Simplifying derivative result

Question

I am doing the derivative of

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

So my result is the following

$$f'(x) = \frac{4x^2 -8x +4}{(x^2-1)^2}$$

I am sure the answer is correct, but in my solutions book and In Wolfram Alpha they simplify until

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

And I don't know why, which steps are they doing?

The result of wolfram alpha is $f'(x)=\large{\frac4{(x+1)^2}}$. See here: http://www.wolframalpha.com/input/?i=differentiate+%5Cfrac%7Bx%5E2+-4x+%2B3%7D%7Bx%5E2-1%7D — callculus42, Jun 04 '16 at 15:42

score 2 · Answer 1 · answered Jun 04 '16 at 15:33

2

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

This is undefined if $x-1=0$

Otherwise. the expression reduces to

$$\dfrac{x-3}{x+1}=\dfrac{x+1-4}{x+1}=?$$

answered Jun 04 '16 at 15:33

lab bhattacharjee

274,582

score 1 · Accepted Answer · answered Jun 04 '16 at 15:37

1

You are correct. Now note that $$\frac{4x^2-8x+4}{(x^2-1)^2}=\frac{4(x^2-2x+1)}{((x-1)(x+1))^2}=\frac{4(x-1)^2}{(x-1)^2(x+1)^2}=\frac{4}{\color{red}{(x+1)^2}}$$ This is not $$\frac{4}{(x^2-1)^2}$$

answered Jun 04 '16 at 15:37

mathlove

139,939

1

sorry for my transcription error, and thanks a lot for all answers. – jsertx Jun 04 '16 at 16:06

score 0 · Answer 3 · answered Jun 04 '16 at 15:35

0

Hint. Observe that, for $x\neq1$, $$ f(x) = \frac{x^2 -4x +3}{x^2-1}= \frac{(x-1)(x-3)}{(x-1)(x+1)}=\frac{x-3}{x+1}. $$

answered Jun 04 '16 at 15:35

Olivier Oloa

120,989

Simplifying derivative result

3 Answers3