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

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

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

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

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

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

Solution steps

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

\frac{d}{d x} (2 x^{2} - 1) = 4 x

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

= \frac{4 x ( x - 1 ) ^{2} - 2 ( x - 1 ) ( 2 x ^{2} - 1 )}{( ( x - 1 ) ^{2} ) ^{2}}

Simplify

\frac{4 x ( x - 1 ) ^{2} - 2 ( x - 1 ) ( 2 x ^{2} - 1 )}{( ( x - 1 ) ^{2} ) ^{2}} : \frac{2 ( - 2 x + 1 )}{( x - 1 ) ^{3}}

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

\frac{\partial}{\partial x} (x^{8} + 2^{y} + x^{y})

y^{^{'}} = \frac{x}{y ^{2}}

t \frac{d y}{d t} + 2 y = t^{2}

\int_{1}^{9} \frac{3 x - 2}{x} d x

ln (e^{- x} + x e^{- x})

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