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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derivative of

\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}}

\int \frac{y}{x z} d x

\frac{s + 2}{( s - 2 ) ( s + 1 )}

\int 2 x y - sin (x) d x

f (x) = (x + 1) (3 x^{2} - 5 x + 2)

y^{^{''}} - y^{^{'}} - y = 0

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)