Is there a trick for higher derivatives?

**A Beautiful Mind** · April 25th 2010, 05:21 PM

Wow. I clicked quote instead of edit. My bad.

**A Beautiful Mind** · April 25th 2010, 05:22 PM

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

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

After I take the first derivative and I need the second, I can get a second derivative but it takes freakin' forever and my test only lasts for about an hour. I was wondering if anyone knew of a fast trick or some creative ways to deal with this problem and minimize time computing it? 'Cause I swear, I don't want to waste forever just getting this when I have to sketch 10 other functions on my examination.

**maddas** · April 25th 2010, 05:24 PM

$\frac{x^2-2x+4}{x-2} = \frac{x(x-2)+4}{x-2} = x + \frac{4}{x-2}$ , if that helps.

Math Help - Is there a trick for higher derivatives?

LinkBack

Thread Tools

Display

Is there a trick for higher derivatives?

Similar Math Help Forum Discussions

Higher Derivatives Question

Higher Derivatives r''(q)

higher-order derivatives

Higher Derivatives

Higher Derivatives

Search Tags