   Chapter 9.7, Problem 14E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Find the derivatives of the functions in Problems 1-32. Simplify and express the answer using positive exponents only. y = 3 x 4 ( 2 x 5 + 1 ) 7

To determine

To calculate: The simplified form of the derivative of y=3x4(2x5+1)7.

Explanation

Given Information:

The function is y=3x4(2x5+1)7.

Formula used:

According to the power rule, if f(x)=xn, then,

f(x)=nxn1

According to the property of differentiation, if a function is of the form, g(x)=cf(x), then,

g(x)=cf(x)

According to the property of differentiation, if a function is of the form f(x)=u(x)+v(x), then,

f(x)=u(x)+v(x)

According to the product rule, if f(x)=u(x)v(x), then

f(x)=u(x)v(x)+v(x)u(x)

The derivative of a constant value, k, is

ddx(k)=0

According to the property of differentiation, if a function is of the form y=un, where u=g(x),

dydx=nun1dudx

Calculation:

Consider the provided function,

y=3x4(2x5+1)7

Consider (2x5+1) to be u,

y=3x4u7

Differentiate both sides with respect to x,

y=ddx(3x4u7)=3ddx(x4u7)

Simplify by the use of the product rule,

y=3((ddx(x4))u7+(ddx(u7))x4)

Simplify by the use of the power rule,

y=3((4x41)u7+(7u71dudx)x4)=3((4x3)u7+(7u6dudx

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Find more solutions based on key concepts 