How do you find the first and second derivative of #y=e^(e^x)#?

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Answer 1 · 2016-10-06T03:22:05

#dy/dx = = e^((e^x + x))#

#(d^2y)/(dx^2) = e^((e^x + x))(e^x + 1)#

#dy/dx = dy/(du)(du)/dx#

Let #u = e^x#, then #y(u) = e^u#, #dy/(du) = e^u#, and #(du)/(dx) = e^x#

Substitute into the chain rule

#dy/dx = (e^u)(e^x)#

Reverse the u substitution:

#dy/dx = (e^(e^x))(e^x)#

This can be simplified into:

#dy/dx = = e^((e^x + x))#

But I will use the previous form when computing the second derivative.

To compute the second derivative, we use the product rule, #(gh)' = g'h + gh'#, on the first derivative:

let #g = (e^(e^x))#, then #h = e^x, g' = (e^(e^x))(e^x)#, and #h' = e^x#

Substituting into the product rule:

#((e^(e^x))(e^x))' = (e^(e^x))(e^x)(e^x) + (e^(e^x))(e^x)#

#(d^2y)/(dx^2) = e^((e^x + x))(e^x + 1)#