   Chapter 3.2, Problem 29E ### 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 f'(x) and f"(x). f ( x ) = x 2 1 + e x

To determine

To find: The derivatives f(x) and f(x).

Explanation

Given:

The function f(x)=x21+ex.

Derivative rule:

(1) Product Rule: ddx[f1(x)f2(x)]=f1(x)ddx[f2(x)]+f2(x)ddx[f1(x)]

(2) Quotient Rule: If f1(x) and f2(x) are both differentiable, then

ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]f1(x)ddx[f2(x)][f2x]2

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

(4) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

(5) Constant multiple rule: ddx(cf)=cddx(f)

(6) Natural exponential function: ddx(ex)=ex

Calculation:

The first derivative of the function f(x)=x21+ex is f(x), which is obtained as follows,

f(x)=ddx(f(x))=ddx(x21+ex)

Apply the quotient rule (2),

f(x)=(1+ex)ddt(x2)x2ddx(1+ex)(1+ex)2

Apply the derivative rule (4),

f(x)=[(1+ex)ddt(x2)]x2[ddx(1)+ddx(ex)](1+ex)2

Apply the derivative rules (3),(6) and simplify the terms.

f(x)=[(1+ex)(2x)]x2[0+ex](1+ex)2=2x+ex(2xx2)(1+ex)2

Therefore, the first derivative of the function f(x)=x21+ex is f(x)=1(1+ex)2[2x+ex(2xx2)].

The second derivative of the function f(x)=x21+ex is f(x), which is obtained as follows,

f(x)=ddx(f(x))=ddx(2x+ex(2xx2)(1+ex)2)

Apply the quotient rule (2) and simplify the terms,

f(x)=(1+ex)2ddx(2x+ex(2xx2))(2x+ex(2xx2))ddx((1+ex)2)((1+ex)2)2=(1+ex)2ddx(2x+ex(2xx2))(2x+ex(2xx2))ddx(1+(ex)2+2ex)(1+ex)4=(1+ex)2ddx(2x+ex(2xx2))(2x+ex(2xx2))ddx(1+exex+2ex)(1+ex)4

Apply the sum rule (4) and the constant multiple rule (5),

f(x)=1(1+ex)4[(1+ex)2[ddx(2x)+ddx(ex(2xx2))](2x+ex(2xx2))[ddx(1)+ddx(exex)+ddx(2ex)]]

f(x)=1(1+ex)4[(1+ex)2[2ddx(x)+ddx(ex(2xx2))](2x+ex(2xx2))[ddx(1)+ddx(exex)2+ddx(ex)]] (1)

Obtain the derivative ddx(ex(2xx2))

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