Let's do some calculus! (17)

Let $f^{(0)}=x^{x^x}$ . We can find the first derivative by taking the natural log on both sides, i.e. $ln(f^{(0)})=ln(x^{x^x})=x^xln(x)$

If we let $g^{(0)}=x^xln(x)$ and using the chain rule we have $\frac{f^{(1)}}{f^{(0)}}=\frac{d}{dx}g^{(0)}$ , i.e. $f^{(1)}=f^{(0)}g^{(1)}$

If we also let $y^{(0)}=x^x$ and $z^{(0)}=ln(x)$ then we can see that $g^{(0)}=y^{(0)}z^{(0)}$ . Furthermore, we can find the first derivative of $y^{(0)}$ by again taking the natural log on both sides and using the chain rule. Thus $ln(y^{(0)})=ln(x^x)=xln(x)$ . i.e. $\frac{y^{(1)}}{y^{(0)}}=\frac{d}{dx}xln(x)=x×\frac{1}{x}+ln(x)=1+z^{(0)}$ , i.e. $y^{(1)}=y^{(0)}(1+z^{(0)})$

So we can see that all of the derivatives of $f^{(0)}$ are dependent on higher derivatives and derivatives of $z^{(0)}$ . So the first 7 derivatives of $z^{(0)}$ are $z^{(0)}=ln(x)$ , $z^{(1)}=\frac{1}{x}$ , $z^{(2)}=-\frac{1}{x^2}$ , $z^{(3)}=\frac{2}{x^3}$ , $z^{(4)}=-\frac{6}{x^4}$ , $z^{(5)}=\frac{24}{x^5}$ , $z^{(6)}=-\frac{120}{x^6}$ and $z^{(7)}=\frac{720}{x^7}$ .

And letting $x=1$ gives $z^{(0)}(1)=0$ , $z^{(1)}(1)=1$ , $z^{(2)}(1)=-1$ , $z^{(3)}(1)=2$ , $z^{(4)}(1)=-6$ , $z^{(5)}(1)=24$ , $z^{(6)}(1)=-120$ and $z^{(7)}(1)=720$ .

Now we can get the derivatives of $y^{(0)}$ :

$y^{(0)}(1)=1^1=1$

$y^{(1)}(1)=y^{(0)}(1)(1+z^{(0)}(1))$ $=1×(1+0)=1$

$y^{(2)}(1)=y^{(1)}(1)(1+z^{(0)}(1))+y^{(0)}(1)z^{1}(1)=$ $1×(1+0)+1×1=2$

$y^{(3)}(1)=y^{(2)}(1)(1+z^{(0)}(1))+2y^{(1)}(1)z^{(1)}(1)+y^{(0)}(1)z^{(2)}(1)$ $=2×(1+0)+2×1×1+1×-1=3$

$y^{(4)}(1)=y^{(3)}(1)(1+z^{(0)}(1))+3y^{(2)}(1)z^{(1)}(1)+3y^{(1)}(1)z^{(2)}(1)+y^{(0)}(1)z^{(3)}(1)$ $=3×(1+0)+3×2×1+3×1×-1+1×2$ $=8$

$y^{(5)}(1)=y^{(4)}(1)(1+z^{(0)}(1))+4y^{(3)}(1)z^{(1)}(1)+6y^{(2)}(1)z^{(2)}(1)+4y^{(1)}(1)z^{(3)}(1)+y^{(0)}(1)z^{(4)}(1)$ $=8×(1+0)+4×3×1+6×2×-1+4×1×2+1×-6$ $=10$

$y^{(6)}(1)=y^{(5)}(1)(1+z^{(0)}(1))+5y^{(4)}(1)z^{(1)}(1)+10y^{(3)}(1)z^{(2)}(1)+10y^{(2)}(1)z^{(3)}(1)+5y^{(1)}(1)z^{(4)}(1)+y^{(0)}(1)z^{(5)}(1)$ $=10×(1+0)+5×8×1+10×3×-1+10×2×2+5×1×-6+1×24$ $=54$

$y^{(7)}(1)=y^{(6)}(1)(1+z^{(0)}(1))+6y^{(5)}(1)z^{(1)}(1)+15y^{(4)}(1)z^{(2)}(1)+20y^{(3)}(1)z^{(3)}(1)+15y^{(2)}(1)z^{(4)}(1)+6y^{(1)}(1)z^{(5)}(1)+y^{(0)}(1)z^{(6)}(1)$ $=54×(1+0)+6×10×1+15×8×-1+20×3×2+15×2×-6+6×1×24+1×-120$ $=-42$

And now for $g^{(0)}$ : $g^{(0)}(1)=y^{(0)}(1)z^{(0)}(1)=1×0=0$

$g^{(1)}(1)=y^{(1)}(1)z^{(0)}(1)+y^{(0)}(1)z^{(1)}(1)$ $=1×0+1×1=1$

$g^{(2)}(1)=y^{(2)}(1)z^{(0)}(1)+2y^{(1)}(1)z^{(1)}(1)+y^{(0)}(1)z^{(2)}(1)$ $=2×0+2×1×1+1×-1=1$

