Differentiating unusual

Here is a generalised solution

Consider the function $f(x) = \frac{1}{x^2+a^2}$

Noting that the $n^{\text{th}}$ derivative of $f(x+c) \quad \forall c \in \mathbb{R}$ is essentially identical to the $n^{\text{th}}$ derivative of $f(x)$ , we proceed without loss of generality.

Let $x = a\cot{\theta} \Leftrightarrow \theta = \tan^{-1}{\frac{a}{x}}$ and differentiate both sides implicitly to obtain $\frac{\text{d}\theta}{\text{d}x} = \frac{\sin^2{\theta}}{-a}$

Rewrite $f(x)$ in terms of $\theta$ to obtain $f(x) = \frac{\sin^2{\theta}}{a^2}$ and differentiate using the chain rule to obtain

$\frac{\text{d}(f(x))}{\text{d}x} = \frac{2\sin{\theta}\cos{\theta}\times \sin^2{\theta}}{-a^3} = \frac{1}{(-a)^3} \times \sin{2\theta} \times \sin^2{\theta}$

The pattern is apparent at this stage, if you do not yet see it, differentiate once more and simplify.

Out of brevity, I will jump straight to the propositional statement for the $n^{\text{th}}$ derivative of $f(x)$ .

$P(n): \frac{\text{d}^n(f(x))}{(\text{d}x)^n} = \frac{n!}{(-a)^{n+2}} \times \sin{(n+1)\theta} \times \sin^{n+1}{\theta} \quad \forall n \in \mathbb{N}$

The case $n=0$ is a trivial case, and we have obtained $n=1$ , so we may take the inductive step.

Assume the statement holds true for some $n \in \mathbb{N}$ .

$P(n): \frac{\text{d}^n(f(x))}{(\text{d}x)^n} = \frac{n!}{(-a)^{n+2}} \times \sin{(n+1)\theta} \times \sin^{n+1}{\theta}$

Then we differentiate both sides w.r.t. $x$ using the product and chain rules to obtain the next derivative.

$\frac{\text{d}^n(f(x))}{(\text{d}x)^n} = \frac{n!}{(-a)^{n+2}} \times \left[ (n+1)\cos{(n+1)\theta} \times \sin^{n+1}{\theta} + (n+1)\sin{(n+1)\theta} \times \sin^n{\theta} \cos{\theta} \right] \times \frac{\sin^2{\theta}}{-a}$

Factor out $n+1$ and merge it with the already existing $n!$ , and likewise for $\sin^n{\theta}$ with $\sin^2{\theta}$ and $(-a)^{n+2}$ with $-a$

$\frac{\text{d}^n(f(x))}{(\text{d}x)^n} = \frac{(n+1)!}{(-a)^{n+3}} \times \left[ \cos{(n+1)\theta} \times \sin{\theta} + \sin{(n+1)\theta} \times \cos{\theta} \right] \times \sin^{n+2}{\theta}$

Use the reverse angle sum expansion identity $\sin{a}\cos{b} + \sin{b}\cos{a} \equiv \sin{(a+b)}$

$\frac{\text{d}^n(f(x))}{(\text{d}x)^n} = \frac{(n+1)!}{(-a)^{n+3}} \times \sin{(n+2)\theta} \times \sin^{n+2}{\theta} \Rightarrow P(n+1)$

This proves the propositional statement for $n+1$ , provided it is true for $n$ .

As the proposition is true for $n=0$ and $n=1$ , by the Axiom of Induction, the proposition is true $\forall n \in \mathbb{N}$

Setting $c = \frac{1}{2}, a = \frac{\sqrt{3}}{2}, n=6$ , and noting $\theta = \tan^{-1}{\frac{\sqrt{3}}{2x+1}}$ we obtain:

$\frac{\text{d}(f(x+\frac{1}{2}))}{(\text{d}x)^6} = \frac{6! \times 2^8}{3^4} \times \sin{7\left(\tan^{-1}{\frac{\sqrt{3}}{2x+1}}\right)} \times \sin^7{\left(\tan^{-1}{\frac{\sqrt{3}}{2x+1}}\right)} = \frac{20480}{9} \times \sin{7\left(\tan^{-1}{\frac{\sqrt{3}}{2x+1}}\right)} \times \sin^7{\left(\tan^{-1}{\frac{\sqrt{3}}{2x+1}}\right)}$

So we have $a+b+c+d+e+f+g = 20480 + 9 + 7 + 2 + 3 + 1 + 7 = 20509$

Use partial fraction decomposition (PFD) via "Covering Method" on $f(x)$ . We notice it has a complex pole-pair: $\begin{aligned} f(x)&=\frac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{1}{(x-x_0)(x-x_0^*)}\underset{\text{PFD}}{=}\frac{A}{x-x_0}+\frac{A^*}{x-x_0^*} &&&\left| x_0=-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right.,\qquad A=-\frac{i}{\sqrt{3}} \end{aligned}$ To simplify the 6'th derivative into sines and arctangents, we write $x-x_0=|x-x_0|e^{i\varphi}$ in polar coordinates: $\begin{aligned} f^{(6)}(x) &= 6!(-1)^6\left[ \frac{A}{(x-x_0)^7} + \frac{A^*}{(x-x_0^*)^7} \right]=6!\cdot 2\Re\left\{ \frac{A}{(x-x_0)^/} \right\} = \frac{-6!\cdot 2}{\sqrt{3}|x-x_0|^7}\Re\left\{ ie^{-i7\varphi} \right\}\\[1em] \varphi&:=\arctg2\left( -\frac{\sqrt{3}}{2},\:x+\frac{1}{2} \right)=-\arctg2(\sqrt{3},\:2x+1)\qquad\Rightarrow\qquad \sin(\varphi)=\frac{-\frac{\sqrt{3}}{2}}{|x-x_0|} \end{aligned}$ With $\sin(\varphi)$ we can eliminate $|x-x_0|$ from the derivative. Collecting coefficients from $\varphi$ , we get $d=3,\:e=2,\:f=1$ and finally $\begin{aligned} f^{(6)}(x)&=\frac{6!\cdot (-2)^8}{\sqrt{3}^8}\sin^7(\varphi)\sin(7\varphi)=\frac{20480}{9}\sin^7(\varphi)\sin(7\varphi) &&&\Rightarrow &&&& a+\ldots+g=\boxed{20509} \end{aligned}$