Use the quotient rule to find the derivative.
-$y=\frac{x^{3}}{x-1}$
A) $y^{\prime}=\frac{-2 x^{3}-3 x^{2}}{(x-1)^{2}}$
B) $y^{\prime}=\frac{2 x^{3}-3 x^{2}}{(x-1)^{2}}$
C) $y^{\prime}=\frac{2 x^{3}+3 x^{2}}{(x-1)^{2}}$
D) $y^{\prime}=\frac{-2 x^{3}+3 x^{2}}{(x-1)^{2}}$

Question

Quizplus · Accepted Answer

The quotient rule states that if $y = \frac{u}{v}$, then $y' = \frac{u'v - uv'}{v^2}$. Here, $u = x^3$ and $v = x-1$. So, $u' = 3x^2$ and $v' = 1$. Applying the quotient rule gives $y' = \frac{(3x^2)(x-1) - (x^3)(1)}{(x-1)^2} = \frac{3x^3 - 3x^2 - x^3}{(x-1)^2} = \frac{2x^3 - 3x^2}{(x-1)^2}$.

Use the Quotient Rule to Find the Derivative y=x3x−1y=\frac{x^{3}}{x-1}y=x−1x3​ A) y′=−2x3−3x2(x−1)2y^{\prime}=\frac{-2 x^{3}-3 x^{2}}{(x-1)^{2}}y′=(x−1)2−2x3−3x2​

Use the Quotient Rule to Find the Derivative $y=\frac{x^{3}}{x-1}$ A) $y^{\prime}=\frac{-2 x^{3}-3 x^{2}}{(x-1)^{2}}$