Re: how about this one...


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Posted by Soroban on October 14, 2002 at 14:39:02:

In Reply to: how about this one... posted by tommy on October 13, 2002 at 21:46:46:

: z=x^x^y

: dz/dx=?

: would you take ln of the function?
: so it becomes:
: ln z = (x^y)(ln x) -> (e^lnx)^y(ln x) -> e^(ylnx)(lnx)

Why are you making it more complicated?
ln z = (x^y)(ln x)

With respect to x (y is a constant):
(1/z)(dz/dx) = yx^(y-1)*(1/x) = yx^(y-2)
and solve for dz/dx.


With respect to y ~~
x is a constant C, so ln x is a constant K.
We have: lnz = C^y (K) ~ exponential rule!



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