Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

@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

Answer 5

what mimi said

Answer 6

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

Answer 7

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

Answer 8

i think i got it: \[3(2x-3)-2(3x-2)/(2x-3)^2\]

Answer 9

is that the same as vi3tbloodxx?

Answer 10

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

Answer 11

oh then thanks just verifying.

Answer 12

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