# struggling to find a dirivative

• Nov 9th 2007, 04:14 AM
james jarvis
struggling to find a dirivative
How do i find the dirivative of y=x-1/2x^2.
Thats. y=x - 1 divided by 2x to the power 2.

Any help would be greatful.
Thanks, James.

(where does the math notations everyone seems to use come from? )
• Nov 9th 2007, 04:26 AM
Krizalid
Split the fraction in two pieces, then use the power rule for derivatives.

--

LaTeX stuff can be found here.
• Nov 9th 2007, 04:27 AM
janvdl
Quote:

Originally Posted by james jarvis
How do i find the dirivative of y=x-1/2x^2.
Thats. y=x - 1 divided by 2x to the power 2.

Any help would be greatful.
Thanks, James.

(where does the math notations everyone seems to use come from? )

We use Latex. (I'll go search for the tutorial and then edit my post and paste it here okay?)

$( \frac{x - 1}{2x} )^{2}$

$= (x(2x)^{-1} - 1(2x)^{-1})^{2}$

$= \frac{1}{4} - \frac{1}{2} x^{-1} + \frac{1}{4}x^{-2}$

$\frac{dy}{dx} = 0 + \frac{1}{2} x^{-2} - \frac{1}{2}x^{-3}$

$= \frac{1}{2x^2} - \frac{1}{2x^3}$

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EDIT: Krizalid gave you the link :D
• Nov 9th 2007, 04:34 AM
james jarvis
Thanks for reply. Always makes sense when you see it! :)
• Nov 9th 2007, 04:35 AM
janvdl
Quote:

Originally Posted by james jarvis
Always makes sense when you see it! :)

You bet! (Nod)
• Nov 9th 2007, 06:49 AM
jackddog
How to find derivative
Your question is not clear.

a) Is the 2x under the 1 only or under x as well?

b) Is it (2)(x^2) or is it (2x)^2. That is, is it just the x that is raised to power 2 or the 2x?

Tell me which is which and I will tell you......how
• Nov 9th 2007, 06:51 AM
janvdl
Quote:

Originally Posted by jackddog
Your question is not clear.

a) Is the 2x under the 1 only or under x as well?

b) Is it (2)(x^2) or is it (2x)^2. That is, is it just the x that is raised to power 2 or the 2x?

Tell me which is which and I will tell you......how

Well my approach seems to be correct as he seems happy with it :rolleyes:
• Nov 9th 2007, 02:08 PM
2taall
I think you use the Quotient Rule.