anonymous
  • anonymous
y=e^e^x find y'
Mathematics
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions.

anonymous
  • anonymous
y=e^e^x find y'
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
y' = u' e^u = e^x * e^(e^x)
anonymous
  • anonymous
xe^ex?
anonymous
  • anonymous
NO

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
can u go through the steps? i dont get it.
anonymous
  • anonymous
The formula: y' = u' e^u
anonymous
  • anonymous
u = e^ --> u' = ??
anonymous
  • anonymous
Then just plug it into the formula :)
anonymous
  • anonymous
i dont get it..
anonymous
  • anonymous
mind going thru the steps?
anonymous
  • anonymous
In differentiating a function that contains another function (or a series of functions for some), you apply the chain rule. Actually, you perform the chain rule when you differentiate and differentiate the different layers of functions until you reach a point when you are differentiating the most basic function x and you just get 1. To get the y' of e^e^x, we should note that the first layer is e^x where x here is e^x. The deriv of e^x = e^x, so the deriv of the first layer is e^x, but since x here is e^x, we make it e^(e^x). Now for the second layer, it's just e^x where x here is still x. So the deriv of the second layer is e^x where x is x. Now for the third layer, the function is just x, and its deriv is just 1. This is where we stop. Deriv of first layer: e^(e^x) Deriv of second layer: e^x Derive of third layer: 1 (stopped here) Then we just multiply everything, as this is what the chain rule states. So the derivative of the e^(e^x) = (e^(e^x))(e^x)(1) or simply (e^(e^x))(e^x).
anonymous
  • anonymous
where did u get the second layer as e^x as?
anonymous
  • anonymous
Because usually it is just e^x, right? That's the most basic form of that function. In this case, however, e was raised to another e^x, which is another function in itself. That's the second layer.
anonymous
  • anonymous
then what is the first layer?
anonymous
  • anonymous
The first layer is e^(e^x) as a whole. The second layer is the e^x inside the first layer (the power).
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
Is it clear already? :)
anonymous
  • anonymous
somewhat...

Looking for something else?

Not the answer you are looking for? Search for more explanations.