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4 answers

I'll write the function as x^(x^x) to avoid confusion.
{note: if you meant (x^x)^x this should be written as x^x^2}

x^(x^x) = e^[ln(x^(x^x))]
= e^[x^x * lnx]
and
d/dx e^u = e^u * du/dx = x^(x^x) * du/dx

here u = x^x * lnx
du/dx = x^x (1/x) + lnx * d/dx[x^x] (by the chain rule)
= x^x (1/x) + lnx * d/dx[e^(ln(x^x)]
= x^x (1/x) + lnx * d/dx[e^(xln(x)]
= x^(x-1) + lnx [x^x(1+lnx)] (using the same method as above with u=xlnx)

so d/dx(x^x^x) = x^(x^x) * {x^(x-1) + lnx [x^x(1+lnx)] }

Whoever voted thumbs down doesnt have a clue...
This answer is correct.

2007-03-09 04:03:36 · answer #1 · answered by Scott R 6 · 1 1

This question is equivalent to solving for dy/dx where

y = x^(x^x)

To solve this, we use logarithmic differentiation. Take the ln of both sides to obtain

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

Now, use the log property that allows us to bring powers inside logs in front of the log itself.

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

Let z = x^x. Then

ln(y) = z ln(x)

Now, differentiate implicitly with respect to x.

(1/y)(dy/dx) = (dz/dx) ln(x) + z (1/x)

Multiply both sides by y, to get

dy/dx = y [ (dz/dx) ln(x) + z (1/x) ]

dy/dx = (dz/dx) y ln(x) + yz (1/x)

Here's where we solve for dz/dx separately.

z = x^x
ln(z) = ln(x^x)
ln(z) = x ln(x)

Implicitly differentiating,

(1/z)(dz/dx) = ln(x) + x(1/x)
(1/z)(dz/dx) = ln(x) + 1

Multiply both sides by z,

dz/dx = z(ln(x) + 1)

But z = x^x, so

dz/dx = (x^x) (ln(x) + 1)

Plugging dz/dx into the original equation,

dy/dx = (dz/dx) y ln(x) + yz (1/x)

dy/dx = [(x^x) (ln(x) + 1)] y ln(x) + yz (1/x)

Plugging in z = x^x into the orignal equation,

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

Plugging in y = x^(x^x),

dy/dx = [(x^x) (ln(x) + 1)] [x^(x^x)] ln(x) + [x^(x^x)] (x^x) (1/x)

Which I can leave for you to simplify.

2007-03-09 04:19:53 · answer #2 · answered by Puggy 7 · 0 0

Derivative= x(x^x)(x.x)= (x^3)(x^x)=x^(x+3)

2007-03-09 04:09:36 · answer #3 · answered by Masry_c777 2 · 0 2

i have d solution but have no idea how to post it here... it's quite long and complicated with all the log's and what nots... and if you really need it, let me know where i can send it...

2007-03-09 04:04:19 · answer #4 · answered by schatz101 3 · 0 2

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