Expert: Paul Klarreich - 1/7/2008

Question

I am trying to take the derivative of

y=x^(x^x)

Here goes.

ln(y)=ln(x^(x^x))=(x^x) ln(x)

y’/y=(x^x)(1/x) + D(x^x) ln(x)

I now have to find D(x^x) and then I’ll plug this answer back in.
--------------------------------------------
Finding D(x^x)
y=(x^x)
ln(y)=ln(x^x)=x ln(x)
y’/y=x(1/x)+ln(x)=(1+ln(x)
y’=(x^x) (1+ln(x))
-------------------------------------
Plugging back in to original equation:

y’/y=(x^x)(1/x)+(x^x)(1+ln(x))ln(x)=(x^x)/x+(x^x)(1+ln(x))ln(x)


Now: recalling y=x^(x^x)

y’=[(x^x)(1/x) + (x^x) (1+ln(x)) ln(x)=  (x^x)/x + (x^x) (1+ln(x)) ln(x)] y
y’=[(x^x)(1/x) + (x^x) (1+ln(x)) ln(x)=  (x^x)/x + (x^x) (1+ln(x)) ln(x)] x(x^x)

any idea where I went wrong?
Thanks


Answer
Questioner:   Mark
Category:  Calculus
Private:  No
 
Subject:  Differentiation
Question:  
I am trying to take the derivative of

y=x^(x^x)

Here goes.

ln(y)=ln(x^(x^x))=(x^x) ln(x)

y’/y=(x^x)(1/x) + D(x^x) ln(x)

I now have to find D(x^x) and then I’ll plug this answer back in.
--------------------------------------------
Finding D(x^x)
y=(x^x)
ln(y)=ln(x^x)=x ln(x)
y’/y=x(1/x)+ln(x)=(1+ln(x)
y’=(x^x) (1+ln(x))
-------------------------------------
Plugging back in to original equation:

y’/y=(x^x)(1/x)+(x^x)(1+ln(x))ln(x)=(x^x)/x+(x^x)(1+ln(x))ln(x)


Now: recalling y=x^(x^x)

y’=[(x^x)(1/x) + (x^x) (1+ln(x)) ln(x)]y =  (x^x)/x + (x^x) (1+ln(x)) ln(x)] y
y’=                                         [(x^x)/x + (x^x) (1+ln(x)) ln(x)]x(x^x)
---------------------------Left out a '^' here ------------------------------>^
any idea where I went wrong?
Thanks
...................................................
Hi, Mark,
Sorry, but it all looks OK to me, except for that detail.
==========================================================
As a check, I tried it this way:

y = x^(x^x)

ln y = x^x ln x

ln ln y = ln(x^x ln x)

ln ln y = x ln x + ln ln x
 1   1                   1    1
---- --- y' = ln x + 1 + ---- ---
ln y  y                  ln x  x

 1   1                     1
---- --- y' = ln x + 1 + -------
ln y  y                  x ln x


y ln y = x^(x^x) x^x ln x
                                  x^(x^x) x^x ln x
y' = x^(x^x) x^x ln x(ln x + 1) +  -------------------
                                       x ln x

                                  x^(x^x) x^x
y' = x^(x^x) x^x ln x(ln x + 1) +  ------------
                                       x
which, I think, is the same thing you have.