What is the nth derivative of $\dfrac{1}{\sqrt{1 + x^2}}$

Question

I'm trying to find a general formula for the $n$th derivative of $$\dfrac{1}{\sqrt{1 + x^2}}$$ I got up to, \begin{eqnarray*} g^{(0)}(x) &=& g(x) \\ g^{(1)}(x) &=& \dfrac{1}{(1 + x^2)^{1/2}} \\ g^{(2)}(x) &=& \dfrac{-x}{(1 + x^2)^{3/2}} \\ g^{(3)}(x) &=& \dfrac{2x^2 - 1}{(x^2 + 1)^{5/2}} \\ g^{(4)}(x) &=& \dfrac{-6x^3 + 9x}{(x^2 + 1)^{7/2}} \\ g^{(5)}(x) &=& \dfrac{24x^4 - 72x^2 + 9}{(x^2 + 1)^{9/2}} \\ g^{(6)}(x) &=& \dfrac{-120x^5 + 600x^3 - 225x}{(x^2 + 1)^{11/2}} \\ g^{(7)}(x) &=& \dfrac{720x^6 - 5400x^4 + 4050x^2 -225}{(x^2 + 1)^{13/2}} \\ g^{(8)}(x) &=& \dfrac{-5040x^7 + 52920x^5 - 66150x^3 + 11025x}{(x^2 + 1)^{15/2}} \\ \end{eqnarray*}

Except for the first term in the numerator ($n!$), and the power in the denominator, I couldn't find a general pattern for the rest of the coefficients in the numerator. Could anyone shed me some light on this problem? Any idea would be greatly appreciated. Thanks.

Answer 1

Since this is part of finding the Maclaurin Series for your given function $g(x)$, you can use the well-known fact for the Maclaurin Series for a binomial expansion:

$$(1+x)^k = \sum_{n=0}^{\infty} {k\choose n}x^n = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 + \cdots$$

In your case,

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

So, $k = -\frac{1}{2}$ and $x = x^2$ per the definition of the expansion for $(1+x)^k$

If you need further help writing out the binomial expansion for your function, let me know and I'll fill in some more details.

Note that this only holds when our Taylor Series is about a point $a = 0$. That is, it is a Maclaurin Series. If $a \ne 0$, only then do we need to compute the nth order derivative about $a$. For a given Taylor Series, we can express a function $f$ as:

$$f(x) = \sum_{n=0}^{N} \frac{f^n(a) (x-a)^n}{n!}$$