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March 31, 2015

March 31, 2015

Posted by **hi** on Sunday, July 1, 2012 at 9:59pm.

x^2-3/x-2.

- calculus -
**Steve**, Monday, July 2, 2012 at 12:16amif y=u/v,

y' = (u'v-uv')/v^2

so, let u = x^2-3, u'=2x

let v = x-2, v'=1

y' = ((2x)(x-2)-(x^2-3)(1))/(x-2)^2

= (2x^2-2x-x^2+3)/(x-2)^2

= (x^2-4x+3)/(x-2)^2

= (x-3)(x-1)/(x-2)^2

or, you can always think of it as a product rule:

y = (x^2-3) * (x-2)^-1

y' = (2x-3)/(x-2) - (x^2-3)*(x-2)^-2

which works out the same

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