   Chapter 9.7, Problem 28E Mathematical Applications for the ...

12th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781337625340

Solutions

Chapter
Section Mathematical Applications for the ...

12th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781337625340
Textbook Problem

Find the derivatives of the functions in Problems 1-32. Simplify and express the answer using positive exponents only. 28.   g ( x ) = 2 x − 1 3 2 x + 1

To determine

To calculate: The simplified form of the derivative of g(x)=2x132x+1.

Explanation

Given Information:

The function is g(x)=2x132x+1.

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 quotient rule, if f(x)=u(x)v(x), then,

f(x)=u(x)v(x)v(x)u(x)[v(x)]2

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,

g(x)=2x132x+1

Rewrite the function,

g(x)=(2x1)132x+1

Consider (2x1) to be u and (2x+1) to be v,

g(x)=u13v

Differentiate both sides with respect to x,

f(x)=ddx(u13v)

Simplify using the quotient rule,

f(x)=((ddx(u13))v)(ddx(v))(u13)(v)2

Simplify using the power rule,

f(x)=((13u131dudx)v)(v11dvdx)(u13)v2=((13u133dudx)v)(dvdx)u13v2=((13u23dudx)v)(dvdx)u13v2

Substitute (2x1) for u and (2x+1) for v,

y=(13(2x+1)(2x1)23(ddx(2x1)))(ddx

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