how do i find the derivative of y=(x^3) / x-1. Do i use the quotient rule?

Question

how do i find the derivative of y=(x^3) / x-1.  Do i use the quotient rule?

anonymous · Answer

Yep. For u/v, the derivative is $$\frac{ v(u')-u(v')}{ v^2 }$$
with u' = derivative of u.

anonymous · Answer

i dont know if i did this right but i got (3x^2-x^3) / x-1

phi · Answer

in your case u is x^3   and  v is (x-1)

u'  is 3x^2   and v'  is 1

v u'  is (x-1) * 3x^2= 3x^3 - 3x^2

u v'  = x^3*1 = x^3

v u' - u v' = 3x^3 - 3x^2 - x^3
that simplifies to 2x^3 - 3x^2

so the final answer is 
$$ \frac{x^2(2x-3)}{(x-1)^2} $$

anonymous · Answer

so i wouldnt use the equation g(x)*f'(x)-f(x)*g'(x) / (g(x))^2 or does both equations give me the same ans

phi · Answer

I used that formula, but I used  u and v for the names rather than f and g

anonymous · Answer

o ok