   Chapter 3.1, Problem 45E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Find the first and second derivatives of the function. f ( x ) = 0.001 x 5 − 0.02 x 3

To determine

To find: The first and second derivatives of the function.

Explanation

Given:

The function f(x)=0.001x50.02x3

Derivative rules:

(1) Constant Multiple Rule: ddx[cf(x)]=cddxf(x)

(2) Difference Rule: ddx[f(x)g(x)]=ddx(f(x))ddx(g(x))

(3) Power Rule: ddx(xn)=nxn1

Calculation:

The first derivative of f(x) is f(x), which is obtained as follows,

f(x)=ddx(f(x)) =ddx(0.001x50.02x3)

Apply the difference rule (2),

f(x)=ddx(0.001x5)ddx(0.02x3)

Apply the constant multiple rule (1),

f(x)=0.001ddx(x5)0.02ddx(x3)

Apply the power rule (3) and simplify the expressions.

f(x)=0.001(5x51)0.02(3x31)=0

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