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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 $y$?

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  • $\begingroup$ Did you try anything to see if that is true? $\endgroup$
    – acat3
    Commented Aug 8, 2021 at 5:09
  • $\begingroup$ @RezhaAdrianTanuharja I did take the LCM to see, ofc, and it checked out. But that still sheds no light on how we get to the decomposed $y$. $\endgroup$
    – Aadi Prasad
    Commented Aug 8, 2021 at 5:14
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    $\begingroup$ @copper.hat Thanks for clarifying that this is not a standard method (but a cleverer one!) $\endgroup$
    – Aadi Prasad
    Commented Aug 8, 2021 at 5:31
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    $\begingroup$ @Aadi Intuition, primarily. If I were to see this in an integral, this would be the first thing I try. Why? Because I've seen similar things before. I know that, if a linear expression appears in a square root, but otherwise there are just polynomial terms, it is often helpful to turn the square root into a $\sqrt{u}$. From there, it might take some partial fractions, or some rationalising the denominator, but you should be able to resolve it as a sum of powers of $u$. $\endgroup$
    – Theo Bendit
    Commented Aug 8, 2021 at 11:46
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    $\begingroup$ In response to Aadi’s question: I don’t like fractions. If I can manipulate the numerator to reduce a fraction, then my first step will be to follow through on this and hope it leads somewhere. In this case, the denominator had $(x-1)^{1/2}$, so I “made” an $x-1$ in the numerator. $\endgroup$
    – Clayton
    Commented Aug 8, 2021 at 13:25

2 Answers 2

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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 $f^{(100)}(x)$.

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

$y’=\frac{(3-x)}{2(1-x)^{3/2}}$,

$y’’=\frac{(7-x)}{2^{2}(1-x)^{5/2}}$,

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

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

… … … … …

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

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

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