   Chapter 9.7, Problem 32E ### 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. R ( x ) = x 3 x 3 + 2 3

To determine

To calculate: The simplified form of the derivative of R(x)=x3x3+23.

Explanation

Given Information:

The function is R(x)=x3x3+23.

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,

R(x)=x3x3+23

Rewrite the function,

R(x)=x(3x3+2)13

Consider (3x3+2) to be u,

R(x)=xu13

Differentiate both sides with respect to x,

R(x)=ddx(3xu13)=3ddx(xu13)

Simplify using the product rule,

R(x)=(ddx(x))u13+(ddx(u13))x

Simplify using the power rule,

R(x)=(x11)u13+(13u131)x=u13+(13u23)x

Take u13 common,

R(x)=u13(1+13u23u13ddx(u)x)=u13(1+13u1ddx(u)x)=u13(

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