How to decompose $\frac{1+x}{\sqrt{(1-x)}}$ into partial fractions?

Question

Basically homework help. The question (Problems of Calculus in One Variable, IA Maron, number 2.3.9(b)) is to find the derivative of the 100th order of the function $$ y = \frac{1+x}{\sqrt{(1-x)}} $$ by 'expansion into a linear combination of simpler functions'.

I can't find any help online. There is a hint at the back, which says that $y$ can be written as $2(\sqrt{1-x})^{-1} - \sqrt{1-x}$, but how we get that, I have no idea.

How am I supposed to decompose $y$?

Answer 1

Hint: Rewrite the numerator: $1+x = 2 - (1 - x)$. Thus $f(x)=\dfrac{1+x}{\sqrt{1-x}}= \dfrac{2}{\sqrt{1-x}} - \sqrt{1-x} = 2(1-x)^{-\frac{1}{2}} - (1-x)^{\frac{1}{2}}$. From this you can find the first few derivatives and see an inductive pattern that leads to a formula for the $f^{(100)}(x)$.

Answer 2

$y=\frac{1+x}{(1-x)^{1/2}}$,

$y’=\frac{(3-x)}{2(1-x)^{3/2}}$,

$y’’=\frac{(7-x)}{2^{2}(1-x)^{5/2}}$,

$y’’’=\frac{3(11-x)}{2^{3}(1-x)^{7/2}}$,

$y’’’’=\frac{3.5(15-x)}{ 2^{4}(1-x)^{9/2}}$,

… … … … …

$y^{(n)}=\frac{(4n-1-x).\prod_{k=2}^n\left(2k-3\right)}{2^{n}(1-x)^{\frac{2n+1}{2}}}$.

Just substitute for n, 100 and you're done.