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January 31, 2015

January 31, 2015

Posted by **Anonymous** on Tuesday, January 29, 2013 at 7:32am.

using first principle method.

- maths -
**Reiny**, Tuesday, January 29, 2013 at 8:19amyou meant:

f(x) = y = (2x+3)/(3x-2)

so f(x+h) = (2x + 2h + 3)/(3x + 3h - 2)

dy/dx = lim [ (2x+2h+3)/(3x+3h-2) - (2x+3)/(3x-2) ]/h , as h --->0

= lim [ ((2x + 2h + 3)(3x-2) - (2x+3)(3x+3h+2))/((3x+3h-2)(3x+h))]/h

= lim [ (5h/((3x+3h-2)(3x-2))]/h , as h -->0

= lim 5/((3x+3h-2)(3x-2)) , as h -->0

= 5/(3x-2)^2

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