BuyFind

Single Variable Calculus: Concepts...

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

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

Answer to Problem 39E

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.

  df(a)dx=dfdadadx

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.

  df(a)dx=dfdadadx

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] .

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!