I am supposed to find the nth order derivative of:
$$\frac{1}{x^4+4}$$
I tried to resolve into partial fractions. But it didn't work out for me.
Edit- where I am stuck
$$\frac{1}{x^4+4}=\frac{1}{(x-1+i)(x-1-i)(x+1+i)(x+1-i)}= \frac{A}{(x-1+i)} + \frac{B}{(x-1-i)} + \frac{C}{(x+1+i)} + \frac{D}{(x+1-i)} $$
where A,B,C,D are to to be found.
I am unable to proceed further.
If I am able to find these values, the rest is easy to handle.
I am a beginner in this subject. Please help.
Best Answer
Hint
As the roots of $p(x)=x^4+4$ are $\xi_k = \sqrt 2 e^{i(\frac{\pi}{4} + k \frac{\pi}{2})}$ where $k \in \{0, 1, 2, 3\}$ you can write
$$\frac{1}{x^4+4} = \sum_{k=0}^3 \frac{a_k}{x-\xi_k}$$
where you have to find the $a_k$ using partial fraction decomposition. Based on the formula $a_k = \frac{1}{p^\prime(\xi_k)} = \frac{1}{4\xi_k^3}$, you get
$$\frac{1}{x^4+4} = \frac{1}{16}\sum_{k=0}^3 \frac{\xi_k}{x-\xi_k}$$ Then the $n$th-derivative of $\frac{a_k}{x-\xi_k}$ is easy to find. And you're done.