Nicolas,

You might well have received an answer to this question, but anyway:

The approach to differentiating equations of this kind is to take logs, so:

y=2^x

becomes

ln y = ln (2^x)

or

ln y = x ln 2

Differentiating (implicitly)gives:

(1/y).(dy/dx) = ln 2

and finally

dy/dx = 2^x. ln 2

As regards the solution to 6.02 and the comment re. solutions of exp(-1/x): the method above will show that the first derivative is

(1/x^2).exp(-1/x)

and all subsequent derivatives will have the form as stated of

P(1/x).exp(-1/x)

where P(1/x) is a polynomial in 1/x.

Hope that helps.