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

1. Jul 14, 2017

### NotASmurf

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))

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

Last edited: Jul 14, 2017
2. Jul 14, 2017

### jbriggs444

If f(x) = g(x) = $x^x$ then f(g(x)) = $g(x)^{g(x)}$ = $(x^x)^{(x^x)}$

3. Jul 14, 2017

### NotASmurf

ah, damn, should have noticed that. Thanks so much.

4. Jul 14, 2017

### NotASmurf

How would i make it into a chain rule form? x^x^x not (x^x)^(x^x)?

5. Jul 14, 2017

For $dy/dx$ where $y=x^{(x^x)}$, I do not get the answer you quoted from Wolfram, but the calculation should be straightforward. Write $y=x^{(x^x)}$ and take the natural twice on both sides of the equation. Then simply apply the chain rule and implicit differentiation of both sides and solve for $dy/dx$. $\\$ In fact I am in agreement with what you quoted from Wolfram other than the factor $e^{(x^{(x^x)})}$ which I have as simply being 1.