# 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 39E
To determine

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

Expert Solution

Thefirst derivative is y'=eαx(βcosβx+αsinβx) and the second derivative is y"=eαx[(α2β2)sinβx+2αβcosβx] .

### Explanation of Solution

Given:

The given function is y=eαxsinβx .

Calculation:

y=eαxsinβx

Use product rule.

(fg)'=fg'+gf'

y'=eαxddx(sinβx)+sinβxddx(eαx)

Apply chain rule.

Let f=sina,a=βx

y'=eαxdda(sina)ddx(βx)+sinβxddx(eαx)

Use derivative rule.

ddx(sinx)=cosx and ddx(xn)=nxn1

y'=eαxcosaβ+sinβxddx(eαx)

Substitute the value of a=βx .

y'=βeαxcosβx+sinβxddx(eαx)

Apply chain rule.

Let f=ea,a=αx

y'=βeαxcosβx+sinβxdda(ea)ddx(αx)

Use derivative rule.

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

y'=βeαxcosβx+sinβxeaα

Substitute the value of a=αx .

y'=βeαxcosβx+sinβxeαxαy'=eαx(βcosβx+αsinβx)

Use product rule.

(fg)'=fg'+gf'

y'=eαxddx(βcosβx+αsinβx)+(βcosβx+αsinβx)ddx(eαx)

Use derivative rule.

ddx(sinkx)=kcoskx , ddx(coskx)=ksinkx and ddx(ekx)=kekx

y"=eαxddx(βcosβx+αsinβx)+(βcosβx+αsinβx)ddx(eαx)y"=eαx[ddx(βcosβx)+ddx(αsinβx)]+(βcosβx+αsinβx)ddx(eαx)y"=eαx[β2sinβx+αβcosβx]+(βcosβx+αsinβx)(αeαx)y"=eαx(β2sinβx+αβcosβx+αβcosβx+α2sinβx)y"=eαx[(α2β2)sinβx+2αβcosβx]

Hence the first derivativeis y'=eαx(βcosβx+αsinβx) and the second derivative is y"=eαx[(α2β2)sinβx+2αβcosβx] .

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