Chapter 9.7, Problem 28E

Mathematical Applications for the ...

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

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)=2xâˆ’132x+1.

Formula used:

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

fâ€²(x)=nxnâˆ’1

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

gâ€²(x)=cfâ€²(x)

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

fâ€²(x)=uâ€²(x)+vâ€²(x)

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

fâ€²(x)=uâ€²(x)v(x)âˆ’vâ€²(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=nunâˆ’1dudx

Calculation:

Consider the provided function,

g(x)=2xâˆ’132x+1

Rewrite the function,

g(x)=(2xâˆ’1)132x+1

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

g(x)=u13v

Differentiate both sides with respect to x,

fâ€²(x)=ddx(u13v)

Simplify using the quotient rule,

fâ€²(x)=((ddx(u13))â‹…v)âˆ’(ddx(v))â‹…(u13)(v)2

Simplify using the power rule,

fâ€²(x)=((13â‹…u13âˆ’1â‹…dudx)â‹…v)âˆ’(v1âˆ’1â‹…dvdx)â‹…(u13)v2=((13u1âˆ’33dudx)v)âˆ’(dvdx)u13v2=((13uâˆ’23dudx)v)âˆ’(dvdx)u13v2

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

yâ€²=(13(2x+1)(2xâˆ’1)âˆ’23(ddx(2xâˆ’1)))âˆ’(ddx

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