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y=e^e^x find y'

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

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Other answers:

can u go through the steps? i dont get it.
The formula: y' = u' e^u
u = e^ --> u' = ??
Then just plug it into the formula :)
i dont get it..
mind going thru the steps?
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).
where did u get the second layer as e^x as?
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.
then what is the first layer?
The first layer is e^(e^x) as a whole. The second layer is the e^x inside the first layer (the power).
oh ok
Is it clear already? :)
somewhat...

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