logarithmic differentaition

[dans]

I need to find Y prime using logarithmic differentiation.

y=x^x^x

can you show me the work?

Thanks

 

[soroban]

Hello, dans!

I assume you've done <u>some</u> logarithmic differentiation before.
. . . This is a killer problem . . .

y = x^{x^x}
.

Take logs: . ln(y) . = . ln(x^{x^x}) . = . x^x ln(x)

Take logs again: . ln[ln(y)] . = . ln[x^x ln(x)] . = . ln(x^x) + ln[ln(x)]

And we have: . ln[ln(y)] . = . x ln(x) + ln[ln(x)]


Differentiate implicitly:
. . . . .1 . . . 1 . . . . . . . . . . 1 . . . . . . . . . . . 1 . . . 1
. . . ------- · -- · y' . .= . .x · -- .+ .ln(x) .+ .------- · --
. . . .ln(y) . .y . . . . . . . . . . x . . . . . . . . . . ln(x) . . x


And I'll let you solve for y' . . .

 

[Gene]

I used a somewhat different method.
I approve of Soroban's leaving you something to do, but it won't make sense without almost finishing it. You have to substitute the defined values for the y & u terms.
y=(x^x)^x
u = x^x
ln(u) = x*ln(x)
d(ln(u)) = xd(ln(x))+ln(x)dx
=dx+ln(x)dx
y=u^x
ln(y) = x(ln(u))
(1/y)(dy) = x(d(ln(u))+ln(u)dx

If you don't end up with
(x^x^x)(x+2xln(x))
try again.