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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

x^{2} - 2 x + 3, x = - 2, x = 1

\frac{d}{d x} ((2 x^{2} - 6)^{- 10})

y^{^{''}} - 8 y^{^{'}} + 16 y = t - 2 e^{4 t}

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

\frac{d y}{d x} = cos (x) \cdot e^{y + s i n (x)}

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