Find f' using definition of the derivative for f=1/(x^2+1)

[dangerous_dave]

I'm having trouble with this question:

Let f(x) = 1/(x^2+1)
Use the definition of the derivative to find f'(x)
Thats as in 1 on the top line, x squared + 1 on the bottom line.

I have the definition as: f'(x) = lim/x->0 (f(x+h)-f(x))/h

I do as it says and get (after a few too hard to write here steps)

(-h^2+h-2xh-1)/(x^4+x^3+x^2+2x^3h+x^2h+2xh^2+xh+h)

And then I'm stuck. And I'm not 100% sure that I am right so far.

Thanks

 

[galactus]

\(\displaystyle \frac{\frac{1}{(x+h)^{2}+1}-\frac{1}{x^{2}+1}}{h}\)

\(\displaystyle =\frac{-2x-h}{x^{4}+2hx^{3}+h^{2}x^{2}+2x^{2}+2hx+h^{2}+1}\)

Now, you should get your derivative by just letting h=0.

 

[dangerous_dave]

Sorry but could you explain how you got that? The first equation I got, but how did you get from that to the second?

 

[stapel]

Let f(x) = 1/(x^2+1)
Use the definition of the derivative to find f'(x)

Let's do this in small chunks.

. . . . .\(\displaystyle f(x)\, =\, \frac{1}{x^2\, +\, 1}\)

. . . . .\(\displaystyle f(x\, +\, h)\, =\, \frac{1}{(x\, +\, h)^2\, +\, 1}\, =\, \frac{1}{x^2\, +\, 2hx\, +\, h^2\, +\, 1}\)

. . . . .\(\displaystyle f(x\, +\, h)\, -\, f(x)\, =\, \frac{1}{x^2\, +\, 2hx\, +\, h^2\, +\, 1}\, - \, \frac{1}{x^2\, +\, 1}\)

. . . . . . . . . . . . . . . .. . . . . .\(\displaystyle =\, \frac{x^2\, +\, 1}{(x^2\, +\, 1)(x^2\, +\, 2hx\, +\, h^2\, +\, 1)}\, -\, \frac{x^2\, +\, 2hx\, +\, h^2\, +\, 1}{(x^2\, +\, 1)(x^2\, +\, 2hx\, +\, h^2\, +\, 1)}\)

. . . . . . . . . . . . . . . .. . . . . .\(\displaystyle =\, \frac{x^2\, +\, 1\, -\, x^2\, -\, 2hx\, -\, h^2\, -\, 1}{(x^2\, +\, 1)(x^2\, +\, 2hx\, +\, h^2\, +\, 1)}\)

. . . . . . . . . . . . . . . .. . . . . .\(\displaystyle =\, \frac{-2hx\, -\, h^2}{(x^2\, +\, 1)(x^2\, +\, 2hx\, +\, h^2\, +\, 1)}\)

. . . . .\(\displaystyle \frac{f(x\, +\, h)\, -\, f(x)}{h}\, =\, \frac{\left(\frac{-2hx\, -\, h^2}{(x^2\, +\, 1)(x^2\, +\, 2hx\, +\, h^2\, +\, 1)}\right)}{h}\)

. . . . . . . . . . . . . . . .. . . . . .\(\displaystyle =\, \frac{-2x\, -\, h}{(x^2\, +\, 1)(x^2\, +\, 2hx\, +\, h^2\, +\, 1)}\)

Now take the limit as h goes to zero. :wink:

Eliz.

 

[dangerous_dave]

ok I followed it all the way up to when you removed the divided by h. How does that dissappear?

Why does it affect the top but not the bottom? (sounds like I need to touch up on my algebra...)

 

[pka]

ok I followed it all the way up to when you removed the divided by h. How does that dissappear?

Do you know how to divide out a common factor?

 

[dangerous_dave]

Well I know that, say, (x^2 + x)/x = x+1, but Im not sure how to do it when dividing by a division...

 

[pka]

Re:

. . . . .\(\displaystyle \frac{f(x\, +\, h)\, -\, f(x)}{h}\, =\, \frac{\left(\frac{-2hx\, -\, h^2}{(x^2\, +\, 1)(x^2\, +\, 2hx\, +\, h^2\, +\, 1)}\right)}{h}\)

. . . . .\(\displaystyle \frac{f(x\, +\, h)\, -\, f(x)}{h}\, =\, \frac{\not h \left(\frac{-2x\, -\, h}{(x^2\, +\, 1)(x^2\, +\, 2hx\, +\, h^2\, +\, 1)}\right)}{\not h}\)

 

[dangerous_dave]

Well thats not quite what I was getting at. I was wondering why you only take the h out of the top and not the bottom. But I understand now anyway . Thanks heaps to all of you

 

[pka]

I was wondering why you only take the h out of the top and not the bottom.

There is no common factor of h in the "bottom".

 

[Dr. Flim-Flam]

f' '(x) = lim [f(x+h)-f(x)]/h; f(x) = 1/(x^2+1); f '(x) = lim (1/h) {1/[(x+h)^2+1] - 1/(x^2+1)}
h->0 h->0

f '(x) = lim (1/h){x^2+1-[(x+h)^2+1]}/[(x^2+1)((x+h)^2+1)] = (1/h){(-2xh-h^2)/(x^2+1)[(x+h)^2+1]}
h->0
f '(x) = lim {(-2x-h)/[(x^2+1)((x+h)^2+1) = -2x/(x^2+1)^2.
h->0