Question

nth order derivative of y= (x^2 -x)/((x^2-4))^2

Answer 1

I think partial fraction decomposition may make finding a formula for the nth order derivative easier:
\[y=\frac{x^2-x}{(x^2-4)^2}=\frac{x^2-x}{(x-2)^2(x+2)^2}=\frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{C}{x+2}+\frac{D}{(x+2)^2}\]
\[x^2-x=A(x-2)(x+2)^2+B(x+2)^2+C(x+2)(x-2)^2+D(x-2)^2\\
x^2-x=A(x^3+2x^2-4x-8)+B(x^2+4x+4)+C(x^3-2x^2-4x+8)\\
~~~~~~~~~~~~~~~+D(x^2-4x+4)\\
x^2-x=(A+C)x^3+(2A+B-2C+D)x^2+(-4A+4B-4C-4D)x\\
~~~~~~~~~~~~~~~+(-8A+4B+8C+4D)\]
Matching up coefficents yields the system
\[\begin{cases}A+C=0\\2A+B-2C+D=1\\-4A+4B-4C-4D=-1\\-8A+4B+8C+4D=0\end{cases}\]
Solving (I'll leave the work to you) for the unknowns yields \(A=\dfrac{1}{8},B=\dfrac{1}{8},C=-\dfrac{1}{8},D=\dfrac{3}{8}\).

So,
\[y=\frac{1}{8(x-2)}+\frac{1}{8(x-2)^2}-\frac{1}{8(x+2)}+\frac{3}{8(x+2)^2}\]
And from here, you'd take a few derivatives and try to find a pattern:
\[y^{(1)}=-\frac{1}{8(x-2)^2}-\frac{1\cdot2}{8(x-2)^3}+\frac{1}{8(x+2)^2}-\frac{3\cdot2}{8(x+2)^3}\]
\[y^{(2)}=\frac{2}{8(x-2)^3}+\frac{1\cdot2\cdot3}{8(x-2)^4}-\frac{2}{8(x+2)^3}+\frac{3\cdot2\cdot3}{8(x+2)^4}\]
\[~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\vdots\]
\[y^{(n)}=(-1)^n\frac{n!}{8(x-2)^{n+1}}+(-1)^n\frac{(n+1)!}{8(x-2)^{n+2}}\\
~~~~~~~~~~~~~~~~-(-1)^{n+1}\frac{n!}{8(x+2)^{n+1}}+(-1)^n\frac{3(n+1)!}{8(x+2)^{n+2}}\]

Answer 2

thanks a lot!! @SithsAndGiggles