# another algebra slip up

• Sep 22nd 2013, 10:56 AM
Jonroberts74
another algebra slip up
I have to find the derivative of y=x + 1/x. I understand finding the derivative.

I split up the two terms and I am using [f(a+h) - f(a)]/h.

so for the first one it was [(x+h) - (x)]/h = h/h = 1

the second part was {[1/(x+h)] - [1/x]}/h = {[x/(x(x+h))] - [(x+h)/(x(x+h)]} = {[h/(x(x+h))]}/h then I go to [h/(x(x+h))] * 1/h but from there it messes me up because I get [h/(x^2h +xh^2)]

I know the final answer is 1-(1/x^2)

thank you
• Sep 22nd 2013, 11:02 AM
topsquark
Re: another algebra slip up
Quote:

Originally Posted by Jonroberts74
I have to find the derivative of y=x + 1/x. I understand finding the derivative.

I split up the two terms and I am using [f(a+h) - f(a)]/h.

so for the first one it was [(x+h) - (x)]/h = h/h = 1

the second part was {[1/(x+h)] - [1/x]}/h = {[x/(x(x+h))] - [(x+h)/(x(x+h)]} = {[h/(x(x+h))]}/h then I go to [h/(x(x+h))] * 1/h but from there it messes me up because I get [h/(x^2h +xh^2)]

I know the final answer is 1-(1/x^2)

thank you

Let's write this out first:
$\lim_{h \to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h} = \text{ ... } = \lim_{h \to 0} \frac{\frac{h}{x(x + h)}}{h}$

$= \lim_{h \to 0} \frac{h}{x(x + h)} \cdot \frac{1}{h}$

Can you finish it now?

-Dan
• Sep 22nd 2013, 11:15 AM
Jonroberts74
Re: another algebra slip up
then it would be 1h/[x(x+h)h] cancel the h's for 1/x(x+h) then 1/x(x+0) = 1/x^2 ? should be -1/(x^2) thought, right? because of it being 1/x+h) - 1/(x)......
• Sep 22nd 2013, 11:30 AM
Plato
Re: another algebra slip up
Quote:

Originally Posted by Jonroberts74
then it would be 1h/[x(x+h)h] cancel the h's for 1/x(x+h) then 1/x(x+0) = 1/x^2 ? should be -1/(x^2) thought, right? because of it being 1/x+h) - 1/(x)......

There is a missing sign above. It should be
$\lim_{h \to 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h} = \text{ ... } = \lim_{h \to 0} \frac{\frac{-h}{x(x + h)}}{h}$
• Sep 22nd 2013, 11:39 AM
Jonroberts74
Re: another algebra slip up
thanks, it is very appreciated.