§
Download
1 / 8

§ 4.3 - PowerPoint PPT Presentation


  • 94 Views
  • Uploaded on

§ 4.3. Differentiation of Exponential Functions. Section Outline. Chain Rule for e g ( x ) Working With Differential Equations Solving Differential Equations at Initial Values Functions of the form e kx. Chain Rule for e g ( x ). Chain Rule for e g ( x ). EXAMPLE. Differentiate.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about ' § 4.3' - honorato-valenzuela


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

§4.3

Differentiation of Exponential Functions


Section outline
Section Outline

  • Chain Rule for eg(x)

  • Working With Differential Equations

  • Solving Differential Equations at Initial Values

  • Functions of the form ekx



Chain Rule for eg(x)

EXAMPLE

Differentiate.

SOLUTION

This is the given function.

Use the chain rule.

Remove parentheses.

Use the chain rule for exponential functions.


Working With Differential Equations

Generally speaking, a differential equation is an equation that contains a derivative.


Solving Differential Equations

EXAMPLE

Determine all solutions of the differential equation

SOLUTION

The equation has the form y΄ = ky with k = 1/3. Therefore, any

solution of the equation has the form

where C is a constant.


Solving Differential Equations at Initial Values

EXAMPLE

Determine all functions y = f(x) such that y΄ = 3y and f(0) = ½.

SOLUTION

The equation has the form y΄ = ky with k = 3. Therefore,

for some constant C. We also require that f(0) = ½. That is,

So C = ½ and



ad