What is the derivative of y=(x^3)/(x-1) ?

The derivative of y=(x^3)/(x-1) is (2x^3-3x^2)/((x-1)^2)

What is the first derivative of y=(x^3)/(x-1) ?

The first derivative of y=(x^3)/(x-1) is (2x^3-3x^2)/((x-1)^2)

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derivative of

y = \frac{x ^{3}}{x - 1}

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

Solution steps

\frac{d}{d x} (\frac{x ^{3}}{x - 1})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( x ^{3} ) ( x - 1 ) - \frac{d}{d x} ( x - 1 ) x ^{3}}{( x - 1 ) ^{2}}

\frac{d}{d x} (x^{3}) = 3 x^{2}

\frac{d}{d x} (x - 1) = 1

= \frac{3 x ^{2} ( x - 1 ) - 1 \cdot x ^{3}}{( x - 1 ) ^{2}}

Simplify

\frac{3 x ^{2} ( x - 1 ) - 1 \cdot x ^{3}}{( x - 1 ) ^{2}} : \frac{2 x ^{3} - 3 x ^{2}}{( x - 1 ) ^{2}}

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

\frac{\partial}{\partial x} (y e^{x} + sin (y))

\frac{\partial}{\partial x} (y ln (x^{2} + y^{2} + 5) - 7)

\int \frac{- 1}{x - 2} d x

\int 4 x^{3} + e^{x} + 1 d x

F (x) = 4 x^{6} (x^{4} - 6 x)

What is the derivative of y=(x^3)/(x-1) ?
The derivative of y=(x^3)/(x-1) is (2x^3-3x^2)/((x-1)^2)
What is the first derivative of y=(x^3)/(x-1) ?
The first derivative of y=(x^3)/(x-1) is (2x^3-3x^2)/((x-1)^2)