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 40E
To determine

Find the first and second derivative of the given function.

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.

  df(a)dx=dfdadadx

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.

  df(a)dx=dfdadadx

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] .

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!