# Nth derivative of a function: I don't know where to start

I'm practicing some problems from past exams and found this one: Find nth derivative of this function:
$f(x)=\frac {x} {x^2-1}$.
I have no idea how to start solving this problems. Is there any theorem for finding nth derivative?

• Do you need the exact derivative, or just its shape?
– user365
Aug 30 '10 at 11:24
• @Marco Castronovo It doesn't say. :( Aug 30 '10 at 11:42
• More generally, the partial fraction decomposition of a complex rational fraction $f(X)$ can be computed as follows. The singular part of $f(X)$ at an order $m$ pole $a$ is obtained by dividing by $(X-a)^m$ the degree $<m$ Taylor approximation of $(X-a)^m f(X)$ at $a$. Aug 31 '10 at 9:13

Maybe we can add some more help -just in case you didn't succeed to find the answer yourself yet. Let

$$\frac{x}{x^2 - 1} = \frac{A}{x-1} + \frac{B}{x+1}$$

be the splitting into partial fractions. (I'm too lazy to compute the coeffitients $A$ and $B$.) Then

$$\frac{d}{dx} \frac{x}{x^2 - 1} = -\frac{A}{(x-1)^2} - \frac{B}{(x+1)^2} \ .$$

Differentiating again,

$$\frac{d^2}{dx^2}\frac{x}{x^2 - 1} = \frac{2A}{(x-1)^3} + \frac{2B}{(x+1)^3} \ .$$

One more time:

$$\frac{d^3}{dx^3} \frac{x}{x^2 - 1} = - \frac{3\cdot 2 A}{(x-1)^4} - \frac{3\cdot 2 B}{(x+1)^4} \ .$$

And sure enough you can find the general pattern now, can't you? Then, use induction to prove your guess.

• As usual, lots and lots of great answers, so big thank you to everyone! I selected this one because it helped me the most. Sep 1 '10 at 21:56

HINT $\;\;\;$ Upon employing partial fractions it reduces to $\;\;\;\rm D^n \:\frac{1}{x+1}. \;\;$ Now employ the Taylor series

$$\rm\; f(t+x) = \sum_{k=0}^\infty \;\; D^n \: f(x) \; \frac{t^n}{n!}$$

and note that for this problem we've $\rm\; f(t+x) = \frac{1}{t+x+1}$ is a geometric series with known coefficients.

Such "generating function" approaches often work smoothly even in much more complicated problems. Indeed, there is a very powerful Umbral calculus that frequently succeeds in computing such closed form expressions, e.g. see Steven Roman: The Umbral Calculus. 1984. For example, below is a small sample of derivative formulas for the countless number of polynomial sequences amenable to umbral calculus analysis

$$\begin{array}{|r|l|} \hline \rm Name & \rm Derivative \; formula \\ \hline\hline \rm Laguerre & \rm L_n^k(x) = (D+1)^{n+k}(-x)^n \\ \rm Exponential & \rm\;\; e_n(x) = e^{-x}(xD)^n e^x \\ \rm Abel & \rm A_n^k(x) = x \; e^{-knD} x^{n-1} \\ \rm Hermite & \rm H_n^k(x) = (-1)^n e^{x^2/(2n)} (kD)^n e^{-x^2/(2n)} \\ \rm Bernoulli & \rm B_n^k(x) = \left(\frac{D}{e^D-1}\right)^k x^n \\ \rm Euler & \rm E_n^k(x) = \left(\frac{2}{e^D+1}\right)^k x^n \\ \end{array}$$

Split it into partial fractions then differentiate.

• When I do that, I can't find any pattern by which I guess how nth derivative will look like. Aug 30 '10 at 11:43
• @Andrejako, see J. Mangaldan's hint below. Aug 30 '10 at 12:44

To add to Derek's hint: you will have to show the validity of the formula

$\frac{\mathrm{d}^k}{\mathrm{d}x^k}\frac1{x}=\frac{(-1)^k k!}{x^{k+1}}$

According to the Binomial Theorem (or using the usual formula for the sum of a geometric series with initial term 1 and common ratio $x^2$),

$f(x)=\frac {x} {x^2-1} = -x \frac {1} {1 - x^2} = -x \left( 1 + x^2 + x^4 + \cdots + x^{2n} + \cdots \right)$

$= -x - x^3 - x^5 - \cdots - x^{2n+1} - \cdots$.

Because the right side converges absolutely for $|x^2| < 1$ you can differentiate it term by term, introducing a coefficient $(2n+1)(2n) \cdots (2n+1-k+1)$ for $x^{2n+1-k}$; in other words, the coefficient of $x^j$ is $(j+1)(j+2) \cdots (j+k)$. Dividing the entire thing through by $k!$ gives a series you can easily relate to the binomial expansion of $( 1 - x^2 )$ to a negative integral value, yielding a closed form solution.

related problem: (I). I am referring you to this book where you will find the complete answer to the problem of finding the nth derivative of rational polynomials and other classes of functions.

Added Some people suggested the post should be self-contained. Here is the example I am referring to

$$f(x) = \frac{ x^2 + x - 1 } { x^4 + x^3 - x - 1 }\,.$$ The roots of the denominator are $$\nonumber x = -1, 1, -\frac{1}{2} + \frac{ i \sqrt{3}}{2}, -\frac{1}{2} - \frac{ i \sqrt{3}}{2}\,.$$ Expressing the function in the partial fraction form $$\nonumber f(x) = \frac{ x^2 + x - 1 } { x^4 + x^3 - x - 1 } = \frac{a_1}{x-1} + \frac{a_2}{x+1} + \frac{a_3}{ x + \frac{1 - i\sqrt{3}}{2}} + \frac{a_4}{ x + \frac{1 + i\sqrt{3}}{2}}\,.$$ To find $a_i$, one can use the formula

$$a_{ i_{\ell}} = \mathrm{Res}( ( x - \alpha_{\ell} )^{{i_\ell} - 1 } f(x)\,, x = \alpha_\ell) \,,\quad i_\ell = 1,2,...,m_\ell,$$

where $\alpha_{\ell}$ are the roots of the denominator and the formula takes care of the multiplicities $m_{\ell}$ of the roots in case there is any. For instance,

\begin{align}\nonumber a_1 & = \mathrm{Res}\left( \frac{ x^2 + x - 1 } { x^4 + x^3 - x - 1 }, x = 1 \right)& \\ \nonumber \\ \nonumber & = \lim_{x = 1} (x-1)\frac{ x^2 + x - 1 } { x^4 + x^3 - x - 1 }& \\ \nonumber \\ \nonumber & = \frac{1}{6}\,,& \end{align}

Once we have the function in the partial fraction form, the $nth$ derivative can be found directly by using the formula $$D^n ( x - \alpha )^{(-k)} = {\frac { \left( -1 \right) ^{n}\Gamma \left( n+k \right) }{\Gamma \left( k \right) }}(x-\alpha)^{-k-n}.$$

Following the above techniques, the final answer is

\begin{align}\nonumber f^{(n)}(x) = (-1)^n \Gamma(n+1) \, &\left( \frac{1}{2\,(x+1)^{n+1}} + \frac{1}{6\,(x-1)^{n+1}} \right. & \\ \nonumber \\ \nonumber & \left. + \frac{\frac{ - 1 - i\sqrt{3}}{3}}{ \left(x + \frac{1 - i\sqrt{3}}{2}\right)^{n+1} } + \frac{\frac{ - 1 + i\sqrt{3}}{3}}{ \left(x + \frac{1 + i\sqrt{3}}{2}\right)^{n+1}} \right) \,. & \end{align}.