# Find the first and second derivative of the given function.

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 3.4, Problem 40E
To determine

Expert Solution

## Answer to Problem 40E

The first derivative is y'=eex+x and the second derivative is y"=eex+x[ex+1] .

### Explanation of Solution

Given:

The given function is y=eex .

Calculation:

y=eαxsinβx

Apply chain rule.

Let f=ea,a=ex

y'=dda(ea)ddx(ex)

Use derivative rule.

ddx(ex)=ex

y'=eaex

Substitute the value of a=ex .

y'=eexexy'=eex+x

Apply chain rule.

Let f=ea,a=ex+x

y"=dda(ea)ddx(ex+x)y"=dda(ea)[ddx(ex)+ddx(x)]

Use derivative rule.

ddx(ex)=ex and ddx(xn)=nxn1

y"=ea[ex+1]

Substitute the value of a=ex+x .

y"=eex+x[ex+1]

Hence the first derivativeis y'=eex+x and the second derivative is y"=eex+x[ex+1] .

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