Calculating the $n^\text{th}$ derivative

Question

How do we calculate the $n^{\text{th}}$ derivative for

$$ \frac{x^3}{(x-a)(x-b)(x-c)}? $$

How can I obtain the partial fraction for the given term?

Answer 1

Using Partial fraction

Above Given fraction is a Improper fraction then by actual division, Since same degree in above and below.

I.e $\displaystyle \frac{x^3}{x\cdot x \cdot x}$

Then $$\displaystyle \frac{x^3}{(x-a)(x-b)(x-c)}= 1+\frac{f(x)}{(x-a)(x-b)(x-c)}$$

Where $f(x)$ is a degree of polynomial $2$

So $$\displaystyle \frac{x^3}{(x-a)(x-b)(x-c)}= 1+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}....................(1)$$

So $$x^3 = (x-a)(x-b)(x-c)+A(x-b)(x-c)+B(x-c)(x-a)+C(x-a)(x-b)$$

Now Put $x=a\;,$ We get $\displaystyle a^3 = A(a-b)(a-c)\Rightarrow A = \frac{a^3}{(a-b)(a-c)}$

Similarly Put $x=b\;,$ We get $\displaystyle B = \frac{b^3}{(b-c)(b-a)}$

Similarly Put $x=c\;,$ We get $\displaystyle C = \frac{c^3}{(c-a)(c-b)}$

Now Let $$\displaystyle f(x) = \frac{x^3}{(x-a)(x-b)(x-c)} = 1+\frac{A}{(x-a)}+\frac{B}{(x-b)}+\frac{C}{(x-c)}$$

So $$\displaystyle f'(x) = -\frac{A}{(x-a)^2}-\frac{B}{(x-b)^2}-\frac{C}{(x-c)^2}$$

So $$\displaystyle f''(x) = \frac{2A}{(x-a)^3}+\frac{2B}{(x-b)^3}+\frac{2C}{(x-c)^3}$$

In a Similar fashion, We get $$\displaystyle f^{n}(x) = (-1)^n\frac{n!\cdot A}{(x-a)^{n+1}}+(-1)^n\frac{n!\cdot B}{(x-b)^{n+1}}+(-1)^n\frac{n!\cdot C}{(x-c)^{n+1}}$$

and put values of $A,B,C$ from above.

Answer 2

So since the coefficients of $x^3$ in the numerator and denominator are equal, there is a constant term of $1$, and the form you want is

$$\frac {x^3}{(x-a)(x-b)(x-c)}=1+\frac A{x-a}+\frac B{x-b}+\frac C{x-c}$$

If you clear fractions you obtain $$x^3=(x-a)(x-b)(x-c)+A(x-b)(x-c)+(x-a)B(x-c)+(x-a)(x-b)C$$

Now successively setting $x=a,b,c$ will give you the values for $A,B,C$

And then differentiating as many times as you like is easy.

Answer 3

If we assume that $a,b,c$ are distinct numbers, from: $$ \text{Res}\left(\frac{z^3}{(z-a)(z-b)(z-c)},z=a\right)=\frac{a^3}{(a-b)(a-c)}\tag{1}$$ it follows that: $$\begin{eqnarray*} f(z)&=&\frac{z^3}{(z-a)(z-b)(z-c)}\tag{2}\\&=&1+\frac{a^3}{(a-b)(a-c)(z-a)}+\frac{b^3}{(b-a)(b-c)(z-b)}+\frac{c^3}{(c-a)(c-b)(z-c)}\end{eqnarray*}$$ and we are able to compute the derivatives of $f(z)$ from the derivatives of $\frac{1}{(z-a)}$, that are straightforward to compute.