What is the derivative of (2x^3-3x^2/((x-1)^2)) ?

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (2 x^{3} - 3 x^{2}) = 6 x^{2} - 6 x

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

= \frac{( 6 x ^{2} - 6 x ) ( x - 1 ) ^{2} - 2 ( x - 1 ) ( 2 x ^{3} - 3 x ^{2} )}{( ( x - 1 ) ^{2} ) ^{2}}

Simplify

\frac{( 6 x ^{2} - 6 x ) ( x - 1 ) ^{2} - 2 ( x - 1 ) ( 2 x ^{3} - 3 x ^{2} )}{( ( x - 1 ) ^{2} ) ^{2}} : \frac{2 x ( x ^{2} - 3 x + 3 )}{( x - 1 ) ^{3}}

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

y^{^{'}} = \frac{4 t ^{3} + 1}{2 y - 6}, y (1) = 2

\int \frac{e ^{\frac{6}{t}}}{t ^{2}} d t

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{2}} - 3 x z)

\frac{d}{d x} (x^{2} + x + 2)

\frac{d y}{d x}, 2 x^{2} + y^{2} - 7 = 0

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