Mathamatica, Symbolab, manual differentiation all return diffrent forumals, I cannot see wolfram's intermediary steps, symbolab's intermediary steps seem like they're treating a function like a constant and I can't trust my work when everything else disagrees, but i don't see where my logic would falter as long as x^x^x can be represented by f(x) = x^x, g(x) = x^x, x^x^x = f(g(x))(adsbygoogle = window.adsbygoogle || []).push({});

$$\frac{dy}{dx} of y = e^x^x $$ is

y = e^x^x

y' = q(g(x))' = q'(g(x))*g'

g(x) = x^x, q(x) = e^x

g'(x) = (ln(x) + 1)*x^x, q'(x) = e^x

y' = e^x^x * (ln(x) + 1)*x^x

y' = x^x * (ln(x) + 1)*e^x^x

The above is correct, dy/dx x^x was obtained via:

g(x) = x^x

ln(g) = xln(x)

ln(g)' = (xln(x))'

ln(g)' = 1 * ln(x) + x * 1/x

g'/g = 1 * ln(x) + x * 1/x

g'/g = ln(x) + 1

g' = (ln(x) + 1)*x^x

But when i try continue for e^x^x^x via chain rule

t=x^x^x = f(g(x))

f=x^x g = x^x

f' = (ln(x) + 1)*x^x, g' = (ln(x) + 1)*x^x

(f(g(x)))' = f'(g(x))*g'(x) = (ln(x) + 1)(ln(x^x) + 1)*x^x^x * x^x = (xln(x)^2 + ln(x) + xln(x) + 1) * x^(x + x^x)

dy/dx x^x^x = (xln(x)^2 + ln(x) + xln(x) + 1) * x^(x + x^x)

I obtain the incorrect answer, as wolfram says dy/dx of x^x^x = e^x^x^x * (xln(x)^2 + xln(x) + 1) * x^(x + x^x -1), but I do not have the professional version of wolfram, thus I cannot see the intermediary steps, and cannot find this information anywhere else,

Any help appreciated.

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# I Why is [x^x^x]' NOT (xln(x)^2+ln(x)+xln(x)+1)*x^(x+x^x)?

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