You Derive Me Crazyyyyyyy

We first find the derivatives of $x^x$ , let $g(x) = x^x$ then

$\ln (g(x)) = x \ln x \Rightarrow \frac {g'(x) }{g(x) } = \ln x + 1 \Rightarrow g'(x) = x^x (\ln x + 1)$

Now let $f(x) = x^{x^x}$ , take the log to both sides of the equation and differentiate them

$\begin{aligned} \ln (f(x)) & = & x^x \cdot \ln x \\ \frac {f'(x)}{f(x)} & = & x^x \cdot \frac {1}{x} + \ln x \cdot \left ( x^x ( \ln x + 1) \right ) \\ & = & x^x \left ( \frac {1}{x} + (\ln x)^2 + \ln x \right ) \\ \end{aligned}$

Set $x = 1$ , we get $f'(1) = 1$ , now differentiate the equation once more

$\LARGE \frac { f(x) f''(x) - (f'(x))^2 }{(f(x))^2}$

$\large = x^x \left (-\frac {1}{x^2} + \frac {2 \ln x }{x} + \frac {1}{x} \right )$

$\large \space \space \space + \left ( \frac {1}{x} + (\ln x)^2 + \ln x \right ) \cdot x^x ( \ln x + 1)$

$= x^x \left ( -\frac {1}{x^2} + \frac {3 \ln x }{x} + \frac {2}{x} + (\ln x)^3 + 2 (\ln x)^2 + \ln x \right )$

Set $x = 1$ , we get $f''(1) = 2$ , now differentiate the equation one last time

$\LARGE \frac { f(x) \cdot \left ( f(x) f'''(x) - f'(x) f''(x) \right ) - 2f'(x) \cdot \left ( f(x) f''(x) - (f'(x))^2 \right ) }{(f(x))^3}$

$\large = x^x \left ( \frac {2}{x^3} + \frac {3 - 3\ln x}{x^2} - \frac {2}{x^2} + \frac {3 (\ln x)^2}{x} + \frac {4 \ln x}{x} + \frac {1}{x} \right)$

$\large \space \space \space + \left ( -\frac {1}{x^2} + \frac {3 \ln x }{x} + \frac {2}{x} + (\ln x)^3 + 2 (\ln x)^2 + \ln x \right ) \cdot x^x ( \ln x + 1)$

Set $x = 1$ , simplify everything, we get $f'''(1) = \boxed{9}$

To avoid "differentiation overload" , let's use Taylor series to do this problem! To simplify the exposition, we will find the Taylor series of $f(x)=(x+1)^{(x+1)^{x+1}}$ based at 0 rather than the Taylor series of $g(x)=x^{x^x}$ based at 1. Note that $f'''(0)=g'''(1)$ is the answer we seek. My solution requires a few steps, but they involve only the algebra of cubic polynomials. At each stage, we will write the Taylor series based at 0 up to the $x^3$ -term. We start with the known Taylor series $\ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3}+...$ . Next $p(x)=(x+1)\ln(x+1)=x+\frac{x^2}{2}-\frac{x^3}{6}+...$ . Now we exponentiate and use the known Taylor series $e^y=1+y+\frac{y^2}{2}+\frac{y^3}{6}+...$ to find $(x+1)^{x+1}=e^{p(x)}=1+x+\frac{x^2}{2}+\frac{x^3}{2}+...$ . Finally, we have $q(x)=(\ln(x+1))(x+1)^{x+1}=x+\frac{x^2}{2}+\frac{5x^3}{6}+...$ and $f(x)=(x+1)^{(x+1)^{x+1}}=e^{q(x)}=1+x+x^2+\frac{3x^3}{2}+...$ . Our answer $f'''(0)$ is the coefficient of $x^3$ multiplied with 3!, which is $f'''(0)=g'''(1)=9$ .

The answer is 9

$f(x)={ x }^{ { x }^{ x } }\\ \\ f~ '''(x)=\frac { { { x^{ x^{ x }+3\, x+3 }\, \log ^{ 6 }x+x^{ x^{ x } }\, \left( 3\, x^{ 3\, x+3 }+3\, x^{ 2\, x+3 } \right) \, \log ^{ 5 }x+x^{ x^{ x } }\, \left( 9\, x^{ 2\, x+3 }+x^{ x+3 }+x^{ 3\, x }\, \left( 3\, x^{ 3 }+3\, x^{ 2 } \right) \right) \, \log ^{ 4 }x+x^{ x^{ x } }\, \left( 3\, x^{ x+3 }+x^{ 2\, x }\, \left( 9\, x^{ 3 }+12\, x^{ 2 } \right) +x^{ 3\, x }\, \left( x^{ 3 }+6\, x^{ 2 } \right) \right) \, \log ^{ 3 }x+x^{ x^{ x } }\, \left( x^{ 2\, x }\, \left( 3\, x^{ 3 }+21\, x^{ 2 }-3\, x \right) +x^{ x }\, \left( 3\, x^{ 3 }+6\, x^{ 2 } \right) +x^{ 3\, x }\, \left( 3\, x^{ 2 }+3\, x \right) \right) \, \log ^{ 2 }x+x^{ x^{ x } }\, \left( 3\, x^{ 3\, x+1 }+x^{ x }\, \left( x^{ 3 }+9\, x^{ 2 }-4\, x \right) +x^{ 2\, x }\, \left( 9\, x^{ 2 }+6\, x \right) \right) \, \log x+x^{ x^{ x } }\, \left( x^{ 3\, x }+x^{ x }\, \left( 3\, x^{ 2 }+2 \right) +x^{ 2\, x }\, \left( 6\, x-3 \right) \right) } } }{ { x }^{ 3 } } \\ \\ f~ '''\left( 1 \right) =9$

Here is a sketch:

Part 1:

Let $f=x^x$ . Then take log on both sides, and compute f'(1)=1 and f''(1)=2. For instance $\frac{f'}{f} = 1 + \ln(x)$ . Then f'(1) = 1. Differentiate both sides again to get an expression in terms of f''. Again substitute x=1 to get f''(1). Note that you don't need the third derivative of f.

Part 2:

Now let $y = x^{f(x)}$ . Do not substitute for f. Instead treat it as an implicit function. For instance $\frac{y'}{y} = \frac{f}{x} + f'\ln(x)$ . Now by differentiating LHS and RHS again, you get an expression in terms of y'', y', y, f, f', f''. Do it again to get an expression in terms of y'''. (You don't need f''' because the term you will get is $f'''\ln(x)$ which is 0 at x=1.) Thus you will get y'(1) = 1, y''(1) = 2 and finally, y'''(1) = 9. QED.