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

The derivative of (2(-x^2-1)/((x^2-1)^2)) is -(4x(-x^2-3))/((x^2-1)^3)

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

The first derivative of (2(-x^2-1)/((x^2-1)^2)) is -(4x(-x^2-3))/((x^2-1)^3)

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

n = 1 \sum \infty ln (\frac{n}{3 n + 1})

\int \frac{3}{x ^{- 1} + 3} d x

2 x^{2} + x + 1

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

\frac{d y}{d t} = - 0.0077 y

What is the derivative of (2(-x^2-1)/((x^2-1)^2)) ?
The derivative of (2(-x^2-1)/((x^2-1)^2)) is -(4x(-x^2-3))/((x^2-1)^3)
What is the first derivative of (2(-x^2-1)/((x^2-1)^2)) ?
The first derivative of (2(-x^2-1)/((x^2-1)^2)) is -(4x(-x^2-3))/((x^2-1)^3)