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