What is the process for differentiating y=e^x and y=lnx?

[JasonX]

please explain in more detail on how we come up with the answers below. Thanks in advance!
(formulas much appreciated)

Differentiate:

1. 1
   y=e^x
   =e^x

1. 2
   y=lnx
   =1/x

 

[Avodyne]

http://tutorial.math.lamar.edu/Classes/CalcI/DiffExpLogFcns.aspx

 

[danago]

please explain in more detail on how we come up with the answers below. Thanks in advance!
(formulas much appreciated)

Differentiate:

1. 1
   y=e^x
   =e^x

1. 2
   y=lnx
   =1/x

You should be careful of the notation you use. y = ln x = 1/x isn't true. You would be more correct in saying that if y = ln x, then y' = 1/x. Just thought id point that out.

 

[unique_pavadrin]

yes good observation danago