Question

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

Answer 1

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

Answer 2

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

Answer 3

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
\[ \frac{x^2(2x-3)}{(x-1)^2} \]

Answer 4

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

Answer 5

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

Answer 6

o ok