What is the derivative of (4(-x^2+1)/((x^2+1)^2)) ?

The derivative of (4(-x^2+1)/((x^2+1)^2)) is -(8x(-x^2+3))/((x^2+1)^3)

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

The first derivative of (4(-x^2+1)/((x^2+1)^2)) is -(8x(-x^2+3))/((x^2+1)^3)

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

- \frac{8 x ( - x ^{2} + 3 )}{( x ^{2} + 1 ) ^{3}}

Solution steps

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

Take the constant out:

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

= 4 \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}}

= 4 \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)

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

Simplify

4 \cdot \frac{( - 2 x ) ( x ^{2} + 1 ) ^{2} - 4 x ( x ^{2} + 1 ) ( - x ^{2} + 1 )}{( ( x ^{2} + 1 ) ^{2} ) ^{2}} : - \frac{8 x ( - x ^{2} + 3 )}{( x ^{2} + 1 ) ^{3}}

= - \frac{8 x ( - x ^{2} + 3 )}{( x ^{2} + 1 ) ^{3}}

y + y^{^{'}} = sin (t)

t^{9} cos (t)

h \to + 0 lim (\frac{∣ h ∣}{h})

\int \frac{ln ( x ) sin ( ln ( x ) )}{x} d x

x \to 1 lim (\frac{x - 1}{x + 1})

What is the derivative of (4(-x^2+1)/((x^2+1)^2)) ?
The derivative of (4(-x^2+1)/((x^2+1)^2)) is -(8x(-x^2+3))/((x^2+1)^3)
What is the first derivative of (4(-x^2+1)/((x^2+1)^2)) ?
The first derivative of (4(-x^2+1)/((x^2+1)^2)) is -(8x(-x^2+3))/((x^2+1)^3)