# Calculus question

1. Apr 23, 2005

### VietDao29

Hi,
$$f(x) = \frac{5x^{2} - 3x - 20}{x^{2} - 2x - 3}$$
Find:
$$f^{(n)}(x)$$
I try to find its rule, but I fail:
I have:
$$f'(x) = \frac{-7x^{2} + 10x - 31}{(x^{2} - 2x - 3)^{2}}$$
And I have f''(x) is some kind of very very complicated number. What should I do in this kind of problem??
Any help will be appreciated,
Viet Dao,

Last edited: Apr 23, 2005
2. Apr 23, 2005

### Zurtex

Differentiate a few times simplifying as much as you can. Here is what I got:

$$f(x) = 5 + \frac{4}{x - 3} + \frac{3}{x + 1}$$

$$f'(x) = -\frac{4}{(x - 3)^2} - \frac{3}{(1 + x)^2}$$

I feel if I go any further I'll completely give the pattern away, carry on differentiating.

3. Apr 23, 2005

### VietDao29

So the answer is:
$$f^{(n)} = \frac{(-1)^{n}4(n!)}{(x - 3)^{n + 1}} + \frac{(-1)^{n}3(n!)}{(x + 1)^{n + 1}}$$
Am I correct??

4. Apr 23, 2005

### dextercioby

U can use induction method to prove your formula right...

Daniel.

5. Apr 23, 2005

### VietDao29

So...
$$f^{1}(x) = -\frac{4}{(x - 3)^{2}} - \frac{3}{(1 + x)^2}$$
So the formula is correct if n = 1.
Assume it's correct for n = k:
$$f^{(k)}(x) = \frac{(-1)^{k}4(k!)}{(x - 3)^{k + 1}} + \frac{(-1)^{k}3(k!)}{(x + 1)^{k + 1}}$$
Prove it's correct if n = k + 1:
$$f^{(k + 1)}(x) = 4(k!)(-1)^{k}\frac{-(k + 1)(x - 3)^{k}}{(x - 3)^{2k + 2}} + (-1)^{k}3(k!)\frac{-(k + 1)(x + 1)^{k}}{(x + 1)^{2k + 2}}$$
$$= \frac{4(k + 1)!(-1)^{k + 1}}{(x - 3)^{k + 2}} + \frac{(-1)^{k + 1}3(k + 1)!}{(x + 1)^{k + 2}}$$
So the formula is true $\forall n \in N*$
Am I correct?
Viet Dao,

Last edited: Apr 23, 2005