Question

Find the derivative of 2x+3/ 3x-2
using first principle method.

Answer 1

you 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

Answer 2

Ek no. Bekar solution

Answer 3

ou 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

Answer 4

The answer is= -13/(3x-2)^2. But the answer they are post is incorrect answer.