How do you find the derivative of "y = x to the power x to the power x to the power x" (y=x^x^x^x)?

Question

I want to know the steps. I got badly confused after first taking the ln on both sides.

Hopefully the question is clear. There are no parenthesis between the x.
If there is any confusion, this will clear it out.
Eg. 2^2^2^2 = 2^2^4 = 2^16 = 65536

Answer 1

Let say we have y=x^x
You know that we have to take ln once
ln(y)=xln(x)

Let say we have y=x^x^x
In this case we need to take ln twice
ln(ln(y))=x(ln(ln(x))

(1/ln(y)(1/y)dy=ln(ln(x))+ x((1/ln(x))(1/x)
1/(yln(y))dy=l[n(ln(x))+1/ln(x)]dx

and so on

Answer 2

This would take far too long to explain, but the final answer is:

dy/dx = (x^x^x^x)*[x^(x^x+x)*(lnx)^3) + x^(x^x+x)*(lnx)^2 + x^(x^x+x-1)*(lnx) + x^(x^x-1)].

The way to do this is to use logarithmic differention. I think I used it three of four times.

I don't think I agree with Payam in that I do not see where the ambiguity comes in. You didn't include any parentheses; and I don't think you meant to. So there is no ambiguity, as we do read from left from to right. I could be wrong.

Answer 3

U-sub I think. There are going to be longs... lots and lots of logs!
It is actually a really difficult problem I think. Try setting X^X^X= u

so y(x) = U^x and so I would take the derivative of U but in order to do that you have to break X^X^X = A^x and A = X^X, so it is really bad, but I think Chain rule is in order. Look up chain rule online! Thats all i remember from calc I is chain rule.

Answer 4

If you know how x^x is differentiated
diff(x^x) = x^x * (ln(x)+1)

Then you can compute these similarly:
diff(x^(x^x)) = x^(x^x) * (x^x * (ln(x)+1) * ln(x)+ x^x /x)

diff(x^(x^(x^x))) =
x^(x^(x^x)) * (x^(x^x) * (x^x * (ln(x)+1) * ln(x) + x^x /x)
* ln(x) + x^(x^x)/x)

Note that since ^ lacks the associative law the expression x^x^x^x is ambiguous and could be parenthesised differently

diff((x^x)^(x^x))=
(x^x)^(x^x) * (x^x*(ln(x)+1) * ln(x^x) + x^x * (ln(x)+1))

and in other forms

Answer 5

Are you talking about y=x^x^x^x^x... when it goes on forever or is it only 4 x's? It turns out that actually you can still differentiate it even if it goes on forever.

y=x^x^x^x...
ln(y)=(x^x^x^x...)*(ln(x))
but, we already know that y=x^x^x^x...so just substitute it in.
ln(y)=y*ln(x)
Now just differentiate and solve for dy/dx
dy/y=dy*ln(x)+ydx/x

Answer 6

I created a PDF file for you with each step. Email me at j underscore c underscore rader at yahoo dot com if you want it.

Answer 7

f(x) = x^x^x^x
This is a nested function. You must start with the innermost function and apply the nested function rule:
d [f(g(x))]dx=g'(x)f'(x).

Answer 8

ERWIN C is right.