Find the $n$-th derivative of $f(x)=\frac{x}{\sqrt{1-x}}$

Question

Find the $n$-th derivative of $$f(x)=\frac{x}{\sqrt{1-x}}$$ First I just calculated the first, second and 3-th, 4-th derivatives and now I want to summarize the general formula. But it seems too complicated. Then I want to use binomial theorem or Taylor expansion... Also got no more clues.

Answer 1

This can be made a lot simpler by changing variables. (Changing variables is commonly taught as a technique for integration, but it can also be handy for differentiation.)

Introduce the new variable $u=1-x$. Then $x=1-u$, and $$f(x) = \frac{1-u}{\sqrt{u}} = u^{-1/2} - u^{1/2}$$ If we define a new function $g(x)=x^{-1/2} - x^{1/2}$ then this tells us that $$f(x) = g(1-x),$$ and therefore on taking derivatives we have $$f^{(n)}(x) = (-1)^n g^{(n)}(1-x)$$

This change of variables allows you to essentially swap out the problem of computing derivatives of $f(x)$ and trade it for computing derivatives of the (much simpler) function $g(x)$.

Now, the derivatives of $g(x)$ are $$g'(x) = \left( - \frac{1}{2}\right)x^{-3/2} - \left(\frac{1}{2}\right)x^{-1/2} $$ $$g''(x) = \left( - \frac{1}{2}\right)\left( - \frac{3}{2}\right)x^{-5/2} - \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)x^{-3/2} $$ $$g'''(x) = \left( - \frac{1}{2}\right)\left( - \frac{3}{2}\right)\left( - \frac{5}{2}\right)x^{-7/2} - \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)x^{-5/2} $$

and in general if we introduce the notation $A_n$ to denote the product of the first $n$ odd numbers (for example, $A_1=1$, $A_2 = 1\cdot 3$, $A_3 = 1\cdot 3 \cdot 5$, etc.) then

$$g^{(n)}(x)=(-1)^n \frac{A_n}{2^n}x^{-(2n+1)/2} + (-1)^n \frac{A_{n-1}}{2^n}x^{-(2n-1)/2}$$

Now we recall that $f^{(n)}(x) = (-1)^n g^{(n)}(1-x)$, so that

$$f^{(n)}(x)=\frac{A_n}{2^n}(1-x)^{-(2n+1)/2} + \frac{A_{n-1}}{2^n}(1-x)^{-(2n-1)/2}$$

The only thing left is to express the coefficients $A_n$ in a more convenient closed form; for that, see Proving formula for product of first n odd numbers.

Answer 2

Let $g(x)=x$, $h(x)=(1-x)^{-1/2}$. We can then say that $g^{[n]}=\begin{cases}x&n=0\\1&n=1\\0&\text{else}\end{cases}$ and that $h^{[n]}=(-1)^n\left(-\frac12\right)^{(n)}(1-x)^{-n-1/2}$, where $x^{(n)}$ is the falling factorial. Then, by the General Leibniz rule,

$$\begin{aligned}\left(g\cdot h\right)^{[n]}&=\sum_{k=0}^n\binom{n}{k}h^{[n-k]}g^{[k]}\\ &=g^{[0]}h^{[n]}+ng^{[1]}h^{[n-1]}+\sum_{k=2}^n\binom{n}{k}h^{[n-k]}g^{[k]}\\ &=\frac{x\cdot(-1)^n\left(-\frac12\right)^{(n)}}{(1-x)^{n+1/2}}+n\frac{(-1)^{n-1}\left(-\frac12\right)^{(n-1)}}{(1-x)^{n-1/2}}\\ &=\frac{(-1)^n\left(-\frac12\right)^{(n-1)}\left(x-2n\right)}{2(1-x)^{n+1/2}}\\ \end{aligned}$$

As per @Luca_Bressan's answer, we may also express the coefficients as $\frac12(-1)^n\left(-\frac12\right)^{(n-1)}=-\frac{(2n-3)!!}{2^n}$.

Answer 3

We have $f(x)=g(x)h(x)$ where $g(x)=x$ and $h(x)=(1-x)^{-1/2}$. Hence using Leibniz's rule we have

$$f^{(n)}(x)=\sum_{k=0}^n{n\choose k}h^{(k)}(x)g^{(n-k)}(x)={n\choose n}h^{(n)}(x)g(x)+{n\choose n-1}h^{(n-1)}(x)g'(x) \quad(*)$$ It remains to compute

$$h^{(n)}(x)=\frac{1}{2}\frac{3}{2}\cdots \frac{2n-1}{2}(1-x)^{(2n-3)/2} =\frac{(2n)!}{2^{2n}n!}(1-x)^{(2n-3)/2}$$ and then substitute it in $(*)$ and simplify.

Answer 4

$x(1-x)^{-\frac 12}$

Generalized binomial theorem.

$x(1-x)^{-\frac 12} = x(1^{-\frac {1}{2}} + \frac {-1}{ 2}1^{-\frac {3}{2}}(-x) + \frac {-1}{ 2}\frac {-3}{2}\frac {1}{2}1^{-\frac {5}{2}}(-x)^2+\frac {-1}{ 2}\frac {-3}{2}\frac {1}{2}\frac {-5}{2}\frac {1}{3}1^{-\frac {5}{2}}(-x)^3\cdots$

$x(1-x)^{-\frac 12} = x + \frac {1}{ 2}x^2 + \frac {3}{8}x^3+\frac {5}{16}x^4\cdots$

$a_1 = 1\\ a_{n+1} = a_n{\frac {2n-1}{2n}}\\ a_n= \frac {(2n-1)!}{2^{2n}((n-1)!)^2}$

Answer 5

We can prove that $$f^{(n)}(x) = - \frac {(2n - 3)!!\, (x - 2n)} {2^n (1 - x)^{(2n + 1)/2}}$$ for $n \ge 2$ by induction on $n$.

The base case is easy. For the inductive step, $$\begin{align*} \frac d {dx} f^{(n)} (x) & = - \frac {(2n - 3)!!\, 2^n (1 - x)^{(2n + 1)/2} + (2n - 3)!!\, (x - 2n) 2^n \frac {2n+1} 2 (1 - x)^{(2n - 1)/2}} {2^{2n} (1 - x)^{2n + 1}} \\ & = - \frac {(2n - 3)!!\, 2^{n-1} (1 - x)^{(2n - 1)/2} \, [2 (1 - x) + (x - 2n) (2n + 1)]} {2^{2n} (1 - x)^{2n + 1}} \\ & = - \frac {(2n - 3)!!\, (2n - 1)(x - 2n - 2)} {2^{n+1} (1 - x)^{(2n + 3)/2}} \\ & = - \frac {[2(n + 1) - 1]!!\, [x - 2(n + 1)]} {2^{n+1} (1 - x)^{[2(n + 1) + 1]/2}}. \end{align*}$$