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

Limits, differentiation, related rates, integration, trig integrals, etc.
nona.m.nona
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### Find dy/dx for y = x^(x^x) ...using log differentiation?

Find dy/dx for y = xx^x.

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

But then what?

stapel_eliz
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Joined: Mon Dec 08, 2008 4:22 pm
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Find dy/dx for y = xx^x.

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

But then what?
Give it another round of logarithmic differentiation!

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

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

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!

nona.m.nona
Posts: 288
Joined: Sun Dec 14, 2008 11:07 pm
Contact:

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

I finally got it. Thank you!