maths
posted by Anonymous .
Find the derivative of 2x+3/ 3x2
using first principle method.

you meant:
f(x) = y = (2x+3)/(3x2)
so f(x+h) = (2x + 2h + 3)/(3x + 3h  2)
dy/dx = lim [ (2x+2h+3)/(3x+3h2)  (2x+3)/(3x2) ]/h , as h >0
= lim [ ((2x + 2h + 3)(3x2)  (2x+3)(3x+3h+2))/((3x+3h2)(3x+h))]/h
= lim [ (5h/((3x+3h2)(3x2))]/h , as h >0
= lim 5/((3x+3h2)(3x2)) , as h >0
= 5/(3x2)^2
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