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**Soroban** Hello, Dan!

It seems that you are seeking an x-intercept of: .y .= .x^x

. . (If I'm wrong, disregard this entire post.)

Is your derivative correct? . . . It's tricky.

Take logs: . ln y .= .ln(x^x) .= .x·ln(x)

Differentiate: . y'/y .= .x(1/x) + ln(x)

Then: . y' .= .y[(1 + ln(x)] .= .x^x[1 + ln(x)]

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .x^x

The interation becomes: . x_(n+1) . = . x_n - -----------------

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x^x[1 + ln(x)]

. . . . . . . . . . . . . . . . . . . . . . . . .1

Therefore: .x_(n+1) . = . x_n - ------------

. . . . . . . . . . . . . . . . . . . . . . 1 + ln(x)