Inspired by Pi Han Goh

Based on Pi Han Goh's brilliant note, we can generalise:

Let $T_n(f(x))$ be the Taylor polynomial of $f(x)$ at $x=0$ .

If $T_n(f(x))=T_n(g(x))$ for two functions, then $T_{n+1}(\ln(x+1)f(x))$ $=T_{n+1}(\ln(x+1)g(x))$ ; since $\ln(x+1)$ does not have a constant term, the degree gets "pushed up". This in turn implies that $T_{n+1}((x+1)^{f(x)})=T_{n+1}((x+1)^{g(x)})$ , by exponentiation.

Now $T_1(x+1)=T_1\left((x+1)^{x+1}\right)$ , by inspection ; applying the above observation $n-1$ times, we find that $T_n\left(^n(x+1)\right)=T_{n}\left(^{n+1}(x+1)\right)$ and therefore $T_n\left(^n(x+1)\right)=T_{n}\left(^m(x+1)\right)$ for all $m\geq{n}$ .

Let $D_n(f(x))$ be the nth derivative of $f(x)$ at $x=0$ . Since $D_n$ is determined by $T_n$ , we have $D_n\left(^n(x+1)\right)=D_{n}\left(^m(x+1)\right)$ for all $m\geq{n}$ .

In particular, $D_3\left(^7(x+1)\right)=D_{3}\left(^3(x+1)\right)=\boxed{9}$ , as we saw in Pi Han Goh's earlier problem.

We use the concept from Otto Bretscher's solution with the tetration notation applied. We want to find the 3rd derivative of $\ ^7 (x+1)$ at $x=0$ . Consider its Maclaurin Series.

Brief synopsis:

Step 1 : Start with the number 1.

Step 2 : Multiply it by the series of $\ln(x+1)$ .

Step 3 : Multiply it by $(x+1)$ ,

Step 4 : Exponentiate it: $e^{\text{previous expression}}$ .

Step 5 : Repeat steps 2 to 4 for another 6 times.

Step 6 : Terminate algorithm. Find the coefficient of $x^3$ and multiply it by $3!=6$ . Do victory dance.

Okay. Here's the full working. Since we're only interested in the 3rd derivative, we only need to expand the series up to the $x^3$ -term.

$\Rightarrow \ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} - O(x^4)$

$\Rightarrow (x+1) \ln(x+1) = x + \frac {x^2}{2} - \frac{x^3}{6} + O(x^4)$ .

Expontentiate it: $\Rightarrow e^y = 1 + y + \frac{y^2}2 + \frac{y^3}6 + O(y^4)$ .

$\Rightarrow e^{(x+1) \ln(x+1) = x + \frac {x^2}{2} - \frac{x^3}{6} + \ldots} = (x+1)^{(x+1)} = 1 + x + \frac{x^2}{2} + \frac{x^3}{2} + O(x^4)$

$\Rightarrow \ln(x+1) \cdot (x+1)^{(x+1)} = x + \frac{x^2}{2} + \frac{5x^3}6 + O(x^4)$ .

$\Rightarrow e^{\ln(x+1) \cdot (x+1)^{(x+1)}} = (x+1)^{(x+1)^{(x+1)}} = 1 + x + x^2 + \frac{3x^3}{2} + O(x^4)$

$\Rightarrow \ln(x+1) \cdot (x+1)^{(x+1)^{(x+1)}} = x + \frac{x^2}{2} + \frac{5x^3}{6} + O(x^4)$ .

$\Rightarrow e^{ \ln(x+1) \cdot (x+1)^{(x+1)^{(x+1)}} } = \ ^4 (x+1) = 1 + x + x^2 + \frac{3x^3}{2} + O(x^4)$

The first 4 terms of the series of $\ ^4 (x+1)$ is the same as $\ ^3 (x+1)$ . If we repeat steps 2 to 4 another 3 more times, we will still conclude that sum of the first four terms of $\ ^7 (x+1)$ as $1 + x + x^2 + \frac{3x^3}{2}$ .

By Maclaurin Series, the third derivative of $\ ^7(x+1)$ satisfies $\frac{3}{2} = \frac{ \frac{d^3}{dx^3} \left [ \ ^7 (x+1) \right ] }{3!} \Rightarrow \frac{d^3}{dx^3} \left [ \ ^7 (x+1) \right ] = \frac 32 \times 3! = \boxed{9}$ .