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by **nona.m.nona** on Thu Feb 26, 2009 3:34 pm

Find dy/dx for y = x^{x^x}.

ln(y) = ln(x^{x^x}) = x^x ln(x)

But then what?

**Sponsor**

by **stapel_eliz** on Thu Feb 26, 2009 6:03 pm

nona.m.nona wrote:Find dy/dx for y = x^{x^x}.

ln(y) = ln(x^{x^x}) = x^x ln(x)

But then what?

Give it another round of logarithmic differentiation! :wink:

. . . . . $\ln{(y)}\, =\, x^x \ln{(x)}$

. . . . . $\ln{(\ln{(y)})}\, =\, \ln{\left(x^x \ln{(x)}\right)}$

Apply a log rule to get:

. . . . . $\ln{(\ln{(y)})}\, =\, \ln{(x^x)}\, +\, \ln{(x)}$

. . . . . $\ln{(\ln{(y)})}\, =\, x \ln{(x)}\, +\, \ln{(x)}$

Now differentiate:

. . . . . $\left(\frac{1}{\ln{(y)}}\right)\,\left(\frac{1}{y}\right)\, \frac{dy}{dx}\, =\, \ln{(x)}\, +\, \frac{x}{x}\, +\, \frac{1}{x}$

Simplify, and isolate the dy/dx. Make sure to substitute for "y" and "ln(y)" in the result.

Check my work!

by **nona.m.nona** on Mon Mar 02, 2009 6:49 pm

I finally got it. Thank you!

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Find dy/dx for y = x^(x^x) ...using log differentiation?

Find dy/dx for y = x^(x^x) ...using log differentiation?

Re: Find dy/dx for y = x^(x^x) ...using log differentiation?