   Chapter 9.7, Problem 25E ### 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. g ( x ) = ( 8 x 4 + 3 ) 2 ( x 3 − 4 x ) 3

To determine

To calculate: The simplified form of the derivative of g(x)=(8x4+3)2(x34x)3.

Explanation

Given Information:

The function is g(x)=(8x4+3)2(x34x)3.

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,

g(x)=(8x4+3)2(x34x)3

Consider (8x4+3) to be u and (x34x) to be v,

g(x)=u2v3

Differentiate both sides with respect to x,

g(x)=ddx(u2v3)

Simplify using the product rule,

g(x)=((ddx(u2))v3+(ddx(v3))u2)

Simplify using the power rule,

g(x)=((2u21dudx)v3+(3v31dvdx)u2)=2uv3dudx+3v2u2dvdx

Take uv2 common,

g(x)=uv2(2vddx(u)+3uddx(v))

Substitute (8x4+3) for u and (x34x) for v,

g(x)=(8x4+3)(x34x)2(2(x34x)ddx(8x4+3)+3(8x4+3)ddx(x34x))=(8x4+3

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