$g^{(3)}(1)=y^{(3)}(1)z^{(0)}(1)+3y^{(2)}(1)z^{(1)}(1)+3y^{(1)}(1)z^{(2)}(1)+y^{(0)}(1)z^{(3)}(1)$ $=3×0+3×2×1+3×1×-1+1×2=5$

$g^{(4)}(1)=y^{(4)}(1)z^{(0)}(1)+4y^{(3)}(1)z^{(1)}(1)+6y^{(2)}(1)z^{(2)}(1)+4y^{(1)}(1)z^{(3)}(1)+y^{(0)}(1)z^{(4)}(1)$ $=8×0+4×3×1+6×2×-1+4×1×2+1×-6$ $=2$

$g^{(5)}(1)=y^{(5)}(1)z^{(0)}(1)+5y^{(4)}(1)z^{(1)}(1)+10y^{(3)}(1)z^{(2)}(1)+10y^{(2)}(1)z^{(3)}(1)+5y^{(1)}(1)z^{(4)}(1)+y^{(0)}(1)z^{(5)}(1)$ $=10×0+5×8×1+10×3×-1+10×2×2+5×1×-6+1×24$ $=44$

$g^{(6)}(1)=y^{(6)}(1)z^{(0)}(1)+6y^{(5)}(1)z^{(1)}(1)+15y^{(4)}(1)z^{(2)}(1)+20y^{(3)}(1)z^{(3)}(1)+15y^{(2)}(1)z^{(4)}(1)+6y^{(1)}(1)z^{(5)}(1)+y^{(0)}(1)z^{(6)}(1)$ $=54×0+6×10×1+15×8×-1+20×3×2+15×2×-6+6×1×24+1×-120$ $=-96$

$g^{(7)}(1)=y^{(7)}(1)z^{(0)}(1)+7y^{(6)}(1)z^{(1)}(1)+21y^{(5)}(1)z^{(2)}(1)+35y^{(4)}(1)z^{(3)}(1)+35y^{(3)}(1)z^{(4)}(1)+21y^{(2)}(1)z^{(5)}(1)+7y^{(1)}(1)z^{(6)}(1)+y^{(0)}(1)z^{(7)}(1)$ $=-42×0+7×54×1+21×10×-1+35×8×2+35×3×-6+21×2×24+7×1×-120+1×720$ $=986$

And so we can now solve for $f^{(7)}(1)$ :

$f^{(1)}(1)=f^{(0)}(1)g^{(1)}(1)=1×1=1$

$f^{(2)}(1)=f^{(1)}(1)g^{(1)}(1)+f^{(0)}(1)g^{(2)}(1)$ $=1×1+1×1=2$

$f^{(3)}(1)=f^{(2)}(1)g^{(1)}(1)+2f^{(1)}(1)g^{(2)}(1)+f^{(0)}(1)g^{(3)}(1)$ $=2×1+2×1×1+1×5=9$

$f^{(4)}(1)=f^{(3)}(1)g^{(1)}(1)+3f^{2}(1)g^{2}(1)+3f^{(1)}(1)g^{(3)}(1)+f^{(0)}(1)g^{(4)}(1)$ $=9×1+3×2×1+3×1×5+1×2=32$

$f^{(5)}(1)=f^{(4)}(1)g^{(1)}(1)+4f^{(3)}(1)g^{(2)}(1)+6f^{(2)}(1)g^{(3)}(1)+4f^{(1)}(1)g^{(4)}(1)+f^{(0)}(1)g^{(5)}(1)$ $=32×1+4×9×1+6×2×5+4×1×2+1×44$ $=180$

$f^{(6)}(1)=f^{(5)}(1)g^{(1)}(1)+5f^{(4)}(1)g^{(2)}(1)+10f^{(3)}(1)g^{(3)}(1)+10f^{(2)}(1)g^{(4)}(1)+5f^{(1)}(1)g^{(5)}(1)+f^{(0)}(1)g^{(6)}(1)$ $=180×1+5×32×1+10×9×5+10×2×2+5×1×44+1×-96$ $=954$

$f^{(7)}(1)=f^{(6)}(1)g^{(1)}(1)+6f^{(5)}(1)g^{(2)}(1)+15f^{(4)}(1)g^{(3)}(1)+20f^{(3)}(1)g^{(4)}(1)+15f^{(2)}(1)g^{(5)}(1)+6f^{(1)}(1)g^{(6)}(1)+f^{(0)}(1)g^{(7)}(1)$ $=954×1+6×180×1+15×32×5+20×9×2+15×2×44+6×1×-96+1×986$ $\boxed{=6524}$

Am sure there is a way to simplify this solution by proving that $g^{(n)}=y^{(n+1)}-y^{(n)}$ , but I couldn't figure out how to do that. If you can work it out, please let me know!