Find the derivative of the given function. y=3x-2/2x-3

Question

mimi_x3 · Answer

Is it: 
$$\large \frac{3x-2}{2x-3}  $$
?

anonymous · Answer

quotient rule:  Lod Hi - Hi d Lo over Lo Lo

anonymous · Answer

Quotient Rule: 
f'(x) = g(x)f'(x) - f(x)g'(x) / (g(x))²

hero · Answer

@vi3tbloodxx , don't forget to put proper parentheses to distinguish the numerator: 

f'(x) = (g(x)f'(x) - f(x)g'(x)) / (g(x))²

anonymous · Answer

what mimi said

anonymous · Answer

yea that's what i meant sorry...

((3x-2)'(2x-3) - (2x-3)'(3x-2)) / (2x-3)²
= (3(2x-3)-2(3x-2)) / (2x-3)²
= (6x-9-6x+4) / (2x-3)²
= -5/(2x-3)²

mimi_x3 · Answer

or..
$$\large y' = \frac{u'v-uv'}{v^{2}}  $$
u = 3x-2 => u' =3 
v = 2x-3 => v'=2

anonymous · Answer

i think i got it: $$3(2x-3)-2(3x-2)/(2x-3)^2$$

anonymous · Answer

is that the same as vi3tbloodxx?

anonymous · Answer

yes with an extra set of parentheses around like this (3(2x-3)-2(3x-2))

anonymous · Answer

oh then thanks just verifying.

anonymous · Answer

You can also use logarithmic differentiation ln(3x-2) - ln(2x-3) = (3/(3x-2)) - (2/(2x-3))