Chapter 9.7, Problem 29E

### 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. y = x 2 4 x − 3 4

To determine

To calculate: The simplified form of the derivative of y=x24x34.

Explanation

Given Information:

The function is y=x24xâˆ’34.

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 product rule, if f(x)=u(x)â‹…v(x), then

fâ€²(x)=uâ€²(x)â‹…v(x)+vâ€²(x)â‹…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=nunâˆ’1dudx

Calculation:

Consider the provided function,

y=x24xâˆ’34

Rewrite the function,

y=x2(4xâˆ’3)14

Consider (4xâˆ’3) to be u,

y=x2u14

Differentiate both sides with respect to x,

yâ€²=ddx(x2u14)

Simplify using the product rule,

yâ€²=((ddx(x2))â‹…u14+(ddx(u14))â‹…x2)

Simplify using the power rule,

yâ€²=((2â‹…x2âˆ’1)â‹…u14+(14u14âˆ’1â‹…dudx)â‹…x2)=(2xu14+(14uâˆ’34â‹…ddx(u))â‹…x2)

Take xu14 common,

yâ€²=xu14(2+14uâˆ’34â‹…uâˆ’14â‹…ddx(u)â‹…x)=xu14(2+14u

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