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

The derivative of (3(-x^2+1)/((x^2+1)^2)) is -(6x(-x^2+3))/((x^2+1)^3)

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

The first derivative of (3(-x^2+1)/((x^2+1)^2)) is -(6x(-x^2+3))/((x^2+1)^3)

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

- \frac{6 x ( - x ^{2} + 3 )}{( x ^{2} + 1 ) ^{3}}

Solution steps

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

Take the constant out:

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

= 3 \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}}

= 3 \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)

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

Simplify

3 \cdot \frac{( - 2 x ) ( x ^{2} + 1 ) ^{2} - 4 x ( x ^{2} + 1 ) ( - x ^{2} + 1 )}{( ( x ^{2} + 1 ) ^{2} ) ^{2}} : - \frac{6 x ( - x ^{2} + 3 )}{( x ^{2} + 1 ) ^{3}}

= - \frac{6 x ( - x ^{2} + 3 )}{( x ^{2} + 1 ) ^{3}}

y^{^{''}} + 2 y^{^{'}} + y = - 2 e^{x}

\frac{d y}{d x} = \frac{x - 4}{y - 6}

\int_{1}^{2} \frac{5}{x ^{3} + 2 x} d x

\int_{0}^{1} (5^{x} - 3^{x}) d x

y = 2 x^{3}, (1, 2)

What is the derivative of (3(-x^2+1)/((x^2+1)^2)) ?
The derivative of (3(-x^2+1)/((x^2+1)^2)) is -(6x(-x^2+3))/((x^2+1)^3)
What is the first derivative of (3(-x^2+1)/((x^2+1)^2)) ?
The first derivative of (3(-x^2+1)/((x^2+1)^2)) is -(6x(-x^2+3))/((x^2+1)^3)