanding of the product rule simplifies the process of differentiation where multiplication of functions is involved, which is common in calculus.

Derivative

A derivative represents the rate at which one quantity changes in relation to another, encapsulating the concept of instantaneous rate of change. In simpler terms, it measures how a function is changing at any given point. For a function $f(x)$, the derivative is often denoted as $f'(x)$.

It is essential in determining the slope of a function at any point on its curve.
A derivative can be calculated for any kind of function, including polynomials.
Derivative calculations form the foundation of other differentiation techniques like the product rule.

By mastering derivatives, you gain the ability to analyze and interpret curves, which is pivotal in both theoretical and applied calculus.

Polynomial Differentiation

Polynomial differentiation is the process of finding the derivative of a polynomial function. A polynomial is typically expressed as a sum of terms consisting of variables raised to a power and multiplied by coefficients. The differentiation process involves applying some straightforward rules to each term:

For a term like $ax^n$, the derivative is $n \cdot ax^{n-1}$.
Constant terms, such as 5 or 3, will have a derivative of 0, since they do not change.
Derivatives are taken term by term in a polynomial expression.

Polynomial differentiation is an essential building block that enables you to handle more complex functions using rules like the product rule. It's the base on which more complicated differentiation techniques build.

Function Expansion

Function expansion involves taking a function and expressing it in a more explicit or broken-down form. In calculus, this often means performing algebraic manipulations, such as expanding products of terms. Expanding a function helps in simplifying the expressions which become tractable for further differentiation or integration.

Expansion works by distributing terms, ensuring each factor multiplies every other part.
When using the product rule, it's common to expand the expressions of $u'v$ and $uv'$ separately.
This practice eases the process of combining and simplifying terms.

Through function expansion, we ensure a complete understanding of the function structure, which is integral in applying calculus operations efficiently.

91影视

Find \(f^{\prime}(x).\) $$f(x)=\left(2-x-3 x^{3}\right)\left(7+x^{5}\right)$$

Short Answer

Step by step solution

Identify the Product Rule

Differentiate \(u(x)\)

Differentiate \(v(x)\)

Apply the Product Rule

Expand \(u'v\)

Expand \(uv'\)

Combine Terms

Key Concepts

Product Rule

Derivative

Polynomial Differentiation

Function Expansion

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Logic and Functions

Pure Maths

Decision Maths

Mechanics Maths

Probability and Statistics

Study anywhere. Anytime. Across all devices.